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Answers 1. y + 1 = 3(x β 3) y = 3x β 10 3. y + 1 = β(x + 7) y = βx β 8 169 2. y β 8 = β2(x + 5) y = β2x β 2 4x + 2) 5x β 10) 7. y β 117 = 0 y = 117 β 9. y β 2 3 = β5(x β β y = β5x + 7 3 11. y = β 5 3 x 13. y = 8 5 x β 8 15. y = 5 17. y = β 5 4 x + 11 8 19. y = βx 21. f (x) = 2x β 1 slope: m = 2 y-intercept: (0, β1) 2, 0 x-intercept: 1 22. f (x) = 3 β x slope: m = β1 y-intercept: (0, 3) x-intercept: (3, 0) β 3) 10. y + 12 = 678(x + 1) y = 678x + 666 12. y = β2 6x + 1) 7 x + 29 7 β 8. y + 3 = β y = β β 2x β 3 2(x β 0) 14. y = 9 4 x β 47 4 16. y = β8 18. y = 2x + 13 6 202 β1 1 2 x β1 β2 β3 y 4 3 2 1 β1 1 2 3 4 x β1 170 Linear and Quadratic Functions 23. f (x) = 3 slope: m = 0 y-intercept: (0, 3) x-intercept: none 24. f (x) = 0 slope: m = 0 y-intercept: (0, 0) x-intercept: {(x, 0) | x is a real number} 25. f (x) = 2 3 3 x + 1 slope: m = 2 3 y-intercept: 0, 1 3 x-intercept: β 1 2, 0 26. f (x) = 1 β x 2 slope: m = β 1 2 y-intercept: 0, 1 2 x-intercept: (1, 0) y y 4 3 2 1 1 β2 β1 1 2 x β2 β1 1 2 x β1 2 1 β1 2 1 y y |
β2 1 2 x β2 β1 1 2 x β1 27. (β1, β1) and 11 5, 27 5 28. d(t) = 3t, t β₯ 0. 29. E(t) = 360t, t β₯ 0. 30. C(x) = 45x + 20, x β₯ 0. 31. C(t) = 80t + 50, 0 β€ t β€ 8. 32. W (x) = 200 +.05x, x β₯ 0 She must make $5500 in weekly sales. 33. C(p) = 0.035p + 1.5 The slope 0.035 means it costs 3.5Β’ per page. C(0) = 1.5 means there is a ο¬xed, or start-up, cost of $1.50 to make each book. 34. F (m) = 2.25m + 2.05 The slope 2.25 means it costs an additional $2.25 for each mile beyond the ο¬rst 0.2 miles. F (0) = 2.05, so according to the model, it would cost $2.05 for a trip of 0 miles. Would this ever really happen? Depends on the driver and the passenger, we suppose. 2.1 Linear Functions 171 35. (a) F (C) = 9 5 C + 32 (c) F (β40) = β40 = C(β40). (b) C(F ) = 5 9 (F β 32) = 5 9 F β 160 9 36. N (T ) = β 2 15 T + 43 3 and N (20) = 35 3 β 12 howls per hour. Having a negative number of howls makes no sense and since N (107.5) = 0 we can put an upper bound of 107.5β¦F on the domain. The lower bound is trickier because thereβs nothing other than common sense to go on. As it gets colder, he howls more often. At some point it will either be so cold that he freezes to death or heβs howling non-stop. So weβre going to say that he can withstand temperatures no lower than β60β¦F so that the applied domain is [β60, 107.5]. 39. C(p) = 40. T (n) = 41. C(m) = 42. |
P (c) = 43. (a) (b) (c) 6p + 1.5 5.5p if if 1 β€ p β€ 5 p β₯ 6 15n 12.5n if if 1 β€ n β€ 9 n β₯ 10 10 10 + 0.15(m β 500) 0 β€ m β€ 500 if if m > 500 0.12c 12 + 0.1(c β 100) if if 1 β€ c β€ 100 c > 100 D(d) = ο£±   β 1 8 2 d + 31 2 2 if if if 0 β€ d β€ 15 15 β€ d β€ 27 27 β€ d β€ 37 D(s) = ο£±   2 1 2 s β 3 8 if if if 0 β€ s β€ 10 10 β€ s β€ 22 22 β€ s β€ 37 8 2 8 2 15 27 37 10 22 37 y = D(d) y = D(s) 172 44. 46. 48 Linear and Quadratic Functions 45. 47 32 β (β3)2 3 β (β3) = 0 49. (3(2)2 + 2(2) β 7) β (3(β4)2 + 2(β4) β 7) 2 β (β4) = β4 23 β (β1)3 2 β (β1) β β 0 16 β 16 β 0 7β3 β 5+4 7 β 5 7+4 5β3 50. 3x2 + 3xh + h2 52. β7 (x β 3)(x + h β 3) 51. β1 x(x + h) 53. 6x + 3h + 2 54. The average rate of change is h(2)βh(0) = β32. During the ο¬rst two seconds after it is dropped, the object has fallen at an average rate of 32 feet per second. (This is called the average velocity of the object.) 2β0 55. The average rate of change is F (28)βF (0) = 0.2372. During the years from 1980 to 2008, the average fuel economy of passenger cars in the US increased, on average, at a rate of 0.2372 miles per gallon per year. 28β0 56. (a) T (4) = 56, so at 10 AM (4 hours after 6 AM |
), it is 56β¦F. T (8) = 64, so at 2 PM (8 hours after 6 AM), it is 64β¦F. T (12) = 56, so at 6 PM (12 hours after 6 AM), it is 56β¦F. (b) The average rate of change is T (8)βT (4) 8β4 = 2. Between 10 AM and 2 PM, the temperature increases, on average, at a rate of 2β¦F per hour. (c) The average rate of change is T (12)βT (8) 12β8 = β2. Between 2 PM and 6 PM, the temperature decreases, on average, at a rate of 2β¦F per hour. (d) The average rate of change is T (12)βT (4) ture, on average, remains constant. 12β4 = 0. Between 10 AM and 6 PM, the tempera- 57. The average rate of change is C(5)βC(3) 5β3 = β2. As production is increased from 3000 to 5000 pens, the cost decreases at an average rate of $200 per 1000 pens produced (20Β’ per pen.) 59. y = 3x 60. y = β6x + 20 62 65. y = β3x 63. y = β2 66. y = 1 6 x + 3 2 68. y = 3x β 4 69. x = 3 61. y = 2 3 x β 4 64. x = β5 67. y = β 3 2 x + 9 70. y = 0 2.2 Absolute Value Functions 173 2.2 Absolute Value Functions There are a few ways to describe what is meant by the absolute value |x| of a real number x. You may have been taught that |x| is the distance from the real number x to 0 on the number line. So, for example, |5| = 5 and | β 5| = 5, since each is 5 units from 0 on the number line. distance is 5 units distance is 5 units β5 β4 β3 β2 β5)2 = x2. Using this deο¬nition, we have Another way to deο¬ne absolute value is by the equation |x| = |5| = 25 = 5. The long and short of both of these procedures is that |x| takes negative real numbers and assigns them to their |
positive counterparts while it leaves positive numbers alone. This last description is the one we shall adopt, and is summarized in the following deο¬nition. 25 = 5 and | β 5| = (β5)2 = β Deο¬nition 2.4. The absolute value of a real number x, denoted |x|, is given by |x| = βx, x, if x < 0 if x β₯ 0 In Deο¬nition 2.4, we deο¬ne |x| using a piecewise-deο¬ned function. (See page 62 in Section 1.4.) To check that this deο¬nition agrees with what we previously understood as absolute value, note that since 5 β₯ 0, to ο¬nd |5| we use the rule |x| = x, so |5| = 5. Similarly, since β5 < 0, we use the rule |x| = βx, so that | β 5| = β(β5) = 5. This is one of the times when itβs best to interpret the expression ββxβ as βthe opposite of xβ as opposed to βnegative xβ. Before we begin studying absolute value functions, we remind ourselves of the properties of absolute value. Theorem 2.1. Properties of Absolute Value: Let a, b and x be real numbers and let n be an integer.a Then Product Rule: |ab| = |a||b| Power Rule: |an| = |a|n whenever an is deο¬ned Quotient Rule: a b = |a| |b|, provided b = 0 Equality Properties: |x| = 0 if and only if x = 0. For c > 0, |x| = c if and only if x = c or βx = c. For c < 0, |x| = c has no solution. aSee page 2 if you donβt remember what an integer is. 174 Linear and Quadratic Functions The proofs of the Product and Quotient Rules in Theorem 2.1 boil down to checking four cases: when both a and b are positive; when they are both negative; when one is positive and the other is negative; and when one or both are zero. For example, suppose we wish to show that |ab| = |a||b|. We need |
to show that this equation is true for all real numbers a and b. If a and b are both positive, then so is ab. Hence, |a| = a, |b| = b and |ab| = ab. Hence, the equation |ab| = |a||b| is the same as ab = ab which is true. If both a and b are negative, then ab is positive. Hence, |a| = βa, |b| = βb and |ab| = ab. The equation |ab| = |a||b| becomes ab = (βa)(βb), which is true. Suppose a is positive and b is negative. Then ab is negative, and we have |ab| = βab, |a| = a and |b| = βb. The equation |ab| = |a||b| reduces to βab = a(βb) which is true. A symmetric argument shows the equation |ab| = |a||b| holds when a is negative and b is positive. Finally, if either a or b (or both) are zero, then both sides of |ab| = |a||b| are zero, so the equation holds in this case, too. All of this rhetoric has shown that the equation |ab| = |a||b| holds true in all cases. The proof of the Quotient Rule is very similar, with the exception that b = 0. The Power Rule can be shown by repeated application of the Product Rule. The βEquality Propertiesβ can be proved using Deο¬nition 2.4 and by looking at the cases when x β₯ 0, in which case |x| = x, or when x < 0, in which case |x| = βx. For example, if c > 0, and |x| = c, then if x β₯ 0, we have x = |x| = c. If, on the other hand, x < 0, then βx = |x| = c, so x = βc. The remaining properties are proved similarly and are left for the Exercises. Our ο¬rst example reviews how to solve basic equations involving absolute value using the properties listed in Theorem 2.1. Example 2.2.1. Solve each of the following equations. 1. |3x β 1| = 6 2. 3 β |x + 5| = 1 |
3. 3|2x + 1| β 5 = 0 4. 4 β |5x + 3| = 5 5. |x| = x2 β 6 6. |x β 2| + 1 = x Solution. 1. The equation |3x β 1| = 6 is of the form |x| = c for c > 0, so by the Equality Properties, |3x β 1| = 6 is equivalent to 3x β 1 = 6 or 3x β 1 = β6. Solving the former, we arrive at x = 7 3, and solving the latter, we get x = β 5 3. We may check both of these solutions by substituting them into the original equation and showing that the arithmetic works out. 2. To use the Equality Properties to solve 3 β |x + 5| = 1, we ο¬rst isolate the absolute value. 3 β |x + 5| = 1 β|x + 5| = β2 |x + 5| = 2 subtract 3 divide by β1 From the Equality Properties, we have x + 5 = 2 or x + 5 = β2, and get our solutions to be x = β3 or x = β7. We leave it to the reader to check both answers in the original equation. 2.2 Absolute Value Functions 175 3. As in the previous example, we ο¬rst isolate the absolute value in the equation 3|2x+1|β5 = 0 3 or 2x + 1 = β 5 3. 3. As usual, we may and get |2x + 1| = 5 Solving the former gives x = 1 substitute both answers in the original equation to check. 3. Using the Equality Properties, we have 2x + 1 = 5 3 and solving the latter gives x = β 4 4. Upon isolating the absolute value in the equation 4 β |5x + 3| = 5, we get |5x + 3| = β1. At this point, we know there cannot be any real solution, since, by deο¬nition, the absolute value of anything is never negative. We are done. 5. The equation |x| = x2 β 6 presents us with some diο¬culty, since x appears both inside and outside of the absolute value. Moreover, there are values of x for which x2 β 6 is positive, negative and zero, so we cannot use the Equality Properties without the |
risk of introducing extraneous solutions, or worse, losing solutions. For this reason, we break equations like this into cases by rewriting the term in absolute values, |x|, using Deο¬nition 2.4. For x < 0, |x| = βx, so for x < 0, the equation |x| = x2 β 6 is equivalent to βx = x2 β 6. Rearranging this gives us x2 + x β 6 = 0, or (x + 3)(x β 2) = 0. We get x = β3 or x = 2. Since only x = β3 satisο¬es x < 0, this is the answer we keep. For x β₯ 0, |x| = x, so the equation |x| = x2 β 6 becomes x = x2 β 6. From this, we get x2 β x β 6 = 0 or (x β 3)(x + 2) = 0. Our solutions are x = 3 or x = β2, and since only x = 3 satisο¬es x β₯ 0, this is the one we keep. Hence, our two solutions to |x| = x2 β 6 are x = β3 and x = 3. 6. To solve |x β 2| + 1 = x, we ο¬rst isolate the absolute value and get |x β 2| = x β 1. Since we see x both inside and outside of the absolute value, we break the equation into cases. The term with absolute values here is |x β 2|, so we replace βxβ with the quantity β(x β 2)β in Deο¬nition 2.4 to get |x β 2| = β(x β 2), (x β 2), if if (x β 2) < 0 (x β 2) β₯ 0 Simplifying yields |x β 2| = βx + 2, x β 2, if x < 2 if x β₯ 2 So, for x < 2, |x β 2| = βx + 2 and our equation |x β 2| = x β 1 becomes βx + 2 = x β 1, which gives x = 3 2. Since this solution satisο¬es x < 2, we keep it. Next, for x β₯ 2, |x β 2| = x β 2, so the equation |x β 2| = |
x β 1 becomes x β 2 = x β 1. Here, the equation reduces to β2 = β1, which signiο¬es we have no solutions here. Hence, our only solution is x = 3 2. Next, we turn our attention to graphing absolute value functions. Our strategy in the next example is to make liberal use of Deο¬nition 2.4 along with what we know about graphing linear functions (from Section 2.1) and piecewise-deο¬ned functions (from Section 1.4). Example 2.2.2. Graph each of the following functions. 1. f (x) = |x| 2. g(x) = |x β 3| 3. h(x) = |x| β 3 4. i(x) = 4 β 2|3x + 1| 176 Linear and Quadratic Functions Find the zeros of each function and the x- and y-intercepts of each graph, if any exist. From the graph, determine the domain and range of each function, list the intervals on which the function is increasing, decreasing or constant, and ο¬nd the relative and absolute extrema, if they exist. Solution. 1. To ο¬nd the zeros of f, we set f (x) = 0. We get |x| = 0, which, by Theorem 2.1 gives us x = 0. Since the zeros of f are the x-coordinates of the x-intercepts of the graph of y = f (x), we get (0, 0) as our only x-intercept. To ο¬nd the y-intercept, we set x = 0, and ο¬nd y = f (0) = 0, so that (0, 0) is our y-intercept as well.1 Using Deο¬nition 2.4, we get f (x) = |x| = βx, x, if x < 0 if x β₯ 0 Hence, for x < 0, we are graphing the line y = βx; for x β₯ 0, we have the line y = x. Proceeding as we did in Section 1.6, we get 3 β2 β1 1 2 3 x f (x) = |x|, x < 0 β3 β2 β1 1 2 3 x f (x) = |x |
|, x β₯ 0 Notice that we have an βopen circleβ at (0, 0) in the graph when x < 0. As we have seen before, this is due to the fact that the points on y = βx approach (0, 0) as the x-values approach 0. Since x is required to be strictly less than zero on this stretch, the open circle is drawn at the origin. However, notice that when x β₯ 0, we get to ο¬ll in the point at (0, 0), which eο¬ectively βplugsβ the hole indicated by the open circle. Thus we get, y 4 3 2 1 β3 β2 β1 1 2 3 x f (x) = |x| 1Actually, since functions can have at most one y-intercept (Do you know why?), as soon as we found (0, 0) as the x-intercept, we knew this was also the y-intercept. 2.2 Absolute Value Functions 177 By projecting the graph to the x-axis, we see that the domain is (ββ, β). Projecting to the y-axis gives us the range [0, β). The function is increasing on [0, β) and decreasing on (ββ, 0]. The relative minimum value of f is the same as the absolute minimum, namely 0 which occurs at (0, 0). There is no relative maximum value of f. There is also no absolute maximum value of f, since the y values on the graph extend inο¬nitely upwards. 2. To ο¬nd the zeros of g, we set g(x) = |x β 3| = 0. By Theorem 2.1, we get x β 3 = 0 so that x = 3. Hence, the x-intercept is (3, 0). To ο¬nd our y-intercept, we set x = 0 so that y = g(0) = |0 β 3| = 3, which yields (0, 3) as our y-intercept. To graph g(x) = |x β 3|, we use Deο¬nition 2.4 to rewrite g as g(x) = |x β 3| = β(x β 3), (x β 3), if if (x β 3) < 0 (x β 3) β₯ 0 Simplifying, |
we get g(x) = βx + 3, x β 3, if x < 3 if x β₯ 3 As before, the open circle we introduce at (3, 0) from the graph of y = βx + 3 is ο¬lled by the point (3, 0) from the line y = x β 3. We determine the domain as (ββ, β) and the range as [0, β). The function g is increasing on [3, β) and decreasing on (ββ, 3]. The relative and absolute minimum value of g is 0 which occurs at (3, 0). As before, there is no relative or absolute maximum value of g. 3. Setting h(x) = 0 to look for zeros gives |x| β 3 = 0. As in Example 2.2.1, we isolate the absolute value to get |x| = 3 so that x = 3 or x = β3. As a result, we have a pair of xintercepts: (β3, 0) and (3, 0). Setting x = 0 gives y = h(0) = |0| β 3 = β3, so our y-intercept is (0, β3). As before, we rewrite the absolute value in h to get h(x) = βx β 3, x β 3, if x < 0 if x β₯ 0 Once again, the open circle at (0, β3) from one piece of the graph of h is ο¬lled by the point (0, β3) from the other piece of h. From the graph, we determine the domain of h is (ββ, β) and the range is [β3, β). On [0, β), h is increasing; on (ββ, 0] it is decreasing. The relative minimum occurs at the point (0, β3) on the graph, and we see β3 is both the relative and absolute minimum value of h. Also, h has no relative or absolute maximum value. 178 Linear and Quadratic Functions (x) = |x β 3| β3 β2 β1 y 1 β1 β2 β3 β4 1 2 3 x h(x) = |x| β 3 x 4. As before, we set i(x) = 0 to ο¬nd the zeros of i and get 4β2|3x+1| |
= 0. Isolating the absolute 3 or x = β1, so 3, 0 and (β1, 0). Substituting x = 0 gives y = i(0) = 4β2|3(0)+1| = 2, value term gives |3x + 1| = 2, so either 3x + 1 = 2 or 3x + 1 = β2. We get x = 1 our x-intercepts are 1 for a y-intercept of (0, 2). Rewriting the formula for i(x) without absolute values gives i(x) = 4 β 2(β(3x + 1)), 4 β 2(3x + 1), if if (3x + 1) < 0 (3x + 1) β₯ 0 = 6x + 6, β6x + 2, if x < β 1 3 if x β₯ β 1 3 The usual analysis near the trouble spot x = β 1 and we get the distinctive ββ¨β shape: 3 gives the βcornerβ of this graph is β 1 3, 4, y 5 3 2 1 β1 β1 1 x i(x) = 4 β 2|3x + 1| and decreasing on β 1 The domain of i is (ββ, β) while the range is (ββ, 4]. The function i is increasing on 3, 4 ββ, β 1 3 and the relative and absolute maximum value of i is 4. Since the graph of i extends downwards forever more, there is no absolute minimum value. As we can see from the graph, there is no relative minimum, either. 3, β. The relative maximum occurs at the point β 1 Note that all of the functions in the previous example bear the characteristic ββ¨β shape of the graph of y = |x|. We could have graphed the functions g, h and i in Example 2.2.2 starting with the graph of f (x) = |x| and applying transformations as in Section 1.7 as our next example illustrates. 2.2 Absolute Value Functions 179 Example 2.2.3. Graph the following functions starting with the graph of f (x) = |x| and using transformations. 1. g(x) = |x β 3| 2. h(x) = |x| β 3 3. i(x) = 4 β 2|3x + 1| Solution. |
We begin by graphing f (x) = |x| and labeling three points, (β1, 1), (0, 0) and (1, 1). y 4 3 2 1 (1, 1) (β1, 1) β3 β2 β1 (0, 0) 1 2 3 x f (x) = |x| 1. Since g(x) = |x β 3| = f (x β 3), Theorem 1.7 tells us to add 3 to each of the x-values of the points on the graph of y = f (x) to obtain the graph of y = g(x). This shifts the graph of y = f (x) to the right 3 units and moves the point (β1, 1) to (2, 1), (0, 0) to (3, 0) and (1, 1) to (4, 1). Connecting these points in the classic ββ¨β fashion produces the graph of y = g(x). y 4 3 2 1 (1, 1) (β1, 1) y 4 3 2 1 (2, 1) (4, 1) β3 β2 β1 (0, 0) 1 2 3 x f (x) = |x| shift right 3 units βββββββββββββ add 3 to each x-coordinate 1 2 (3, 0) 4 5 6 x g(x) = f (x β 3) = |x β 3| 2. For h(x) = |x| β 3 = f (x) β 3, Theorem 1.7 tells us to subtract 3 from each of the y-values of the points on the graph of y = f (x) to obtain the graph of y = h(x). This shifts the graph of y = f (x) down 3 units and moves (β1, 1) to (β1, β2), (0, 0) to (0, β3) and (1, 1) to (1, β2). Connecting these points with the ββ¨β shape produces our graph of y = h(x). y 4 3 2 1 (1, 1) (β1, 1) β3 β2 β1 (0, 0) 1 2 3 x f (x) = |x| shift down 3 units βββββββββββββ |
subtract 3 from each y-coordinate y 1 β1 β3 β2 β1 1 2 3 x (β1, β2) β2 (1, β2) (0, β3) β3 β4 h(x) = f (x) β 3 = |x| β 3 180 Linear and Quadratic Functions 3. We re-write i(x) = 4 β 2|3x + 1| = 4 β 2f (3x + 1) = β2f (3x + 1) + 4 and apply Theorem 1.7. First, we take care of the changes on the βinsideβ of the absolute value. Instead of |x|, we have |3x + 1|, so, in accordance with Theorem 1.7, we ο¬rst subtract 1 from each of the x-values of points on the graph of y = f (x), then divide each of those new values by 3. This eο¬ects a horizontal shift left 1 unit followed by a horizontal shrink by a factor of 3. These transformations move (β1, 1) to β 2 3, 0 and (1, 1) to (0, 1). Next, we take 3, 1, (0, 0) to β 1 care of whatβs happening βoutside ofβ the absolute value. Theorem 1.7 instructs us to ο¬rst multiply each y-value of these new points by β2 then add 4. Geometrically, this corresponds to a vertical stretch by a factor of 2, a reο¬ection across the x-axis and ο¬nally, a vertical shift up 4 units. These transformations move β 2 3, 4, and (0, 1) to (0, 2). Connecting these points with the usual ββ¨β shape produces our graph of y = i(x). 3, 1 to β 2 3, 0 to β 1 3, 2, β 1 y 4 3 2 1 (1, 1) (β1, 10, 2) β1 β1 1 x β3 β2 β1 (0, 0) 1 2 3 x βββββββββββββ f (x) = |x| i(x) = β2f (3x + 1) + 4 = β2|3x + 1| + 4 While the methods |
in Section 1.7 can be used to graph an entire family of absolute value functions, not all functions involving absolute values posses the characteristic ββ¨β shape. As the next example illustrates, often there is no substitute for appealing directly to the deο¬nition. Example 2.2.4. Graph each of the following functions. Find the zeros of each function and the x- and y-intercepts of each graph, if any exist. From the graph, determine the domain and range of each function, list the intervals on which the function is increasing, decreasing or constant, and ο¬nd the relative and absolute extrema, if they exist. 1. f (x) = |x| x Solution. 2. g(x) = |x + 2| β |x β 3| + 1 1. We ο¬rst note that, due to the fraction in the formula of f (x), x = 0. Thus the domain is (ββ, 0) βͺ (0, β). To ο¬nd the zeros of f, we set f (x) = |x| x = 0. This last equation implies |x| = 0, which, from Theorem 2.1, implies x = 0. However, x = 0 is not in the domain of f, 2.2 Absolute Value Functions 181 which means we have, in fact, no x-intercepts. We have no y-intercepts either, since f (0) is undeο¬ned. Re-writing the absolute value in the function gives f (x) = ο£±   βx x x x,, if x < 0 if x > 0 = β1, 1, if x < 0 if x > 0 To graph this function, we graph two horizontal lines: y = β1 for x < 0 and y = 1 for x > 0. We have open circles at (0, β1) and (0, 1) (Can you explain why?) so we get y β3 β2 β1 1 2 3 x f (x) = |x| x As we found earlier, the domain is (ββ, 0) βͺ (0, β). The range consists of just two y-values: {β1, 1}. The function f is constant on (ββ, 0) and ( |
0, β). The local minimum value of f is the absolute minimum value of f, namely β1; the local maximum and absolute maximum values for f also coincide β they both are 1. Every point on the graph of f is simultaneously a relative maximum and a relative minimum. (Can you remember why in light of Deο¬nition 1.11? This was explored in the Exercises in Section 1.6.2.) 2. To ο¬nd the zeros of g, we set g(x) = 0. The result is |x + 2| β |x β 3| + 1 = 0. Attempting to isolate the absolute value term is complicated by the fact that there are two terms with absolute values. In this case, it easier to proceed using cases by re-writing the function g with two separate applications of Deο¬nition 2.4 to remove each instance of the absolute values, one at a time. In the ο¬rst round we get β(x + 2) β |x β 3| + 1, (x + 2) β |x β 3| + 1, if if (x + 2) < 0 (x + 2) β₯ 0 = βx β 1 β |x β 3|, x + 3 β |x β 3|, if x < β2 if x β₯ β2 g(x) = Given that |x β 3| = β(x β 3), x β 3, if if (x β 3) < 0 (x β 3) β₯ 0 = βx + 3, x β 3, if x < 3 if x β₯ 3, we need to break up the domain again at x = 3. Note that if x < β2, then x < 3, so we replace |x β 3| with βx + 3 for that part of the domain, too. Our completed revision of the form of g yields 182 Linear and Quadratic Functions g(x) = ο£±   βx β 1 β (βx + 3), x + 3 β (βx + 3), x + 3 β (x β 3), if x < β2 if x β₯ β2 and x < 3 if x β₯ 3 = ο£±   β4, 2x, 6, if x < β2 if β2 β€ x < 3 if |
x β₯ 3 To solve g(x) = 0, we see that the only piece which contains a variable is g(x) = 2x for β2 β€ x < 3. Solving 2x = 0 gives x = 0. Since x = 0 is in the interval [β2, 3), we keep this solution and have (0, 0) as our only x-intercept. Accordingly, the y-intercept is also (0, 0). To graph g, we start with x < β2 and graph the horizontal line y = β4 with an open circle at (β2, β4). For β2 β€ x < 3, we graph the line y = 2x and the point (β2, β4) patches the hole left by the previous piece. An open circle at (3, 6) completes the graph of this part. Finally, we graph the horizontal line y = 6 for x β₯ 3, and the point (3, 6) ο¬lls in the open circle left by the previous part of the graph. The ο¬nished graph is y 6 5 4 3 2 1 β1 β2 β3 β4 β4 β3 β2 β1 1 2 3 4 x g(x) = |x + 2| β |x β 3| + 1 The domain of g is all real numbers, (ββ, β), and the range of g is all real numbers between β4 and 6 inclusive, [β4, 6]. The function is increasing on [β2, 3] and constant on (ββ, β2] and [3, β). The relative minimum value of f is β4 which matches the absolute minimum. The relative and absolute maximum values also coincide at 6. Every point on the graph of y = g(x) for x < β2 and x > 3 yields both a relative minimum and relative maximum. The point (β2, β4), however, gives only a relative minimum and the point (3, 6) yields only a relative maximum. (Recall the Exercises in Section 1.6.2 which dealt with constant functions.) Many of the applications that the authors are aware of involving absolute values also involve absolute value inequalities. For that reason, we save our discussion of applications for Section 2.4. 2.2 Absolute Value Functions 183 2.2.1 Exercises In Exercises 1 - 15, solve the equation. 1. |x| = |
6 4. 4 β |x| = 3 7. 5 β |x| 2 = 1 2. |3x β 1| = 10 3. |4 β x| = 7 5. 2|5x + 1| β 3 = 0 6. |7x β 1| + 2 = 0 8. 2 3 |5 β 2x| β 1 2 = 5 9. |x| = x + 3 10. |2x β 1| = x + 1 11. 4 β |x| = 2x + 1 12. |x β 4| = x β 5 13. |x| = x2 14. |x| = 12 β x2 15. |x2 β 1| = 3 Prove that if |f (x)| = |g(x)| then either f (x) = g(x) or f (x) = βg(x). Use that result to solve the equations in Exercises 16 - 21. 16. |3x β 2| = |2x + 7| 17. |3x + 1| = |4x| 18. |1 β 2x| = |x + 1| 19. |4 β x| β |x + 2| = 0 20. |2 β 5x| = 5|x + 1| 21. 3|x β 1| = 2|x + 1| In Exercises 22 - 33, graph the function. Find the zeros of each function and the x- and y-intercepts of each graph, if any exist. From the graph, determine the domain and range of each function, list the intervals on which the function is increasing, decreasing or constant, and ο¬nd the relative and absolute extrema, if they exist. 22. f (x) = |x + 4| 23. f (x) = |x| + 4 24. f (x) = |4x| 25. f (x) = β3|x| 26. f (x) = 3|x + 4| β 4 27. f (x) = 1 3 |2x β 1| 28. f (x) = |x + 4| x + 4 29. f (x) = |2 β x| 2 β x 30. f (x) = x + |x| β 3 31. f (x) = |x + 2| β x 32. f (x |
) = |x + 2| β |x| 33. f (x) = |x + 4| + |x β 2| 34. With the help of your classmates, ο¬nd an absolute value function whose graph is given below. y 4 3 2 1 β8β7β6β5β4β3β2β 35. With help from your classmates, prove the second, third and ο¬fth parts of Theorem 2.1. 36. Prove The Triangle Inequality: For all real numbers a and b, |a + b| β€ |a| + |b|. 184 Linear and Quadratic Functions 2.2.2 Answers 1. x = β6 or x = 6 4. x = β1 or x = 1 7. x = β3 or x = 3 2. x = β3 or x = 11 3 3. x = β3 or x = 11 5. x = β 1 2 or x = 1 10 8. x = β 13 8 or x = 53 8 6. no solution 9. x = β 3 2 12. no solution 10. x = 0 or x = 2 11. x = 1 13. x = β1, x = 0 or x = 1 14. x = β3 or x = 3 15. x = β2 or x = 2 16. x = β1 or x = 9 17. x = β 1 7 or x = 1 19. x = 1 20. x = β 3 10 18. x = 0 or x = 2 21. x = 1 5 or x = 5 22. f (x) = |x + 4| f (β4) = 0 x-intercept (β4, 0) y-intercept (0, 4) Domain (ββ, β) Range [0, β) Decreasing on (ββ, β4] Increasing on [β4, β) Relative and absolute min. at (β4, 0) No relative or absolute maximum 23. f (x) = |x| + 4 No zeros No x-intercepts y-intercept (0, 4) Domain (ββ, β) Range [4, β) Decreasing on (ββ, 0] Increasing on [0, β) Relative and absolute minimum at (0, 4) No relative or absolute maximum y 4 3 2 1 β8 β7 β6 β5 β4 |
β3 β2 β4 β3 β2 β1 1 2 3 4 x 2.2 Absolute Value Functions 185 24. f (x) = |4x| f (0) = 0 x-intercept (0, 0) y-intercept (0, 0) Domain (ββ, β) Range [0, β) Decreasing on (ββ, 0] Increasing on [0, β) Relative and absolute minimum at (0, 0) No relative or absolute maximum 25. f (x) = β3|x| f (0) = 0 x-intercept (0, 0) y-intercept (0, 0) Domain (ββ, β) Range (ββ, 0] Increasing on (ββ, 0] Decreasing on [0, β) Relative and absolute maximum at (0, 0) No relative or absolute minimum 26. f (x) = 3|x + 4| β 4 = 0, β 8 f β 16 3 x-intercepts β 16 y-intercept (0, 8) Domain (ββ, β) Range [β4, β) Decreasing on (ββ, β4] Increasing on [β4, β) Relative and absolute min. at (β4, β4) No relative or absolute maximum 3 |2x β 1| 2, 0 27. f (x) = 1 f 1 = 0 2 x-intercepts 1 y-intercept 0, 1 3 Domain (ββ, β) Range [0, β) Decreasing on ββ, 1 2 Increasing on 2 β1 1 2 x y 1 2 x β2 β1 β1 β2 β3 β4 β5 β8β7β6β5β4β3β2β1 β1 x 1 β2 β3 β4 Relative and absolute min. at 1 No relative or absolute maximum 2, 0 y 2 1 β3 β2 β1 1 2 3 4 x 186 28. f (x) = |x + 4| x + 4 No zeros No x-intercept y-intercept (0, 1) Domain (ββ, β4) βͺ (β4, β) Range {β1, 1} Constant on (ββ, β4) Constant on (β4, β) Absolute minimum at every point (x, β1) 29. f ( |
x) = |2 β x| 2 β x No zeros No x-intercept y-intercept (0, 1) Domain (ββ, 2) βͺ (2, β) Range {β1, 1} Constant on (ββ, 2) Constant on (2, β) Absolute minimum at every point (x, β1) 30. Re-write f (x) = x + |x| β 3 as x < 0 x β₯ 0 β3 2x β 3 f (x) = if if = 0 f 3 2 x-intercept 3 2, 0 y-intercept (0, β3) Domain (ββ, β) Range [β3, β) Increasing on [0, β) Constant on (ββ, 0] Absolute minimum at every point (x, β3) where x β€ 0 No absolute maximum Linear and Quadratic Functions where x < β4 Absolute maximum at every point (x, 1) where x > β4 Relative maximum AND minimum at every point on the graph y 1 β8 β7 β6 β5 β4 β3 β2 β1 β1 1 x where x > 2 Absolute maximum at every point (x, 1) where x < 2 Relative maximum AND minimum at every point on the graph y 1 β3 β2 β1 β1 1 2 3 4 5 x Relative minimum at every point (x, β3) where x β€ 0 Relative maximum at every point (x, β3) where x < 0 y 2 1 β2 β1 β1 β2 β3 β4 1 2 x 2.2 Absolute Value Functions 187 31. Re-write f (x) = |x + 2| β x as f (x) = β2x β 2 2 if if x < β2 x β₯ β2 No zeros No x-intercepts y-intercept (0, 2) Domain (ββ, β) Range [2, β) Decreasing on (ββ, β2] Constant on [β2, β) Absolute minimum at every point (x, 2) where x β₯ β2 32. Re-write f (x) = |x + 2| β |x| as x < β2 if if β2 β€ x < 0 x β₯ 0 if β2 2x + 2 2 f (x) = ο£±  � |
οΏ½οΏ½ f (β1) = 0 x-intercept (β1, 0) y-intercept (0, 2) Domain (ββ, β) Range [β2, 2] Increasing on [β2, 0] Constant on (ββ, β2] Constant on [0, β) Absolute minimum at every point (x, β2) where x β€ β2 33. Re-write f (x) = |x + 4| + |x β 2| as β2x β 2 6 2x + 2 x < β4 if if β4 β€ x < 2 x β₯ 2 if f (x) = ο£±   No zeros No x-intercept y-intercept (0, 6) Domain (ββ, β) Range [6, β) Decreasing on (ββ, β4] Constant on [β4, 2] Increasing on [2, β) Absolute minimum at every point (x, 6) where β4 β€ x β€ 2 No absolute maximum Relative minimum at every point (x, 6) where 35. f (x) = ||x| β 4| No absolute maximum Relative minimum at every point (x, 2) where x β₯ β2 Relative maximum at every point (x, 2) where x > β2 y 3 2 1 β3 β2 β1 1 2 x Absolute maximum at every point (x, 2) where x β₯ 0 Relative minimum at every point (x, β2) where x β€ β2 and at every point (x, 2) where x > 0 Relative maximum at every point (x, β2) where x < β2 and at every point (x, 2) where 3 β2 β1 β1 β2 β4 β€ x β€ 2 Relative maximum at every point (x, 6) where β5 β4 β3 β2 β1 1 2 3 x 188 Linear and Quadratic Functions 2.3 Quadratic Functions You may recall studying quadratic equations in Intermediate Algebra. In this section, we review those equations in the context of our next family of functions: the quadratic functions. Deο¬nition 2.5. A quadratic function is a function of the form f (x) = ax2 + bx + c, where a, b and c are real numbers with a = 0. The domain of a quad |
ratic function is (ββ, β). The most basic quadratic function is f (x) = x2, whose graph appears below. Its shape should look familiar from Intermediate Algebra β it is called a parabola. The point (0, 0) is called the vertex of the parabola. In this case, the vertex is a relative minimum and is also the where the absolute minimum value of f can be found. y 4 3 2 1 (2, 4) (1, 1) (β2, 4) (β1, 1) β2 β1 1 (0, 0) f (x) = x2 2 x Much like many of the absolute value functions in Section 2.2, knowing the graph of f (x) = x2 enables us to graph an entire family of quadratic functions using transformations. Example 2.3.1. Graph the following functions starting with the graph of f (x) = x2 and using transformations. Find the vertex, state the range and ο¬nd the x- and y-intercepts, if any exist. 1. g(x) = (x + 2)2 β 3 2. h(x) = β2(x β 3)2 + 1 Solution. 1. Since g(x) = (x + 2)2 β 3 = f (x + 2) β 3, Theorem 1.7 instructs us to ο¬rst subtract 2 from each of the x-values of the points on y = f (x). This shifts the graph of y = f (x) to the left 2 units and moves (β2, 4) to (β4, 4), (β1, 1) to (β3, 1), (0, 0) to (β2, 0), (1, 1) to (β1, 1) and (2, 4) to (0, 4). Next, we subtract 3 from each of the y-values of these new points. This moves the graph down 3 units and moves (β4, 4) to (β4, 1), (β3, 1) to (β3, β2), (β2, 0) to (β2, 3), (β1, 1) to (β1, β2) and (0, 4) to (0, 1). We connect the dots in parabolic fashion to get 2.3 Quadratic Functions 189 y 4 3 2 1 (β2, 4) (β1, |
1) y (2, 4) (β4, 1) 1 (0, 1) (1, 1) (β3, β2) (β1, β2) β4 β3 β2 β1 x β1 β2 β1 (0, 0) 1 2 x β3 (β2, β3) f (x) = x2 βββββββββββββ g(x) = f (x + 2) β 3 = (x + 2)2 β 3 From the graph, we see that the vertex has moved from (0, 0) on the graph of y = f (x) to (β2, β3) on the graph of y = g(x). This sets [β3, β) as the range of g. We see that the graph of y = g(x) crosses the x-axis twice, so we expect two x-intercepts. To ο¬nd these, we set y = g(x) = 0 and solve. Doing so yields the equation (x + 2)2 β 3 = 0, or (x + 2)2 = 3. Extracting square roots gives x + 2 = Β± 3. Our x-intercepts are (β2 β 3, 0) β (β0.27, 0). The y-intercept of the graph, 3, 0) β (β3.73, 0) and (β2 + (0, 1) was one of the points we originally plotted, so we are done. 3, or x = β2 Β± β β β β 2. Following Theorem 1.7 once more, to graph h(x) = β2(x β 3)2 + 1 = β2f (x β 3) + 1, we ο¬rst start by adding 3 to each of the x-values of the points on the graph of y = f (x). This eο¬ects a horizontal shift right 3 units and moves (β2, 4) to (1, 4), (β1, 1) to (2, 1), (0, 0) to (3, 0), (1, 1) to (4, 1) and (2, 4) to (5, 4). Next, we multiply each of our y-values ο¬rst by β2 and then add 1 to that result. Geometrically, |
this is a vertical stretch by a factor of 2, followed by a reο¬ection about the x-axis, followed by a vertical shift up 1 unit. This moves (1, 4) to (1, β7), (2, 1) to (2, β1), (3, 0) to (3, 1), (4, 1) to (4, β1) and (5, 4) to (5, β7). y (3, 1) 2 1 (2, β1) 3 4 5 (4, β1) x 1 β1 β2 β3 β4 β5 β6 (2, 4) (1, 1) (1, β7) (5, β7) y 4 3 2 1 (β2, 4) (β1, 1) β2 β1 (0, 0) 1 2 x f (x) = x2 βββββββββββββ h(x) = β2f (x β 3) + 1 = β2(x β 3)2 + 1 The vertex is (3, 1) which makes the range of h (ββ, 1]. From our graph, we know that there are two x-intercepts, so we set y = h(x) = 0 and solve. We get β2(x β 3)2 + 1 = 0 190 Linear and Quadratic Functions which gives (x β 3)2 = 1 add 3 to each side,2 we get x = 6Β± 2 6+ 2, 0 β 2 β 2 2. Extracting square roots1 gives x β 3 = Β± 2, so that when we β 6β 2 β (2.29, 0) and, 0 2. Hence, our x-intercepts are β 2 β (3.71, 0). Although our graph doesnβt show it, there is a y-intercept which can be found by setting x = 0. With h(0) = β2(0 β 3)2 + 1 = β17, we have that our y-intercept is (0, β17). A few remarks about Example 2.3.1 are in order. First note that neither the formula given for g(x) nor the one given for h(x) match the form given in Deο¬nition 2.5. We could, of course, |
convert both g(x) and h(x) into that form by expanding and collecting like terms. Doing so, we ο¬nd g(x) = (x + 2)2 β 3 = x2 + 4x + 1 and h(x) = β2(x β 3)2 + 1 = β2x2 + 12x β 17. While these βsimpliο¬edβ formulas for g(x) and h(x) satisfy Deο¬nition 2.5, they do not lend themselves to graphing easily. For that reason, the form of g and h presented in Example 2.3.2 is given a special name, which we list below, along with the form presented in Deο¬nition 2.5. Deο¬nition 2.6. Standard and General Form of Quadratic Functions: Suppose f is a quadratic function. The general form of the quadratic function f is f (x) = ax2 + bx + c, where a, b and c are real numbers with a = 0. The standard form of the quadratic function f is f (x) = a(x β h)2 + k, where a, h and k are real numbers with a = 0. It is important to note at this stage that we have no guarantees that every quadratic function can be written in standard form. This is actually true, and we prove this later in the exposition, but for now we celebrate the advantages of the standard form, starting with the following theorem. Theorem 2.2. Vertex Formula for Quadratics in Standard Form: For the quadratic function f (x) = a(x β h)2 + k, where a, h and k are real numbers with a = 0, the vertex of the graph of y = f (x) is (h, k). We can readily verify the formula given Theorem 2.2 with the two functions given in Example 2.3.1. After a (slight) rewrite, g(x) = (x + 2)2 β 3 = (x β (β2))2 + (β3), and we identify h = β2 and k = β3. Sure enough, we found the vertex of the graph of y = g(x) to be (β2, β3). For h(x) = β2(x β 3 |
)2 + 1, no rewrite is needed. We can directly identify h = 3 and k = 1 and, sure enough, we found the vertex of the graph of y = h(x) to be (3, 1). To see why the formula in Theorem 2.2 produces the vertex, consider the graph of the equation y = a(x β h)2 + k. When we substitute x = h, we get y = k, so (h, k) is on the graph. If x = h, then x β h = 0 so (x β h)2 is a positive number. If a > 0, then a(x β h)2 is positive, thus y = a(x β h)2 + k is always a number larger than k. This means that when a > 0, (h, k) is the lowest point on the graph and thus the parabola must open upwards, making (h, k) the vertex. A similar argument 1and rationalizing denominators! 2and get common denominators! 2.3 Quadratic Functions 191 shows that if a < 0, (h, k) is the highest point on the graph, so the parabola opens downwards, and (h, k) is also the vertex in this case. Alternatively, we can apply the machinery in Section 1.7. Since the vertex of y = x2 is (0, 0), we can determine the vertex of y = a(xβh)2 +k by determining the ο¬nal destination of (0, 0) as it is moved through each transformation. To obtain the formula f (x) = a(x β h)2 + k, we start with g(x) = x2 and ο¬rst deο¬ne g1(x) = ag(x) = ax2. This is results in a vertical scaling and/or reο¬ection.3 Since we multiply the output by a, we multiply the y-coordinates on the graph of g by a, so the point (0, 0) remains (0, 0) and remains the vertex. Next, we deο¬ne g2(x) = g1(x β h) = a(x β h)2. This induces a horizontal shift right or left h units4 moves the vertex, in either case, to (h, 0). Finally, f (x |
) = g2(x) + k = a(x β h)2 + k which eο¬ects a vertical shift up or down k units5 resulting in the vertex moving from (h, 0) to (h, k). In addition to verifying Theorem 2.2, the arguments in the two preceding paragraphs have also shown us the role of the number a in the graphs of quadratic functions. The graph of y = a(xβh)2+k is a parabola βopening upwardsβ if a > 0, and βopening downwardsβ if a < 0. Moreover, the symmetry enjoyed by the graph of y = x2 about the y-axis is translated to a symmetry about the vertical line x = h which is the vertical line through the vertex.6 This line is called the axis of symmetry of the parabola and is dashed in the ο¬gures below. vertex a < 0 vertex a > 0 Graphs of y = a(x β h)2 + k. Without a doubt, the standard form of a quadratic function, coupled with the machinery in Section 1.7, allows us to list the attributes of the graphs of such functions quickly and elegantly. What remains to be shown, however, is the fact that every quadratic function can be written in standard form. To convert a quadratic function given in general form into standard form, we employ the ancient rite of βCompleting the Squareβ. We remind the reader how this is done in our next example. Example 2.3.2. Convert the functions below from general form to standard form. Find the vertex, axis of symmetry and any x- or y-intercepts. Graph each function and determine its range. 1. f (x) = x2 β 4x + 3. 2. g(x) = 6 β x β x2 3Just a scaling if a > 0. If a < 0, there is a reο¬ection involved. 4Right if h > 0, left if h < 0. 5Up if k > 0, down if k < 0 6You should use transformations to verify this! 192 Solution. Linear and Quadratic Functions 1. To convert from general form to standard form, we complete the square.7 First, we verify that the coeο¬cient of x2 is 1. Next, we ο¬nd the coeοΏ½ |
οΏ½οΏ½cient of x, in this case β4, and take half of it to get 1 2 (β4) = β2. This tells us that our target perfect square quantity is (x β 2)2. To get an expression equivalent to (x β 2)2, we need to add (β2)2 = 4 to the x2 β 4x to create a perfect square trinomial, but to keep the balance, we must also subtract it. We collect the terms which create the perfect square and gather the remaining constant terms. Putting it all together, we get f (x) = x2 β 4x + 3 2 (β4) = β2.) = x2 β 4x + 4 β 4 + 3 (Add and subtract (β2)2 = 4 to (x2 + 4x).) = x2 β 4x + 4 β 4 + 3 (Group the perfect square trinomial.) = (x β 2)2 β 1 (Factor the perfect square trinomial.) (Compute 1 Of course, we can always check our answer by multiplying out f (x) = (x β 2)2 β 1 to see that it simpliο¬es to f (x) = x2 β 4x β 1. In the form f (x) = (x β 2)2 β 1, we readily ο¬nd the vertex to be (2, β1) which makes the axis of symmetry x = 2. To ο¬nd the x-intercepts, we set y = f (x) = 0. We are spoiled for choice, since we have two formulas for f (x). Since we recognize f (x) = x2 β 4x + 3 to be easily factorable,8 we proceed to solve x2 β 4x + 3 = 0. Factoring gives (x β 3)(x β 1) = 0 so that x = 3 or x = 1. The x-intercepts are then (1, 0) and (3, 0). To ο¬nd the y-intercept, we set x = 0. Once again, the general form f (x) = x2 β 4x + 3 is easiest to work with here, and we ο¬nd y = f (0) = 3. Hence, the y-intercept is (0, 3). With the vertex, axis of symmetry and the intercepts, we get a pretty good graph |
without the need to plot additional points. We see that the range of f is [β1, β) and we are done. 2. To get started, we rewrite g(x) = 6 β x β x2 = βx2 β x + 6 and note that the coeο¬cient of x2 is β1, not 1. This means our ο¬rst step is to factor out the (β1) from both the x2 and x terms. We then follow the completing the square recipe as above. g(x) = βx2 β x + 6 = (β1) x2 + x + 6 x2 + x + 1 = (β1) = (β1) x2 + x + 1 4 = β x + 1 + 25 4 2 2 4 4 β 1 + 6 + (β1) β 1 4 (Factor the coeο¬cient of x2 from x2 and x.) + 6 (Group the perfect square trinomial.) 7If you forget why we do what we do to complete the square, start with a(x β h)2 + k, multiply it out, step by step, and then reverse the process. 8Experience pays oο¬, here! 2.3 Quadratic Functions 193 2 + 25 4, we get the vertex to be β 1 From g(x) = β x + 1 and the axis of symmetry to 2 be x = β 1 2. To get the x-intercepts, we opt to set the given formula g(x) = 6 β x β x2 = 0. Solving, we get x = β3 and x = 2, so the x-intercepts are (β3, 0) and (2, 0). Setting x = 0, we ο¬nd g(0) = 6, so the y-intercept is (0, 6). Plotting these points gives us the graph below. We see that the range of g is ββ, 25 4 2, 25. 0, 3) x = 2 (1, 0) (3, 0) β1 1 2 3 4 5 x β1 (2, β1) f (x) = x2 β 4x + 3 y (0, 6) β 1 2, 25 4 6 5 4 3 2 (β3, 0) x = 1 2 (2, 0) β3 β2 β1 1 2 x g(x |
) = 6 β x β x2 With Example 2.3.2 fresh in our minds, we are now in a position to show that every quadratic function can be written in standard form. We begin with f (x) = ax2 + bx + c, assume a = 0, and complete the square in complete generality. f (x) = ax2 + bx + c x + c (Factor out coeο¬cient of x2 from x2 and x.) x2 + = a x2 + = a x2 + = a b a b a b a b2 4a2 + c x + x + b2 4a2 β b2 4a2 β a = a x + 2 b 2a + 4ac β b2 4a b2 4a2 + c (Group the perfect square trinomial.) (Factor and get a common denominator.) so that Comparing this last expression with the standard form, we identify (x β h) with x + b 2a h = β b. As such, we have derived a vertex formula for the general form. We summarize both vertex formulas in the box at the top of the next page. 2a. Instead of memorizing the value k = 4acβb2, we see that f β b 2a = 4acβb2 4a 4a 194 Linear and Quadratic Functions Equation 2.4. Vertex Formulas for Quadratic Functions: Suppose a, b, c, h and k are real numbers with a = 0. If f (x) = a(x β h)2 + k, the vertex of the graph of y = f (x) is the point (h, k). If f (x) = ax2 + bx + c, the vertex of the graph of y = f (x) is the point β b 2a, f β. b 2a There are two more results which can be gleaned from the completed-square form of the general form of a quadratic function, f (x) = ax2 + bx + c = a x + 2 b 2a + 4ac β b2 4a We have seen that the number a in the standard form of a quadratic function determines whether the parabola opens upwards (if a > 0) or downwards (if a < 0). We see here that this number a is none other than the coeοΏ½ |
οΏ½οΏ½cient of x2 in the general form of the quadratic function. In other words, it is the coeο¬cient of x2 alone which determines this behavior β a result that is generalized in Section 3.1. The second treasure is a re-discovery of the quadratic formula. Equation 2.5. The Quadratic Formula: If a, b and c are real numbers with a = 0, then the solutions to ax2 + bx + c = 0 are βb Β± x = β b2 β 4ac 2a. Assuming the conditions of Equation 2.5, the solutions to ax2 + bx + c = 0 are precisely the zeros of f (x) = ax2 + bx + c. Since f (x) = ax2 + bx + c = a x + 2 b 2a + 4ac β b2 4a the equation ax2 + bx + c = 0 is equivalent to a x + 2 b 2a + 4ac β b2 4a = 0. Solving gives 2.3 Quadratic Functions 195 a x + 2 b 2a + 4ac β b2 4a = 0 2 b 2a 2a = β 4ac β b2 4a b2 β 4ac 4a 1 a b2 β 4ac 4a2 b2 β 4ac 4a2 = = = Β± = Β± β b2 β 4ac 2a β b2 β 4ac 2a b 2a Β± x = β x + 2 b 2a x + x + b 2a b 2a extract square roots βb Β± x = β b2 β 4ac 2a In our discussions of domain, we were warned against having negative numbers underneath the b2 β 4ac is part of the Quadratic Formula, we will need to pay special square root. Given that attention to the radicand b2 β 4ac. It turns out that the quantity b2 β 4ac plays a critical role in determining the nature of the solutions to a quadratic equation. It is given a special name. β Deο¬nition 2.7. If a, b and c are real numbers with a = 0, then the discriminant of the quadratic equation ax2 + bx + c = 0 is the quantity b2 β 4ac. The discriminant βdiscriminatesβ between the kinds of |
solutions we get from a quadratic equation. These cases, and their relation to the discriminant, are summarized below. Theorem 2.3. Discriminant Trichotomy: Let a, b and c be real numbers with a = 0. If b2 β 4ac < 0, the equation ax2 + bx + c = 0 has no real solutions. If b2 β 4ac = 0, the equation ax2 + bx + c = 0 has exactly one real solution. If b2 β 4ac > 0, the equation ax2 + bx + c = 0 has exactly two real solutions. The proof of Theorem 2.3 stems from the position of the discriminant in the quadratic equation, and is left as a good mental exercise for the reader. The next example exploits the fruits of all of our labor in this section thus far. 196 Linear and Quadratic Functions Example 2.3.3. Recall that the proο¬t (deο¬ned on page 82) for a product is deο¬ned by the equation Proο¬t = Revenue β Cost, or P (x) = R(x) β C(x). In Example 2.1.7 the weekly revenue, in dollars, made by selling x PortaBoy Game Systems was found to be R(x) = β1.5x2 + 250x with the restriction (carried over from the price-demand function) that 0 β€ x β€ 166. The cost, in dollars, to produce x PortaBoy Game Systems is given in Example 2.1.5 as C(x) = 80x + 150 for x β₯ 0. 1. Determine the weekly proο¬t function P (x). 2. Graph y = P (x). Include the x- and y-intercepts as well as the vertex and axis of symmetry. 3. Interpret the zeros of P. 4. Interpret the vertex of the graph of y = P (x). 5. Recall that the weekly price-demand equation for PortaBoys is p(x) = β1.5x + 250, where p(x) is the price per PortaBoy, in dollars, and x is the weekly sales. What should the price per system be in order to maximize proο¬t? Solution. 1. To ο¬nd the proο¬t function P (x), we subtract P |
(x) = R(x) β C(x) = β1.5x2 + 250x β (80x + 150) = β1.5x2 + 170x β 150. Since the revenue function is valid when 0 β€ x β€ 166, P is also restricted to these values. 2. To ο¬nd the x-intercepts, we set P (x) = 0 and solve β1.5x2 + 170x β 150 = 0. The mere thought of trying to factor the left hand side of this equation could do serious psychological damage, so we resort to the quadratic formula, Equation 2.5. Identifying a = β1.5, b = 170, and c = β150, we obtain x = = = = β βb Β± b2 β 4ac 2a β170 Β± 1702 β 4(β1.5)(β150) β β170 Β± 2(β1.5) 28000 β3 β 170 Β± 20 70 3 170β20 3 β 70 170+20 3 β 70. To ο¬nd the y-intercept, we set We get two x-intercepts: x = 0 and ο¬nd y = P (0) = β150 for a y-intercept of (0, β150). To ο¬nd the vertex, we use the fact that P (x) = β1.5x2 + 170x β 150 is in the general form of a quadratic function and 2(β1.5) = 170 appeal to Equation 2.4. Substituting a = β1.5 and b = 170, we get x = β 170 3. and, 0, 0 2.3 Quadratic Functions 197 3 To ο¬nd the y-coordinate of the vertex, we compute P 170 and ο¬nd that our vertex 3 is 170. The axis of symmetry is the vertical line passing through the vertex so it is 3, 14000 the line x = 170 3. To sketch a reasonable graph, we approximate the x-intercepts, (0.89, 0) and (112.44, 0), and the vertex, (56.67, 4666.67). (Note that in order to get the x-intercepts and the vertex to show up in the same picture, we had to scale the x-axis diοΏ½ |
οΏ½οΏ½erently than the y-axis. This results in the left-hand x-intercept and the y-intercept being uncomfortably close to each other and to the origin in the picture.) = 14000 3 y 4000 3000 2000 1000 10 20 30 40 50 60 70 80 90 100 110 120 x 3. The zeros of P are the solutions to P (x) = 0, which we have found to be approximately 0.89 and 112.44. As we saw in Example 1.5.3, these are the βbreak-evenβ points of the proο¬t function, where enough product is sold to recover the cost spent to make the product. More importantly, we see from the graph that as long as x is between 0.89 and 112.44, the graph y = P (x) is above the x-axis, meaning y = P (x) > 0 there. This means that for these values of x, a proο¬t is being made. Since x represents the weekly sales of PortaBoy Game Systems, we round the zeros to positive integers and have that as long as 1, but no more than 112 game systems are sold weekly, the retailer will make a proο¬t. 4. From the graph, we see that the maximum value of P occurs at the vertex, which is approximately (56.67, 4666.67). As above, x represents the weekly sales of PortaBoy systems, so we canβt sell 56.67 game systems. Comparing P (56) = 4666 and P (57) = 4666.5, we conclude that we will make a maximum proο¬t of $4666.50 if we sell 57 game systems. 5. In the previous part, we found that we need to sell 57 PortaBoys per week to maximize proο¬t. To ο¬nd the price per PortaBoy, we substitute x = 57 into the price-demand function to get p(57) = β1.5(57) + 250 = 164.5. The price should be set at $164.50. Our next example is another classic application of quadratic functions. Example 2.3.4. Much to Donnieβs surprise and delight, he inherits a large parcel of land in Ashtabula County from one of his (e)strange(d) relatives. |
The time is ο¬nally right for him to pursue his dream of farming alpaca. He wishes to build a rectangular pasture, and estimates that he has enough money for 200 linear feet of fencing material. If he makes the pasture adjacent to a stream (so no fencing is required on that side), what are the dimensions of the pasture which maximize the area? What is the maximum area? If an average alpaca needs 25 square feet of grazing area, how many alpaca can Donnie keep in his pasture? 198 Linear and Quadratic Functions Solution. It is always helpful to sketch the problem situation, so we do so below. river w pasture w l We are tasked to ο¬nd the dimensions of the pasture which would give a maximum area. We let w denote the width of the pasture and we let l denote the length of the pasture. Since the units given to us in the statement of the problem are feet, we assume w and l are measured in feet. The area of the pasture, which weβll call A, is related to w and l by the equation A = wl. Since w and l are both measured in feet, A has units of feet2, or square feet. We are given the total amount of fencing available is 200 feet, which means w + l + w = 200, or, l + 2w = 200. We now have two equations, A = wl and l + 2w = 200. In order to use the tools given to us in this section to maximize A, we need to use the information given to write A as a function of just one variable, either w or l. This is where we use the equation l + 2w = 200. Solving for l, we ο¬nd l = 200 β 2w, and we substitute this into our equation for A. We get A = wl = w(200 β 2w) = 200w β 2w2. We now have A as a function of w, A(w) = 200w β 2w2 = β2w2 + 200w. Before we go any further, we need to ο¬nd the applied domain of A so that we know what values of w make sense in this problem situation.9 Since w represents the width of the pasture, w > 0. Likewise, l represents the length of the pasture, so l = 200 β 2w > 0. Solving this latter inequality, we ο¬ |
nd w < 100. Hence, the function we wish to maximize is A(w) = β2w2 +200w for 0 < w < 100. Since A is a quadratic function (of w), we know that the graph of y = A(w) is a parabola. Since the coeο¬cient of w2 is β2, we know that this parabola opens downwards. This means that there is a maximum value to be found, and we know it occurs at the vertex. Using the vertex formula, we ο¬nd w = β 200 2(β2) = 50, and A(50) = β2(50)2 + 200(50) = 5000. Since w = 50 lies in the applied domain, 0 < w < 100, we have that the area of the pasture is maximized when the width is 50 feet. To ο¬nd the length, we use l = 200 β 2w and ο¬nd l = 200 β 2(50) = 100, so the length of the pasture is 100 feet. The maximum area is A(50) = 5000, or 5000 square feet. If an average alpaca requires 25 square feet of pasture, Donnie can raise 5000 25 = 200 average alpaca. We conclude this section with the graph of a more complicated absolute value function. Example 2.3.5. Graph f (x) = |x2 β x β 6|. Solution. Using the deο¬nition of absolute value, Deο¬nition 2.4, we have f (x) = β x2 β x β 6, x2 β x β 6, if x2 β x β 6 < 0 if x2 β x β 6 β₯ 0 The trouble is that we have yet to develop any analytic techniques to solve nonlinear inequalities such as x2 β x β 6 < 0. You wonβt have to wait long; this is one of the main topics of Section 2.4. 9Donnie would be very upset if, for example, we told him the width of the pasture needs to be β50 feet. 2.3 Quadratic Functions 199 2 β 1 2 β 6 = β 25 Nevertheless, we can attack this problem graphically. To that end, we graph y = g(x) = x2 β x β 6 using the intercepts and the vertex. To ο¬nd the x-inter |
cepts, we solve x2 β x β 6 = 0. Factoring gives (x β 3)(x + 2) = 0 so x = β2 or x = 3. Hence, (β2, 0) and (3, 0) are x-intercepts. The y-intercept (0, β6) is found by setting x = 0. To plot the vertex, we ο¬nd x = β b 2, and y = 1 4 = β6.25. Plotting, we get the parabola seen below on the left. To obtain 2 points on the graph of y = f (x) = |x2 β x β 6|, we can take points on the graph of g(x) = x2 β x β 6 and apply the absolute value to each of the y values on the parabola. We see from the graph of g that for x β€ β2 or x β₯ 3, the y values on the parabola are greater than or equal to zero (since the graph is on or above the x-axis), so the absolute value leaves these portions of the graph alone. For x between β2 and 3, however, the y values on the parabola are negative. For example, the point (0, β6) on y = x2 β x β 6 would result in the point (0, | β 6|) = (0, β(β6)) = (0, 6) on the graph of f (x) = |x2 β x β 6|. Proceeding in this manner for all points with x-coordinates between β2 and 3 results in the graph seen below on the right. 2a = β β1 2(13β2β1 β1 1 2 3 x β3β2β1 β1 1 2 3 x β2 β3 β4 β5 β6 β2 β3 β4 β5 β6 y = g(x) = x2 β x β 6 y = f (x) = |x2 β x β 6| If we take a step back and look at the graphs of g and f in the last example, we notice that to obtain the graph of f from the graph of g, we reο¬ect a portion of the graph of g about the x-axis. We can see this analytically by substituting g(x) = x2 β x β 6 into the formula for f |
(x) and calling to mind Theorem 1.4 from Section 1.7. f (x) = βg(x), g(x), if g(x) < 0 if g(x) β₯ 0 The function f is deο¬ned so that when g(x) is negative (i.e., when its graph is below the x-axis), the graph of f is its refection across the x-axis. This is a general template to graph functions of the form f (x) = |g(x)|. From this perspective, the graph of f (x) = |x| can be obtained by reο¬ecting the portion of the line g(x) = x which is below the x-axis back above the x-axis creating the characteristic ββ¨β shape. 200 Linear and Quadratic Functions 2.3.1 Exercises In Exercises 1 - 9, graph the quadratic function. Find the x- and y-intercepts of each graph, if any exist. If it is given in general form, convert it into standard form; if it is given in standard form, convert it into general form. Find the domain and range of the function and list the intervals on which the function is increasing or decreasing. Identify the vertex and the axis of symmetry and determine whether the vertex yields a relative and absolute maximum or minimum. 1. f (x) = x2 + 2 2. f (x) = β(x + 2)2 3. f (x) = x2 β 2x β 8 4. f (x) = β2(x + 1)2 + 4 5. f (x) = 2x2 β 4x β 1 6. f (x) = β3x2 + 4x β 7 7. f (x) = x2 + x + 1 8. f (x) = β3x2 + 5x + 4 9.10 f (x) = x2 β 1 100 x β 1 In Exercises 10 - 14, the cost and price-demand functions are given for diο¬erent scenarios. For each scenario, Find the proο¬t function P (x). Find the number of items which need to be sold in order to maximize proο¬t. Find the maximum proο¬t. Find the price to charge per item in order to maximize proοΏ½ |
οΏ½t. Find and interpret break-even points. 10. The cost, in dollars, to produce x βIβd rather be a Sasquatchβ T-Shirts is C(x) = 2x + 26, x β₯ 0 and the price-demand function, in dollars per shirt, is p(x) = 30 β 2x, 0 β€ x β€ 15. 11. The cost, in dollars, to produce x bottles of 100% All-Natural Certiο¬ed Free-Trade Organic Sasquatch Tonic is C(x) = 10x + 100, x β₯ 0 and the price-demand function, in dollars per bottle, is p(x) = 35 β x, 0 β€ x β€ 35. 12. The cost, in cents, to produce x cups of Mountain Thunder Lemonade at Juniorβs Lemonade Stand is C(x) = 18x + 240, x β₯ 0 and the price-demand function, in cents per cup, is p(x) = 90 β 3x, 0 β€ x β€ 30. 13. The daily cost, in dollars, to produce x Sasquatch Berry Pies is C(x) = 3x + 36, x β₯ 0 and the price-demand function, in dollars per pie, is p(x) = 12 β 0.5x, 0 β€ x β€ 24. 14. The monthly cost, in hundreds of dollars, to produce x custom built electric scooters is C(x) = 20x + 1000, x β₯ 0 and the price-demand function, in hundreds of dollars per scooter, is p(x) = 140 β 2x, 0 β€ x β€ 70. 10We have already seen the graph of this function. It was used as an example in Section 1.6 to show how the graphing calculator can be misleading. 2.3 Quadratic Functions 201 15. The International Silver Strings Submarine Band holds a bake sale each year to fund their trip to the National Sasquatch Convention. It has been determined that the cost in dollars of baking x cookies is C(x) = 0.1x + 25 and that the demand function for their cookies is p = 10 β.01x. How many cookies should they bake in order to maximize their proο¬t? 16. Using data from Bureau of Transportation Statistics, the average fuel economy F in miles per gallon for passenger cars in the US can be modeled by |
F (t) = β0.0076t2 + 0.45t + 16, 0 β€ t β€ 28, where t is the number of years since 1980. Find and interpret the coordinates of the vertex of the graph of y = F (t). 17. The temperature T, in degrees Fahrenheit, t hours after 6 AM is given by: T (t) = β 1 2 t2 + 8t + 32, 0 β€ t β€ 12 What is the warmest temperature of the day? When does this happen? 18. Suppose C(x) = x2 β 10x + 27 represents the costs, in hundreds, to produce x thousand pens. How many pens should be produced to minimize the cost? What is this minimum cost? 19. Skippy wishes to plant a vegetable garden along one side of his house. In his garage, he found 32 linear feet of fencing. Since one side of the garden will border the house, Skippy doesnβt need fencing along that side. What are the dimensions of the garden which will maximize the area of the garden? What is the maximum area of the garden? 20. In the situation of Example 2.3.4, Donnie has a nightmare that one of his alpaca herd fell into the river and drowned. To avoid this, he wants to move his rectangular pasture away from the river. This means that all four sides of the pasture require fencing. If the total amount of fencing available is still 200 linear feet, what dimensions maximize the area of the pasture now? What is the maximum area? Assuming an average alpaca requires 25 square feet of pasture, how many alpaca can he raise now? 21. What is the largest rectangular area one can enclose with 14 inches of string? 22. The height of an object dropped from the roof of an eight story building is modeled by h(t) = β16t2 + 64, 0 β€ t β€ 2. Here, h is the height of the object oο¬ the ground, in feet, t seconds after the object is dropped. How long before the object hits the ground? 23. The height h in feet of a model rocket above the ground t seconds after lift-oο¬ is given by h(t) = β5t2 + 100t, for 0 β€ t β€ 20. When does the rocket reach its maximum height above the ground? What is its maximum height? 24. Carlβs friend Jason participates in the Highland |
Games. In one event, the hammer throw, the height h in feet of the hammer above the ground t seconds after Jason lets it go is modeled by h(t) = β16t2 + 22.08t + 6. What is the hammerβs maximum height? What is the hammerβs total time in the air? Round your answers to two decimal places. 202 Linear and Quadratic Functions 25. Assuming no air resistance or forces other than the Earthβs gravity, the height above the ground at time t of a falling object is given by s(t) = β4.9t2 + v0t + s0 where s is in meters, t is in seconds, v0 is the objectβs initial velocity in meters per second and s0 is its initial position in meters. (a) What is the applied domain of this function? (b) Discuss with your classmates what each of v0 > 0, v0 = 0 and v0 < 0 would mean. (c) Come up with a scenario in which s0 < 0. (d) Letβs say a slingshot is used to shoot a marble straight up from the ground (s0 = 0) with an initial velocity of 15 meters per second. What is the marbleβs maximum height above the ground? At what time will it hit the ground? (e) Now shoot the marble from the top of a tower which is 25 meters tall. When does it hit the ground? (f) What would the height function be if instead of shooting the marble up oο¬ of the tower, you were to shoot it straight DOWN from the top of the tower? 26. The two towers of a suspension bridge are 400 feet apart. The parabolic cable11 attached to the tops of the towers is 10 feet above the point on the bridge deck that is midway between the towers. If the towers are 100 feet tall, ο¬nd the height of the cable directly above a point of the bridge deck that is 50 feet to the right of the left-hand tower. 27. Graph f (x) = |1 β x2| 28. Find all of the points on the line y = 1 β x which are 2 units from (1, β1). 29. Let L be the line y = 2x + 1. Find a function D(x) which measures the distance squared from a point on L to (0, 0). Use this |
to ο¬nd the point on L closest to (0, 0). 30. With the help of your classmates, show that if a quadratic function f (x) = ax2 + bx + c has two real zeros then the x-coordinate of the vertex is the midpoint of the zeros. In Exercises 31 - 36, solve the quadratic equation for the indicated variable. 31. x2 β 10y2 = 0 for x 32. y2 β 4y = x2 β 4 for x 33. x2 β mx = 1 for x 34. y2 β 3y = 4x for y 35. y2 β 4y = x2 β 4 for y 36. βgt2 + v0t + s0 = 0 for t (Assume g = 0.) 11The weight of the bridge deck forces the bridge cable into a parabola and a free hanging cable such as a power line does not form a parabola. We shall see in Exercise 35 in Section 6.5 what shape a free hanging cable makes. 2.3 Quadratic Functions 203 2.3.2 Answers 1. f (x) = x2 + 2 (this is both forms!) No x-intercepts y-intercept (0, 2) Domain: (ββ, β) Range: [2, β) Decreasing on (ββ, 0] Increasing on [0, β) Vertex (0, 2) is a minimum Axis of symmetry x = 0 2. f (x) = β(x + 2)2 = βx2 β 4x β 4 x-intercept (β2, 0) y-intercept (0, β4) Domain: (ββ, β) Range: (ββ, 0] Increasing on (ββ, β2] Decreasing on [β2, β) Vertex (β2, 0) is a maximum Axis of symmetry x = β2 3. f (x) = x2 β 2x β 8 = (x β 1)2 β 9 x-intercepts (β2, 0) and (4, 0) y-intercept (0, β8) Domain: (ββ, β) Range: [β9, β) Decreasing on (ββ, 1] Increasing on [1, β) Vertex (1, β9) is a |
minimum Axis of symmetry x = 1 4. f (x) = β2(x + 1)2 + 4 = β2x2 β 4x + 2 β 2, 0) 2, 0) and (β1 + β x-intercepts (β1 β y-intercept (0, 2) Domain: (ββ, β) Range: (ββ, 4] Increasing on (ββ, β1] Decreasing on [β1, β) Vertex (β1, 4) is a maximum Axis of symmetry x = β1 y 10 2 β1 1 2 x β4 β3 β2 β1 β1 y x β2 β3 β4 β5 β6 β7 β8 y 2 1 β2 β1 β1 1 2 3 4 x β2 β3 β4 β5 β6 β7 β8 β9 y 4 3 2 1 β1 β2 β3 β4 1 x β3 β2 β1 204 Linear and Quadratic Functions 5. f (x) = 2x2 β 4x β 1 = 2(x β 1)2 β 3 β 6 β 6 2+ 2, 0 2β 2, 0 and x-intercepts y-intercept (0, β1) Domain: (ββ, β) Range: [β3, β) Increasing on [1, β) Decreasing on (ββ, 1] Vertex (1, β3) is a minimum Axis of symmetry x = 1 6. f (x) = β3x2 + 4x β 7 = β3 x β 2 3 2 β 17 3 No x-intercepts y-intercept (0, β7) Domain: (ββ, β) Range: ββ, β 17 3 Increasing on ββ, 2 3 Decreasing on 2 3, β Vertex 2 is a maximum 3, β 17 Axis of symmetry 1 1 2 3 x 1 2 x β1 β2 β3 y β1 β2 β3 β4 β5 β6 β7 β8 β9 β10 β11 β12 β13 β14 7. f (x) = x2 + No x-intercepts y-intercept (0, 1) Domain: (ββ, β) 4, β Range: 3 Increasing on β 1 2, β Decreasing on ββ, β |
1 2 Vertex β 1 is a minimum 2, 3 4 Axis of symmetry 2 β1 1 x 2.3 Quadratic Functions 205 8. f (x) = β3x2 + 5x + 4 = β3 x β 5 6 5β 73, 0 5+ and, 0 73 β β 2 + 73 12 6 x-intercepts 6 y-intercept (0, 4) Domain: (ββ, β) Range: ββ, 73 12 Increasing on ββ, 5 6 Decreasing on 5 6, β Vertex 5 6, 73 12 Axis of symmetry x = 5 6 is a maximum 100 x β 1 = x β 1 200 1β 1+ 2 β 40001 40000 β 40001 and 200 β 40001 9. f (x) = x2 β 1 x-intercepts 200 y-intercept (0, β1) Domain: (ββ, β) Range: β 40001 40000, β Decreasing on ββ, 1 200 200, β Increasing on 1 Vertex 1 200, β 40001 Axis of symmetry x = 1 200 40000 is a minimum12 y 6 5 4 3 2 1 β1 1 2 3 x β1 β2 β2 β1 1 2 x 10. P (x) = β2x2 + 28x β 26, for 0 β€ x β€ 15. 7 T-shirts should be made and sold to maximize proο¬t. The maximum proο¬t is $72. The price per T-shirt should be set at $16 to maximize proο¬t. The break even points are x = 1 and x = 13, so to make a proο¬t, between 1 and 13 T-shirts need to be made and sold. 11. P (x) = βx2 + 25x β 100, for 0 β€ x β€ 35 Since the vertex occurs at x = 12.5, and it is impossible to make or sell 12.5 bottles of tonic, maximum proο¬t occurs when either 12 or 13 bottles of tonic are made and sold. The maximum proο¬t is $56. The price per bottle can be either $23 (to sell 12 bottles) or $22 (to sell 13 bottles.) Both will result in the maximum proο¬t. The break even points are x = 5 and x = 20 |
, so to make a proο¬t, between 5 and 20 bottles of tonic need to be made and sold. 12Youβll need to use your calculator to zoom in far enough to see that the vertex is not the y-intercept. 206 12. 13. 14. Linear and Quadratic Functions P (x) = β3x2 + 72x β 240, for 0 β€ x β€ 30 12 cups of lemonade need to be made and sold to maximize proο¬t. The maximum proο¬t is 192Β’ or $1.92. The price per cup should be set at 54Β’ per cup to maximize proο¬t. The break even points are x = 4 and x = 20, so to make a proο¬t, between 4 and 20 cups of lemonade need to be made and sold. P (x) = β0.5x2 + 9x β 36, for 0 β€ x β€ 24 9 pies should be made and sold to maximize the daily proο¬t. The maximum daily proο¬t is $4.50. The price per pie should be set at $7.50 to maximize proο¬t. The break even points are x = 6 and x = 12, so to make a proο¬t, between 6 and 12 pies need to be made and sold daily. P (x) = β2x2 + 120x β 1000, for 0 β€ x β€ 70 30 scooters need to be made and sold to maximize proο¬t. The maximum monthly proο¬t is 800 hundred dollars, or $80,000. The price per scooter should be set at 80 hundred dollars, or $8000 per scooter. The break even points are x = 10 and x = 50, so to make a proο¬t, between 10 and 50 scooters need to be made and sold monthly. 15. 495 cookies 16. The vertex is (approximately) (29.60, 22.66), which corresponds to a maximum fuel economy of 22.66 miles per gallon, reached sometime between 2009 and 2010 (29 β 30 years after 1980.) Unfortunately, the model is only valid up until 2008 (28 years after 1908.) So, at this point, we are using the model to predict the maximum fuel economy. 17. 64β¦ at 2 PM (8 hours after 6 AM.) 18. 5000 pens should |
be produced for a cost of $200. 19. 8 feet by 16 feet; maximum area is 128 square feet. 20. 50 feet by 50 feet; maximum area is 2500 feet; he can raise 100 average alpacas. 21. The largest rectangle has area 12.25 square inches. 22. 2 seconds. 23. The rocket reaches its maximum height of 500 feet 10 seconds after lift-oο¬. 24. The hammer reaches a maximum height of approximately 13.62 feet. The hammer is in the air approximately 1.61 seconds. 2.3 Quadratic Functions 207 25. (a) The applied domain is [0, β). (d) The height function is this case is s(t) = β4.9t2 + 15t. The vertex of this parabola is approximately (1.53, 11.48) so the maximum height reached by the marble is 11.48 meters. It hits the ground again when t β 3.06 seconds. (e) The revised height function is s(t) = β4.9t2 + 15t + 25 which has zeros at t β β1.20 and t β 4.26. We ignore the negative value and claim that the marble will hit the ground after 4.26 seconds. (f) Shooting down means the initial velocity is negative so the height functions becomes s(t) = β4.9t2 β 15t + 25. 26. Make the vertex of the parabola (0, 10) so that the point on the top of the left-hand tower where the cable connects is (β200, 100) and the point on the top of the right-hand tower is (200, 100). Then the parabola is given by p(x) = 9 4000 x2 + 10. Standing 50 feet to the right of the left-hand tower means youβre standing at x = β150 and p(β150) = 60.625. So the cable is 60.625 feet above the bridge deck there. β 3 β 2 7, β1 β 2 β 7 28. 27. y = |1 β x22 β1 1 2 x 29. D(x) = x2 +(2x+1)2 = 5x2 +4x+1, D is minimized when x = β 2 5, so the point on y = 2x+1 closest to (0, 0 |
) is β 2 5, 1 5 31. x = Β±y β 10 34. y = β 3 Β± 16x + 9 2 32. x = Β±(y β 2) 35. y = 2 Β± x 33. x = 36. t = β m Β± m2 + 4 2 v0 Β± v2 0 + 4gs0 2g 208 Linear and Quadratic Functions 2.4 Inequalities with Absolute Value and Quadratic Functions In this section, not only do we develop techniques for solving various classes of inequalities analytically, we also look at them graphically. The ο¬rst example motivates the core ideas. Example 2.4.1. Let f (x) = 2x β 1 and g(x) = 5. 1. Solve f (x) = g(x). 2. Solve f (x) < g(x). 3. Solve f (x) > g(x). 4. Graph y = f (x) and y = g(x) on the same set of axes and interpret your solutions to parts 1 through 3 above. Solution. 1. To solve f (x) = g(x), we replace f (x) with 2x β 1 and g(x) with 5 to get 2x β 1 = 5. Solving for x, we get x = 3. 2. The inequality f (x) < g(x) is equivalent to 2x β 1 < 5. Solving gives x < 3 or (ββ, 3). 3. To ο¬nd where f (x) > g(x), we solve 2x β 1 > 5. We get x > 3, or (3, β). 4. To graph y = f (x), we graph y = 2x β 1, which is a line with a y-intercept of (0, β1) and a slope of 2. The graph of y = g(x) is y = 5 which is a horizontal line through (0, 5). y y = g(x) y = f (x1 To see the connection between the graph and the Algebra, we recall the Fundamental Graphing Principle for Functions in Section 1.6: the point (a, b) is on the graph of f if and only if f (a) = b. In other words, a generic point on the graph of y = f (x) |
is (x, f (x)), and a generic 2.4 Inequalities with Absolute Value and Quadratic Functions 209 point on the graph of y = g(x) is (x, g(x)). When we seek solutions to f (x) = g(x), we are looking for x values whose y values on the graphs of f and g are the same. In part 1, we found x = 3 is the solution to f (x) = g(x). Sure enough, f (3) = 5 and g(3) = 5 so that the point (3, 5) is on both graphs. In other words, the graphs of f and g intersect at (3, 5). In part 2, we set f (x) < g(x) and solved to ο¬nd x < 3. For x < 3, the point (x, f (x)) is below (x, g(x)) since the y values on the graph of f are less than the y values on the graph of g there. Analogously, in part 3, we solved f (x) > g(x) and found x > 3. For x > 3, note that the graph of f is above the graph of g, since the y values on the graph of f are greater than the y values on the graph of g for those values of x. y y = g(x) y = f (xx) y = g(x1 1 2 3 4 x β1 f (x) < g(x) on (ββ, 3) f (x) > g(x) on (3, β) The preceding example demonstrates the following, which is a consequence of the Fundamental Graphing Principle for Functions. Graphical Interpretation of Equations and Inequalities Suppose f and g are functions. The solutions to f (x) = g(x) are the x values where the graphs of y = f (x) and y = g(x) intersect. The solution to f (x) < g(x) is the set of x values where the graph of y = f (x) is below the graph of y = g(x). The solution to f (x) > g(x) is the set of x values where the graph of y = f (x) above the graph of y = g(x). The next example turns the tables and furnishes the graphs of two functions and |
asks for solutions to equations and inequalities. 210 Linear and Quadratic Functions Example 2.4.2. The graphs of f and g are below. (The graph of y = g(x) is bolded.) Use these graphs to answer the following questions. y 4 3 2 1 (β1, 2) y = g(x) (1, 2) β2 β1 1 2 x β1 y = f (x) 1. Solve f (x) = g(x). 2. Solve f (x) < g(x). 3. Solve f (x) β₯ g(x). Solution. 1. To solve f (x) = g(x), we look for where the graphs of f and g intersect. These appear to be at the points (β1, 2) and (1, 2), so our solutions to f (x) = g(x) are x = β1 and x = 1. 2. To solve f (x) < g(x), we look for where the graph of f is below the graph of g. This appears to happen for the x values less than β1 and greater than 1. Our solution is (ββ, β1)βͺ(1, β). 3. To solve f (x) β₯ g(x), we look for solutions to f (x) = g(x) as well as f (x) > g(x). We solved the former equation and found x = Β±1. To solve f (x) > g(x), we look for where the graph of f is above the graph of g. This appears to happen between x = β1 and x = 1, on the interval (β1, 1). Hence, our solution to f (x) β₯ g(x) is [β1, 1]. y 4 3 2 1 (β1, 2) y = g(x) (1, 2) (β1, 2) y 4 3 2 1 y = g(x) (1, 2) β2 β1 1 2 x β2 β1 1 2 x β1 y = f (x) f (x) < g(x) β1 y = f (x) f (x) β₯ g(x) 2.4 Inequalities with Absolute Value and Quadratic Functions 211 We now turn our attention to solving inequalities involving the absolute value. We have the following theorem from Intermediate Algebra to help us. |
Theorem 2.4. Inequalities Involving the Absolute Value: Let c be a real number. For c > 0, |x| < c is equivalent to βc < x < c. For c > 0, |x| β€ c is equivalent to βc β€ x β€ c. For c β€ 0, |x| < c has no solution, and for c < 0, |x| β€ c has no solution. For c β₯ 0, |x| > c is equivalent to x < βc or x > c. For c β₯ 0, |x| β₯ c is equivalent to x β€ βc or x β₯ c. For c < 0, |x| > c and |x| β₯ c are true for all real numbers. As with Theorem 2.1 in Section 2.2, we could argue Theorem 2.4 using cases. However, in light of what we have developed in this section, we can understand these statements graphically. For instance, if c > 0, the graph of y = c is a horizontal line which lies above the x-axis through (0, c). To solve |x| < c, we are looking for the x values where the graph of y = |x| is below the graph of y = c. We know that the graphs intersect when |x| = c, which, from Section 2.2, we know happens when x = c or x = βc. Graphing, we get y (βc, c) (c, c) βc c x We see that the graph of y = |x| is below y = c for x between βc and c, and hence we get |x| < c is equivalent to βc < x < c. The other properties in Theorem 2.4 can be shown similarly. Example 2.4.3. Solve the following inequalities analytically; check your answers graphically. 1. |x β 1| β₯ 3 3. 2 < |x β 1| β€ 5 Solution. 2. 4 β 3|2x + 1| > β2 4. |x + 1| β₯ x + 4 2 1. From Theorem 2.4, |x β 1| β₯ 3 is equivalent to x β 1 β€ β3 or x β 1 β₯ 3. Solving, we get x β€ β2 or x β₯ 4, which, in interval notation is (ββ, β2] βͺ [ |
4, β). Graphically, we have 212 Linear and Quadratic Functions x β 1| β4 β3 β2 β1 1 2 3 4 5 x We see that the graph of y = |x β 1| is above the horizontal line y = 3 for x < β2 and x > 4 hence this is where |x β 1| > 3. The two graphs intersect when x = β2 and x = 4, so we have graphical conο¬rmation of our analytic solution. 2. To solve 4 β 3|2x + 1| > β2 analytically, we ο¬rst isolate the absolute value before applying Theorem 2.4. To that end, we get β3|2x + 1| > β6 or |2x + 1| < 2. Rewriting, we now have. Graphically we β2 < 2x + 1 < 2 so that β 3 2 and 1 see that the graph of y = 4 β 3|2x + 1| is above y = β2 for x values between β 3 2. 2. In interval notation, we write β |2x + 1| β2 β1 y 4 3 2 1 β1 β2 β3 β4 1 2 x y = β2 3. Rewriting the compound inequality 2 < |x β 1| β€ 5 as β2 < |x β 1| and |x β 1| β€ 5β allows us to solve each piece using Theorem 2.4. The ο¬rst inequality, 2 < |x β 1| can be re-written as |x β 1| > 2 so x β 1 < β2 or x β 1 > 2. We get x < β1 or x > 3. Our solution to the ο¬rst inequality is then (ββ, β1) βͺ (3, β). For |x β 1| β€ 5, we combine results in Theorems 2.1 and 2.4 to get β5 β€ x β 1 β€ 5 so that β4 β€ x β€ 6, or [β4, 6]. Our solution to 2 < |x β 1| β€ 5 is comprised of values of x which satisfy both parts of the inequality, so we take the intersection1 of (ββ, β1) βͺ (3, β) and [β4, 6] to get [β4, β1) οΏ½ |
οΏ½ (3, 6]. Graphically, we see that the graph of y = |x β 1| is βbetweenβ the horizontal lines y = 2 and y = 5 for x values between β4 and β1 as well as those between 3 and 6. Including the x values where y = |x β 1| and y = 5 intersect, we get 1See Deο¬nition 1.2 in Section 1.1.1. 2.4 Inequalities with Absolute Value and Quadratic Functions 213 x β 1| y = 5 y = 2 β8 β7 β6 β5 β4 β3 β2 β. We need to exercise some special caution when solving |x + 1| β₯ x+4 2. As we saw in Example 2.2.1 in Section 2.2, when variables are both inside and outside of the absolute value, itβs usually best to refer to the deο¬nition of absolute value, Deο¬nition 2.4, to remove the absolute values and proceed from there. To that end, we have |x + 1| = β(x + 1) if x < β1 and |x + 1| = x + 1 if x β₯ β1. We break the inequality into cases, the ο¬rst case being when x < β1. For these values of x, our inequality becomes β(x + 1) β₯ x+4 2. Solving, we get β2x β 2 β₯ x + 4, so that β3x β₯ 6, which means x β€ β2. Since all of these solutions fall into the category x < β1, we keep them all. For the second case, we assume x β₯ β1. Our inequality becomes x + 1 β₯ x+4 2, which gives 2x + 2 β₯ x + 4 or x β₯ 2. Since all of these values of x are greater than or equal to β1, we accept all of these solutions as well. Our ο¬nal answer is (ββ, β2] βͺ [2, β). y = |x + 1| y 4 3 2 y = x+4 2 β4 β3 β2 β1 1 2 3 4 x We now turn our attention to quadratic inequalities. In the last example of Section 2.3, we needed to determine the solution to x2 β x β 6 < 0. We will now |
re-visit this problem using some of the techniques developed in this section not only to reinforce our solution in Section 2.3, but to also help formulate a general analytic procedure for solving all quadratic inequalities. If we consider f (x) = x2 β x β 6 and g(x) = 0, then solving x2 β x β 6 < 0 corresponds graphically to ο¬nding 214 Linear and Quadratic Functions the values of x for which the graph of y = f (x) = x2 β x β 6 (the parabola) is below the graph of y = g(x) = 0 (the x-axis). Weβve provided the graph again for reference. y 6 5 4 3 2 1 β3β2β1 β1 1 2 3 x β2 β3 β4 β5 β6 y = x2 β x β 6 We can see that the graph of f does dip below the x-axis between its two x-intercepts. The zeros of f are x = β2 and x = 3 in this case and they divide the domain (the x-axis) into three intervals: (ββ, β2), (β2, 3) and (3, β). For every number in (ββ, β2), the graph of f is above the x-axis; in other words, f (x) > 0 for all x in (ββ, β2). Similarly, f (x) < 0 for all x in (β2, 3), and f (x) > 0 for all x in (3, β). We can schematically represent this with the sign diagram below. (+) 0 (β) 0 (+) β2 3 Here, the (+) above a portion of the number line indicates f (x) > 0 for those values of x; the (β) indicates f (x) < 0 there. The numbers labeled on the number line are the zeros of f, so we place 0 above them. We see at once that the solution to f (x) < 0 is (β2, 3). Our next goal is to establish a procedure by which we can generate the sign diagram without graphing the function. An important property2 of quadratic functions is that if the function is positive at one point and negative at another, the function must have at least one zero in between. Graphically, this means that a parabola canβt |
be above the x-axis at one point and below the x-axis at another point without crossing the x-axis. This allows us to determine the sign of all of the function values on a given interval by testing the function at just one value in the interval. This gives us the following. 2We will give this property a name in Chapter 3 and revisit this concept then. 2.4 Inequalities with Absolute Value and Quadratic Functions 215 Steps for Solving a Quadratic Inequality 1. Rewrite the inequality, if necessary, as a quadratic function f (x) on one side of the in- equality and 0 on the other. 2. Find the zeros of f and place them on the number line with the number 0 above them. 3. Choose a real number, called a test value, in each of the intervals determined in step 2. 4. Determine the sign of f (x) for each test value in step 3, and write that sign above the corresponding interval. 5. Choose the intervals which correspond to the correct sign to solve the inequality. Example 2.4.4. Solve the following inequalities analytically using sign diagrams. Verify your answer graphically. 1. 2x2 β€ 3 β x 3. x2 + 1 β€ 2x Solution. 2. x2 β 2x > 1 4. 2x β x2 β₯ |x β 1| β 1 1. To solve 2x2 β€ 3 β x, we ο¬rst get 0 on one side of the inequality which yields 2x2 + x β 3 β€ 0. We ο¬nd the zeros of f (x) = 2x2 + x β 3 by solving 2x2 + x β 3 = 0 for x. Factoring gives (2x + 3)(x β 1) = 0, so x = β 3 2 or x = 1. We place these values on the number line with 0 2, 1 and (1, β). For the above them and choose test values in the intervals ββ, β 3 2 2, 1, we pick x = 0; and for (1, β), x = 2. interval ββ, β 3 2 Evaluating the function at the three test values gives us f (β2) = 3 > 0, so we place (+) above ββ, β 3 2, 1; and, f (2) = 7, 2 which means (+) |
is placed above (1, β). Since we are solving 2x2 + x β 3 β€ 0, we look for solutions to 2x2 + x β 3 < 0 as well as solutions for 2x2 + x β 3 = 0. For 2x2 + x β 3 < 0, we need the intervals which we have a (β). Checking the sign diagram, we see this is β 3 2, 1. We know 2x2 + x β 3 = 0 when x = β 3 ; f (0) = β3 < 0, so (β) goes above the interval β 3, we choose3 x = β2; for β 3 2 and x = 1, so our ο¬nal answer is β 3, β 3 2, 1. To verify our solution graphically, we refer to the original inequality, 2x2 β€ 3 β x. We let g(x) = 2x2 and h(x) = 3 β x. We are looking for the x values where the graph of g is below that of h (the solution to g(x) < h(x)) as well as the points of intersection (the solutions to g(x) = h(x)). The graphs of g and h are given on the right with the sign chart on the left. 3We have to choose something in each interval. If you donβt like our choices, please feel free to choose diο¬erent numbers. Youβll get the same sign chart. 216 Linear and Quadratic Functions (+) 0 (β) 0 (+) β 3 2 β = 2x2 y = 3 β x β2 β1 1 2 x β 2. Once again, we re-write x2 β 2x > 1 as x2 β 2x β 1 > 0 and we identify f (x) = x2 β 2x β 1. When we go to ο¬nd the zeros of f, we ο¬nd, to our chagrin, that the quadratic x2 β 2x β 1 doesnβt factor nicely. Hence, we resort to the quadratic formula to solve x2 β 2x β 1 = 0, and arrive at x = 1 Β± 2. As before, these zeros divide the number line into three pieces. To help us decide on test values, we approximate 1 β 2 β 2.4. We choose x = β1, x = 0 |
and x = 3 as our test values and ο¬nd f (β1) = 2, which is (+); f (0) = β1 which is (β); and f (3) = 2 which is (+) again. Our solution to x2 β 2x β 1 > 0 is where 2, β. To check the inequality we have (+), so, in interval notation ββ, 1 β x2 β 2x > 1 graphically, we set g(x) = x2 β 2x and h(x) = 1. We are looking for the x values where the graph of g is above the graph of h. As before we present the graphs on the right and the sign chart on the left. 2 β β0.4 and 1 + 2 βͺ 1 + β β β β (+) β1 0 β 1 β (β) 2 0 0 β 1 + (+) 3 β2 β1 1 2 3 x y = x2 β 2x 3. To solve x2 + 1 β€ 2x, as before, we solve x2 β 2x + 1 β€ 0. Setting f (x) = x2 β 2x + 1 = 0, we ο¬nd the only one zero of f, x = 1. This one x value divides the number line into two intervals, from which we choose x = 0 and x = 2 as test values. We ο¬nd f (0) = 1 > 0 and f (2) = 1 > 0. Since we are looking for solutions to x2 β 2x + 1 β€ 0, we are looking for x values where x2 β 2x + 1 < 0 as well as where x2 β 2x + 1 = 0. Looking at our sign diagram, there are no places where x2 β 2x + 1 < 0 (there are no (β)), so our solution is only x = 1 (where x2 β 2x + 1 = 0). We write this as {1}. Graphically, we solve x2 + 1 β€ 2x by graphing g(x) = x2 + 1 and h(x) = 2x. We are looking for the x values where the graph of g is below the graph of h (for x2 +1 < 2x) and where the two graphs intersect (x2 +1 = 2x). Notice that the line and the parabola touch at (1, 2 |
), but the parabola is always above the line otherwise.4 4In this case, we say the line y = 2x is tangent to y = x2 + 1 at (1, 2). Finding tangent lines to arbitrary functions is a fundamental problem solved, in general, with Calculus. 2.4 Inequalities with Absolute Value and Quadratic Functions 217 0 1 (+) 0 (+) 2 y 4 3 2 1 y = x2 + 1 y = 2x β1 1 x 4. To solve our last inequality, 2x β x2 β₯ |x β 1| β 1, we re-write the absolute value using cases. For x < 1, |x β 1| = β(x β 1) = 1 β x, so we get 2x β x2 β₯ 1 β x β 1, or x2 β 3x β€ 0. Finding the zeros of f (x) = x2 β 3x, we get x = 0 and x = 3. However, we are only concerned with the portion of the number line where x < 1, so the only zero that we concern ourselves with is x = 0. This divides the interval x < 1 into two intervals: (ββ, 0) and (0, 1). We choose 2 as our test values. We ο¬nd f (β1) = 4 and f 1 x = β1 and x = 1 4. Hence, our solution to x2 β 3x β€ 0 for x < 1 is [0, 1). Next, we turn our attention to the case x β₯ 1. Here, |x β 1| = x β 1, so our original inequality becomes 2x β x2 β₯ x β 1 β 1, or x2 β x β 2 β€ 0. Setting g(x) = x2 β x β 2, we ο¬nd the zeros of g to be x = β1 and x = 2. Of these, only x = 2 lies in the region x β₯ 1, so we ignore x = β1. Our test intervals are now [1, 2) and (2, β). We choose x = 1 and x = 3 as our test values and ο¬nd g(1) = β2 and g(3) = 4. Hence, our solution to g(x) = x2 β x β 2 β€ 0, in this region is [1, 2). = β 5 2 0 0 |
(+) β1 (β) 1 2 1 (β) 0 2 (+) 3 1 Combining these into one sign diagram, we have that our solution is [0, 2]. Graphically, to check 2x β x2 β₯ |x β 1| β 1, we set h(x) = 2x β x2 and i(x) = |x β 1| β 1 and look for the x values where the graph of h is above the the graph of i (the solution of h(x) > i(x)) as well as the x-coordinates of the intersection points of both graphs (where h(x) = i(x)). The combined sign chart is given on the left and the graphs are on the right. (+) 0 (β) 0 (+) 0 2 β1 0 3 y y = 2x β x2 1 β1 1 2 3 x y = |x β 1| β 1 218 Linear and Quadratic Functions One of the classic applications of inequalities is the notion of tolerances.5 Recall that for real numbers x and c, the quantity |x β c| may be interpreted as the distance from x to c. Solving inequalities of the form |x β c| β€ d for d β₯ 0 can then be interpreted as ο¬nding all numbers x which lie within d units of c. We can think of the number d as a βtoleranceβ and our solutions x as being within an accepted tolerance of c. We use this principle in the next example. Example 2.4.5. The area A (in square inches) of a square piece of particle board which measures x inches on each side is A(x) = x2. Suppose a manufacturer needs to produce a 24 inch by 24 inch square piece of particle board as part of a home oο¬ce desk kit. How close does the side of the piece of particle board need to be cut to 24 inches to guarantee that the area of the piece is within a tolerance of 0.25 square inches of the target area of 576 square inches? Solution. Mathematically, we express the desire for the area A(x) to be within 0.25 square inches of 576 as |A β 576| β€ 0.25. Since A(x) = x2, we get |x2 β 576| β€ 0.25, which is equivalent to β0.25 β€ x2 β 576 β€ 0. |
25. One way to proceed at this point is to solve the two inequalities β0.25 β€ x2 β 576 and x2 β 576 β€ 0.25 individually using sign diagrams and then taking the intersection of the solution sets. While this way will (eventually) lead to the correct answer, we take this opportunity to showcase the increasing property of the square root: if 0 β€ a β€ b, then β β a β€ b. To use this property, we proceed as follows β0.25 β€ x2 β 576 β€ 0.25 x2 575.75 β€ β x2 575.75 β€ |x| 575.75 β€ β€ 576.25 β β€ β β€ β β β (add 576 across the inequalities.) (take square roots.) β x2 = |x|) ( 575.75 βͺ β β β 575.75 β€ |x| β€ 576.25 576.25 575.75 β€ |x| to be ββ, β 576.25. To solve 576.25, β β β 575.75] βͺ [ 575.75, β β β β By Theorem 2.4, we ο¬nd the solution to 576.25 to be β the solution to |x| β€ intersect these two sets to get [β length, we discard the negative answers and get [ β the piece of particle board must be cut between tolerance of (approximately) 0.005 inches of the target length of 24 inches. 575.75, β and 576.25, we 576.25]. Since x represents a 576.25]. This means that the side of 576.25 β 24.005 inches, a 575.75 β 23.995 and 576.25, β 575.75, β β β β Our last example in the section demonstrates how inequalities can be used to describe regions in the plane, as we saw earlier in Section 1.2. Example 2.4.6. Sketch the following relations. 1. R = {(x, y) : y > |x|} 2. S = {(x, y) : y β€ 2 β x2} 3. T = {(x, y) : |x| < y β€ 2 β x2} |
5The underlying concept of Calculus can be phrased in terms of tolerances, so this is well worth your attention. 2.4 Inequalities with Absolute Value and Quadratic Functions 219 Solution. 1. The relation R consists of all points (x, y) whose y-coordinate is greater than |x|. If we graph y = |x|, then we want all of the points in the plane above the points on the graph. Dotting the graph of y = |x| as we have done before to indicate that the points on the graph itself are not in the relation, we get the shaded region below on the left. 2. For a point to be in S, its y-coordinate must be less than or equal to the y-coordinate on the parabola y = 2 β x2. This is the set of all points below or on the parabola y = 2 β x2. y 2 1 y 2 1 β2 β1 1 2 x β2 β1 1 2 x β1 β1 The graph of R The graph of S 3. Finally, the relation T takes the points whose y-coordinates satisfy both the conditions given in R and those of S. Thus we shade the region between y = |x| and y = 2 β x2, keeping those points on the parabola, but not the points on y = |x|. To get an accurate graph, we need to ο¬nd where these two graphs intersect, so we set |x| = 2 β x2. Proceeding as before, breaking this equation into cases, we get x = β1, 1. Graphing yields y 2 1 β1 β2 β1 1 2 x The graph of T 220 Linear and Quadratic Functions 2.4.1 Exercises In Exercises 1 - 32, solve the inequality. Write your answer using interval notation. 1. |3x β 5| β€ 4 3. |2x + 1| β 5 < 0 5. |3x + 5| + 2 < 1 7. 2 β€ |4 β x| < 7 2. |7x + 2| > 10 4. |2 β x| β 4 β₯ β3 6. 2|7 β x| + 4 > 1 8. 1 < |2x β 9| β€ 3 9. |x + 3| β₯ |6x + 9| 10. |x β 3| β |
|2x + 1| < 0 11. |1 β 2x| β₯ x + 5 13. x β₯ |x + 1| 15. x + |2x β 3| < 2 17. x2 + 2x β 3 β₯ 0 19. x2 + 9 < 6x 21. x2 + 4 β€ 4x 23. 3x2 β€ 11x + 4 25. 2x2 β 4x β 1 > 0 27. 2 β€ |x2 β 9| < 9 29. x2 + x + 1 β₯ 0 31. x|x + 5| β₯ β6 12. x + 5 < |x + 5| 14. |2x + 1| β€ 6 β x 16. |3 β x| β₯ x β 5 18. 16x2 + 8x + 1 > 0 20. 9x2 + 16 β₯ 24x 22. x2 + 1 < 0 24. x > x2 26. 5x + 4 β€ 3x2 28. x2 β€ |4x β 3| 30. x2 β₯ |x| 32. x|x β 3| < 2 33. The proο¬t, in dollars, made by selling x bottles of 100% All-Natural Certiο¬ed Free-Trade Organic Sasquatch Tonic is given by P (x) = βx2 + 25x β 100, for 0 β€ x β€ 35. How many bottles of tonic must be sold to make at least $50 in proο¬t? 34. Suppose C(x) = x2 β 10x + 27, x β₯ 0 represents the costs, in hundreds of dollars, to produce x thousand pens. Find the number of pens which can be produced for no more than $1100. 35. The temperature T, in degrees Fahrenheit, t hours after 6 AM is given by T (t) = β 1 2 t2+8t+32, for 0 β€ t β€ 12. When is it warmer than 42β¦ Fahrenheit? 2.4 Inequalities with Absolute Value and Quadratic Functions 221 36. The height h in feet of a model rocket above the ground t seconds after lift-oο¬ is given by h(t) = β5t2 + 100t, for 0 β€ t β€ 20. When is the rocket at least 250 feet oο¬ the ground? Round your answer to two decimal places. 37. If a slingshot is |
used to shoot a marble straight up into the air from 2 meters above the ground with an initial velocity of 30 meters per second, for what values of time t will the marble be over 35 meters above the ground? (Refer to Exercise 25 in Section 2.3 for assistance if needed.) Round your answers to two decimal places. 38. What temperature values in degrees Celsius are equivalent to the temperature range 50β¦F to 95β¦F? (Refer to Exercise 35 in Section 2.1 for assistance if needed.) In Exercises 39 - 42, write and solve an inequality involving absolute values for the given statement. 39. Find all real numbers x so that x is within 4 units of 2. 40. Find all real numbers x so that 3x is within 2 units of β1. 41. Find all real numbers x so that x2 is within 1 unit of 3. 42. Find all real numbers x so that x2 is at least 7 units away from 4. 43. The surface area S of a cube with edge length x is given by S(x) = 6x2 for x > 0. Suppose the cubes your company manufactures are supposed to have a surface area of exactly 42 square centimeters, but the machines you own are old and cannot always make a cube with the precise surface area desired. Write an inequality using absolute value that says the surface area of a given cube is no more than 3 square centimeters away (high or low) from the target of 42 square centimeters. Solve the inequality and write your answer using interval notation. 44. Suppose f is a function, L is a real number and Ξ΅ is a positive number. Discuss with your classmates what the inequality |f (x) β L| < Ξ΅ means algebraically and graphically.6 In Exercises 45 - 50, sketch the graph of the relation. 45. R = {(x, y) : y β€ x β 1} 47. R = {(x, y) : β1 < y β€ 2x + 1} 46. R = (x, y) : y > x2 + 1 48. R = (x, y) : x2 β€ y < x + 2 49. R = {(x, y) : |x| β 4 < y < 2 β x} 50. R = (x, y) : x2 < y β€ |4x β 3| 51. Prove the second, third and fourth parts of Theorem 2.4 |
. 6Understanding this type of inequality is really important in Calculus. 222 Linear and Quadratic Functions 2.4.2 Answers 1. 1 3, 3 3. (β3, 2) 5. No solution 7. (β3, 2] βͺ [6, 11) 9. β 12 7, β 6 5 11. ββ, β 4 3 βͺ [6, β) 13. No Solution. 15. 1, 5 3 17. (ββ, β3] βͺ [1, β) 19. No solution 21. {2} 23. β 1 3, 4 25. ββ,, β 27. β β 2, β 11 βͺ β β 7, 0 β 7 0, βͺ βͺ β β 11, 3 2 β3 2. ββ, β 12 7 βͺ 8 7, β 4. (ββ, 1] βͺ [3, β) 6. (ββ, β) 8. [3, 4) βͺ (5, 6] 10. (ββ, β4) βͺ 2 3, β 12. (ββ, β5) 14. β7, 5 3 16. (ββ, β) 18. ββ, β 1 4 βͺ β 1 4, β 20. (ββ, β) 22. No solution 24. (0, 1) ββ, 5β 73 β 6 βͺ β 5+ 73 6, β 26. 28. β2 β β 7, β2 + β 7 βͺ [1, 3] 29. (ββ, β) 30. (ββ, β1] βͺ {0} βͺ [1, β) 31. [β6, β3] βͺ [β2, β) 32. (ββ, 1) βͺ 2, 3+ 17 β 2 33. P (x) β₯ 50 on [10, 15]. This means anywhere between 10 and 15 bottles of tonic need to be sold to earn at least $50 in proο¬t. 34. C(x) β€ 11 on [2, 8]. This means anywhere between 2000 and 8000 pens can be produced and the cost will not exceed $1100. β β 35. T (t) > 42 on (8 β |
2 11, 8 + 2 11) β (1.37, 14.63), which corresponds to between 7:22 AM (1.37 hours after 6 AM) to 8:38 PM (14.63 hours after 6 AM.) However, since the model is valid only for t, 0 β€ t β€ 12, we restrict our answer and ο¬nd it is warmer than 42β¦ Fahrenheit from 7:22 AM to 6 PM. 2.4 Inequalities with Absolute Value and Quadratic Functions 223 36. h(t) β₯ 250 on [10 β 5 β β 2, 10 + 5 2] β [2.93, 17.07]. This means the rocket is at least 250 feet oο¬ the ground between 2.93 and 17.07 seconds after lift oο¬. 37. s(t) = β4.9t2 + 30t + 2. s(t) > 35 on (approximately) (1.44, 4.68). This means between 1.44 and 4.68 seconds after it is launched into the air, the marble is more than 35 feet oο¬ the ground. 38. From our previous work C(F ) = 5 9 (F β 32) so 50 β€ F β€ 95 becomes 10 β€ C β€ 35. 39. |x β 2| β€ 4, [β2, 6] 40. |3x + 1| β€ 2, β1, 1 3 41. |x2 β 3| β€ 1, [β2, 2] β 42. |x2 β 4| β₯ 7, (ββ, β 11 ] βͺ [ 11, β) 43. Solving |S(x) β 42| β€ 3, and disregarding the negative solutions yields [2.550, 2.739]. The edge length must be within 2.550 and 2.739 centimeters. 13 2, 15 2 β 45. 47. y 3 2 1 β2 β1 1 2 3 x β1 β2 β3 y 5 4 3 2 1 β2 β1 1 2 x 46. y 4 3 2 1 β2 β1 1 2 x 48. y 4 3 2 1 β1 1 2 x 224 49. y 4 3 2 1 Linear and Quadratic Functions 50. y 20 15 10 5 β2 β1 1 2 3 x β4 β3 β2 β1 1 2 3 x β |
1 β2 β3 β4 2.5 Regression 2.5 Regression 225 We have seen examples already in the text where linear and quadratic functions are used to model a wide variety of real world phenomena ranging from production costs to the height of a projectile above the ground. In this section, we use some basic tools from statistical analysis to quantify linear and quadratic trends that we may see in real world data in order to generate linear and quadratic models. Our goal is to give the reader an understanding of the basic processes involved, but we are quick to refer the reader to a more advanced course1 for a complete exposition of this material. Suppose we collected three data points: {(1, 2), (3, 1), (4, 3)}. By plotting these points, we can clearly see that they do not lie along the same line. If we pick any two of the points, we can ο¬nd a line containing both which completely misses the third, but our aim is to ο¬nd a line which is in some sense βcloseβ to all the points, even though it may go through none of them. The way we measure βclosenessβ in this case is to ο¬nd the total squared error between the data points and the line. Consider our three data points and the line y = 1 2. For each of our data points, we ο¬nd the vertical distance between the point and the line. To accomplish this, we need to ο¬nd a point on the line directly above or below each data point - in other words, a point on the line with the same x-coordinate as our data point. For example, to ο¬nd the point on the line directly below (1, 2), we plug x = 1 into y = 1 2 and we get the point (1, 1). Similarly, we get (3, 1) to correspond to (3, 2) and 4, 5 2 for (4, 3). We ο¬nd the total squared error E by taking the sum of the squares of the diο¬erences of the ycoordinates of each data point and its corresponding point on the line. For the data and line above E = (2 β 1)2 + (1 β 2)2 + 3 β 5 4. Using advanced mathematical machinery,2 it is possible to 2 ο¬nd the line which results in the lowest value of |
E. This line is called the least squares regression line, or sometimes the βline of best ο¬tβ. The formula for the line of best ο¬t requires notation we wonβt present until Chapter 9.1, so we will revisit it then. The graphing calculator can come to our assistance here, since it has a built-in feature to compute the regression line. We enter the data and perform the Linear Regression feature and we get = 9 1and authors with more expertise in this area, 2Like Calculus and Linear Algebra 226 Linear and Quadratic Functions The calculator tells us that the line of best ο¬t is y = ax + b where the slope is a β 0.214 and the y-coordinate of the y-intercept is b β 1.428. (We will stick to using three decimal places for our approximations.) Using this line, we compute the total squared error for our data to be E β 1.786. The value r is the correlation coeο¬cient and is a measure of how close the data is to being on the same line. The closer |r| is to 1, the better the linear ο¬t. Since r β 0.327, this tells us that the line of best ο¬t doesnβt ο¬t all that well - in other words, our data points arenβt close to being linear. The value r2 is called the coeο¬cient of determination and is also a measure of the goodness of ο¬t.3 Plotting the data with its regression line results in the picture below. Our ο¬rst example looks at energy consumption in the US over the past 50 years.4 Year Energy Usage, in Quads5 34.6 45.1 67.8 78.3 84.6 98.9 1950 1960 1970 1980 1990 2000 Example 2.5.1. Using the energy consumption data given above, 1. Plot the data using a graphing calculator. 3We refer the interested reader to a course in Statistics to explore the signiο¬cance of r and r2. 4See this Department of Energy activity 5The unit 1 Quad is 1 Quadrillion = 1015 BTUs, which is enough heat to raise Lake Erie roughly 1β¦F 2.5 Regression 227 2. Find the least squares regression line and comment on the goodness |
of ο¬t. 3. Interpret the slope of the line of best ο¬t. 4. Use the regression line to predict the annual US energy consumption in the year 2013. 5. Use the regression line to predict when the annual consumption will reach 120 Quads. Solution. 1. Entering the data into the calculator gives The data certainly appears to be linear in nature. 2. Performing a linear regression produces We can tell both from the correlation coeο¬cient as well as the graph that the regression line is a good ο¬t to the data. 3. The slope of the regression line is a β 1.287. To interpret this, recall that the slope is the rate of change of the y-coordinates with respect to the x-coordinates. Since the y-coordinates represent the energy usage in Quads, and the x-coordinates represent years, a slope of positive 1.287 indicates an increase in annual energy usage at the rate of 1.287 Quads per year. 4. To predict the energy needs in 2013, we substitute x = 2013 into the equation of the line of best ο¬t to get y = 1.287(2013) β 2473.890 β 116.841. The predicted annual energy usage of the US in 2013 is approximately 116.841 Quads. 228 Linear and Quadratic Functions 5. To predict when the annual US energy usage will reach 120 Quads, we substitute y = 120 into the equation of the line of best ο¬t to get 120 = 1.287x β 2473.908. Solving for x yields x β 2015.454. Since the regression line is increasing, we interpret this result as saying the annual usage in 2015 wonβt yet be 120 Quads, but that in 2016, the demand will be more than 120 Quads. Our next example gives us an opportunity to ο¬nd a nonlinear model to ο¬t the data. According to the National Weather Service, the predicted hourly temperatures for Painesville on March 3, 2009 were given as summarized below. Time Temperature, β¦F 10AM 11AM 12PM 1PM 2PM 3PM 4PM 17 19 21 23 24 24 23 To enter this data into the calculator, we need to adjust the x values, since just entering the numbers could cause confusion. (Do you see why?) We have a few options available to us. Perhaps the easiest |
is to convert the times into the 24 hour clock time so that 1 PM is 13, 2 PM is 14, etc.. If we enter these data into the graphing calculator and plot the points we get While the beginning of the data looks linear, the temperature begins to fall in the afternoon hours. This sort of behavior reminds us of parabolas, and, sure enough, it is possible to ο¬nd a parabola of best ο¬t in the same way we found a line of best ο¬t. The process is called quadratic regression and its goal is to minimize the least square error of the data with their corresponding points on the parabola. The calculator has a built in feature for this as well which yields 2.5 Regression 229 The coeο¬cient of determination R2 seems reasonably close to 1, and the graph visually seems to be a decent ο¬t. We use this model in our next example. Example 2.5.2. Using the quadratic model for the temperature data above, predict the warmest temperature of the day. When will this occur? Solution. The maximum temperature will occur at the vertex of the parabola. Recalling the Vertex Formula, Equation 2.4, x = β b 2(β0.321) β 14.741. This corresponds to roughly 2 : 45 PM. To ο¬nd the temperature, we substitute x = 14.741 into y = β0.321x2 + 9.464x β 45.857 to get y β 23.899, or 23.899β¦F. 2a β β 9.464 The results of the last example should remind you that regression models are just that, models. Our predicted warmest temperature was found to be 23.899β¦F, but our data says it will warm to 24β¦F. Itβs all well and good to observe trends and guess at a model, but a more thorough investigation into why certain data should be linear or quadratic in nature is usually in order - and that, most often, is the business of scientists. 230 Linear and Quadratic Functions 2.5.1 Exercises 1. According to this website6, the census data for Lake County, Ohio is: Year Population 1970 197200 1980 212801 1990 215499 2000 227511 (a) Find the least squares regression line for these data and comment |
on the goodness of ο¬t.7 Interpret the slope of the line of best ο¬t. (b) Use the regression line to predict the population of Lake County in 2010. (The recorded ο¬gure from the 2010 census is 230,041) (c) Use the regression line to predict when the population of Lake County will reach 250,000. 2. According to this website8, the census data for Lorain County, Ohio is: Year Population 1970 256843 1980 274909 1990 271126 2000 284664 (a) Find the least squares regression line for these data and comment on the goodness of ο¬t. Interpret the slope of the line of best ο¬t. (b) Use the regression line to predict the population of Lorain County in 2010. (The recorded ο¬gure from the 2010 census is 301,356) (c) Use the regression line to predict when the population of Lake County will reach 325,000. 3. Using the energy production data given below Year Production (in Quads) 1950 1960 1970 1980 1990 2000 35.6 42.8 63.5 67.2 70.7 71.2 (a) Plot the data using a graphing calculator and explain why it does not appear to be linear. (b) Discuss with your classmates why ignoring the ο¬rst two data points may be justiο¬ed from a historical perspective. (c) Find the least squares regression line for the last four data points and comment on the goodness of ο¬t. Interpret the slope of the line of best ο¬t. (d) Use the regression line to predict the annual US energy production in the year 2010. (e) Use the regression line to predict when the annual US energy production will reach 100 Quads. 6http://www.ohiobiz.com/census/Lake.pdf 7Weβll develop more sophisticated models for the growth of populations in Chapter 6. For the moment, we use a theorem from Calculus to approximate those functions with lines. 8http://www.ohiobiz.com/census/Lorain.pdf 2.5 Regression 231 4. The chart below contains a portion of the fuel consumption information for a 2002 Toyota Echo that I (Jeο¬) used to own. The ο¬rst row is the cumulative number of gallons of gasoline that I had used and the second |
row is the odometer reading when I reο¬lled the gas tank. So, for example, the fourth entry is the point (28.25, 1051) which says that I had used a total of 28.25 gallons of gasoline when the odometer read 1051 miles. Gasoline Used (Gallons) Odometer (Miles) 0 9.26 19.03 28.25 36.45 44.64 53.57 62.62 71.93 81.69 90.43 41 356 731 1051 1347 1631 1966 2310 2670 3030 3371 Find the least squares line for this data. Is it a good ο¬t? What does the slope of the line represent? Do you and your classmates believe this model would have held for ten years had I not crashed the car on the Turnpike a few years ago? (Iβm keeping a fuel log for my 2006 Scion xA for future College Algebra books so I hope not to crash it, too.) 5. On New Yearβs Day, I (Jeο¬, again) started weighing myself every morning in order to have an interesting data set for this section of the book. (Discuss with your classmates if that makes me a nerd or a geek. Also, the professionals in the ο¬eld of weight management strongly discourage weighing yourself every day. When you focus on the number and not your overall health, you tend to lose sight of your objectives. I was making a noble sacriο¬ce for science, but you should not try this at home.) The whole chart would be too big to put into the book neatly, so Iβve decided to give only a small portion of the data to you. This then becomes a Civics lesson in honesty, as you shall soon see. There are two charts given below. One has my weight for the ο¬rst eight Thursdays of the year (January 1, 2009 was a Thursday and weβll count it as Day 1.) and the other has my weight for the ο¬rst 10 Saturdays of the year. Day # (Thursday) My weight in pounds Day # (Saturday) My weight in pounds 1 8 15 22 29 36 43 50 238.2 237.0 235.6 234.4 233.0 233.8 232.8 232.0 3 10 17 24 31 38 45 52 59 66 238.4 235.8 235.0 234 |
.2 236.2 236.2 235.2 233.2 236.8 238.2 (a) Find the least squares line for the Thursday data and comment on its goodness of ο¬t. (b) Find the least squares line for the Saturday data and comment on its goodness of ο¬t. (c) Use Quadratic Regression to ο¬nd a parabola which models the Saturday data and com- ment on its goodness of ο¬t. (d) Compare and contrast the predictions the three models make for my weight on January 1, 2010 (Day #366). Can any of these models be used to make a prediction of my weight 20 years from now? Explain your answer. 232 Linear and Quadratic Functions (e) Why is this a Civics lesson in honesty? Well, compare the two linear models you obtained above. One was a good ο¬t and the other was not, yet both came from careful selections of real data. In presenting the tables to you, I have not lied about my weight, nor have you used any bad math to falsify the predictions. The word weβre looking for here is βdisingenuousβ. Look it up and then discuss the implications this type of data manipulation could have in a larger, more complex, politically motivated setting. (Even Obi-Wan presented the truth to Luke only βfrom a certain point of view.β) 6. (Data that is neither linear nor quadratic.) Weβll close this exercise set with two data sets that, for reasons presented later in the book, cannot be modeled correctly by lines or parabolas. It is a good exercise, though, to see what happens when you attempt to use a linear or quadratic model when itβs not appropriate. (a) This ο¬rst data set came from a Summer 2003 publication of the Portage County Animal Protective League called βTattle Tailsβ. They make the following statement and then have a chart of data that supports it. βIt doesnβt take long for two cats to turn into 80 million. If two cats and their surviving oο¬spring reproduced for ten years, youβd end up with 80,399,780 cats.β We assume N (0) = 2. Year x Number of Cats N (x 10 12 66 382 2201 12680 73041 4207 |
15 2423316 13968290 80399780 Use Quadratic Regression to ο¬nd a parabola which models this data and comment on its goodness of ο¬t. (Spoiler Alert: Does anyone know what type of function we need here?) (b) This next data set comes from the U.S. Naval Observatory. That site has loads of awesome stuο¬ on it, but for this exercise I used the sunrise/sunset times in Fairbanks, Alaska for 2009 to give you a chart of the number of hours of daylight they get on the 21st of each month. Weβll let x = 1 represent January 21, 2009, x = 2 represent February 21, 2009, and so on. Month Number Hours of Daylight 1 2 3 4 5 6 7 8 9 10 11 12 5.8 9.3 12.4 15.9 19.4 21.8 19.4 15.6 12.4 9.1 5.6 3.3 Use Quadratic Regression to ο¬nd a parabola which models this data and comment on its goodness of ο¬t. (Spoiler Alert: Does anyone know what type of function we need here?) 2.5 Regression 2.5.2 Answers 233 1. (a) y = 936.31x β 1645322.6 with r = 0.9696 which indicates a good ο¬t. The slope 936.31 indicates Lake Countyβs population is increasing at a rate of (approximately) 936 people per year. (b) According to the model, the population in 2010 will be 236,660. (c) According to the model, the population of Lake County will reach 250,000 sometime between 2024 and 2025. 2. (a) y = 796.8x β 1309762.5 with r = 0.8916 which indicates a reasonable ο¬t. The slope 796.8 indicates Lorain Countyβs population is increasing at a rate of (approximately) 797 people per year. (b) According to the model, the population in 2010 will be 291,805. (c) According to the model, the population of Lake County will reach 325,000 sometime between 2051 and 2052. 3. (c) y = 0.266xβ459.86 with r = 0.9607 which indicates a good ο¬t. The slope 0. |
266 indicates the countryβs energy production is increasing at a rate of 0.266 Quad per year. (d) According to the model, the production in 2010 will be 74.8 Quad. (e) According to the model, the production will reach 100 Quad in the year 2105. 4. The line is y = 36.8x + 16.39. We have r =.99987 and r2 =.9997 so this is an excellent ο¬t to the data. The slope 36.8 represents miles per gallon. 5. (a) The line for the Thursday data is y = β.12x + 237.69. We have r = β.9568 and r2 =.9155 so this is a really good ο¬t. (b) The line for the Saturday data is y = β0.000693x + 235.94. We have r = β0.008986 and r2 = 0.0000807 which is horrible. This data is not even close to linear. (c) The parabola for the Saturday data is y = 0.003x2β0.21x+238.30. We have R2 =.47497 which isnβt good. Thus the data isnβt modeled well by a quadratic function, either. (d) The Thursday linear model had my weight on January 1, 2010 at 193.77 pounds. The Saturday models give 235.69 and 563.31 pounds, respectively. The Thursday line has my weight going below 0 pounds in about ο¬ve and a half years, so thatβs no good. The quadratic has a positive leading coeο¬cient which would mean unbounded weight gain for the rest of my life. The Saturday line, which mathematically does not ο¬t the data at all, yields a plausible weight prediction in the end. I think this is why grown-ups talk about βLies, Damned Lies and Statistics.β 6. (a) The quadratic model for the cats in Portage county is y = 1917803.54x2β16036408.29x+ 24094857.7. Although R2 =.70888 this is not a good model because itβs so far oο¬ for small values of x. Case in point, the model gives us 24,094,858 cats when x |
= 0 but we know N (0) = 2. 234 Linear and Quadratic Functions (b) The quadratic model for the hours of daylight in Fairbanks, Alaska is y =.51x2 +6.23xβ.36. Even with R2 =.92295 we should be wary of making predictions beyond the data. Case in point, the model gives β4.84 hours of daylight when x = 13. So January 21, 2010 will be βextra darkβ? Obviously a parabola pointing down isnβt telling us the whole story. Chapter 3 Polynomial Functions 3.1 Graphs of Polynomials Three of the families of functions studied thus far β constant, linear and quadratic β belong to a much larger group of functions called polynomials. We begin our formal study of general polynomials with a deο¬nition and some examples. Deο¬nition 3.1. A polynomial function is a function of the form f (x) = anxn + anβ1xnβ1 +... + a2x2 + a1x + a0, where a0, a1,..., an are real numbers and n β₯ 1 is a natural number. The domain of a polynomial function is (ββ, β). There are several things about Deο¬nition 3.1 that may be oο¬-putting or downright frightening. The best thing to do is look at an example. Consider f (x) = 4x5 β 3x2 + 2x β 5. Is this a polynomial function? We can re-write the formula for f as f (x) = 4x5 + 0x4 + 0x3 + (β3)x2 + 2x + (β5). Comparing this with Deο¬nition 3.1, we identify n = 5, a5 = 4, a4 = 0, a3 = 0, a2 = β3, a1 = 2 and a0 = β5. In other words, a5 is the coeο¬cient of x5, a4 is the coeο¬cient of x4, and so forth; the subscript on the aβs merely indicates to which power of x the coeο¬cient belongs. The business of restricting n to be a |
natural number lets us focus on well-behaved algebraic animals.1 Example 3.1.1. Determine if the following functions are polynomials. Explain your reasoning. 1. g(x) = 4 + x3 x β 4. f (x) = 3 x 2. p(x) = 4x + x3 x 3. q(x) = 4x + x3 x2 + 4 5. h(x) = |x| 6. z(x) = 0 1Enjoy this while it lasts. Before weβre through with the book, youβll have been exposed to the most terrible of algebraic beasts. We will tame them all, in time. 236 Solution. Polynomial Functions 1. We note directly that the domain of g(x) = x3+4 x is x = 0. By deο¬nition, a polynomial has all real numbers as its domain. Hence, g canβt be a polynomial. 2. Even though p(x) = x3+4x simpliο¬es to p(x) = x2 + 4, which certainly looks like the form given in Deο¬nition 3.1, the domain of p, which, as you may recall, we determine before we simplify, excludes 0. Alas, p is not a polynomial function for the same reason g isnβt. x 3. After what happened with p in the previous part, you may be a little shy about simplifying q(x) = x3+4x x2+4 to q(x) = x, which certainly ο¬ts Deο¬nition 3.1. If we look at the domain of q before we simpliο¬ed, we see that it is, indeed, all real numbers. A function which can be written in the form of Deο¬nition 3.1 whose domain is all real numbers is, in fact, a polynomial. β 4. We can rewrite f (x) = 3 x as f (x) = x polynomial. 1 3. Since 1 3 is not a natural number, f is not a 5. The function h(x) = |x| isnβt a polynomial, since it canβt be written as a combination of powers of x even though it can be written |
as a piecewise function involving polynomials. As we shall see in this section, graphs of polynomials possess a quality2 that the graph of h does not. 6. Thereβs nothing in Deο¬nition 3.1 which prevents all the coeο¬cients an, etc., from being 0. Hence, z(x) = 0, is an honest-to-goodness polynomial. Deο¬nition 3.2. Suppose f is a polynomial function. Given f (x) = anxn + anβ1xnβ1 +... + a2x2 + a1x + a0 with an = 0, we say β The natural number n is called the degree of the polynomial f. β The term anxn is called the leading term of the polynomial f. β The real number an is called the leading coeο¬cient of the polynomial f. β The real number a0 is called the constant term of the polynomial f. If f (x) = a0, and a0 = 0, we say f has degree 0. If f (x) = 0, we say f has no degree.a aSome authors say f (x) = 0 has degree ββ for reasons not even we will go into. The reader may well wonder why we have chosen to separate oο¬ constant functions from the other polynomials in Deο¬nition 3.2. Why not just lump them all together and, instead of forcing n to be a natural number, n = 1, 2,..., allow n to be a whole number, n = 0, 1, 2,.... We could unify all 2One which really relies on Calculus to verify. 3.1 Graphs of Polynomials 237 of the cases, since, after all, isnβt a0x0 = a0? The answer is βyes, as long as x = 0.β The function f (x) = 3 and g(x) = 3x0 are diο¬erent, because their domains are diο¬erent. The number f (0) = 3 is deο¬ned, whereas g(0) = 3(0)0 is not.3 Indeed, much of the theory we will develop in this chapter doesnοΏ½ |
οΏ½t include the constant functions, so we might as well treat them as outsiders from the start. One good thing that comes from Deο¬nition 3.2 is that we can now think of linear functions as degree 1 (or βο¬rst degreeβ) polynomial functions and quadratic functions as degree 2 (or βsecond degreeβ) polynomial functions. Example 3.1.2. Find the degree, leading term, leading coeο¬cient and constant term of the following polynomial functions. 1. f (x) = 4x5 β 3x2 + 2x β 5 2. g(x) = 12x + x3 3. h(x) = 4 β x 5 Solution. 4. p(x) = (2x β 1)3(x β 2)(3x + 2) 1. There are no surprises with f (x) = 4x5 β 3x2 + 2x β 5. It is written in the form of Deο¬nition 3.2, and we see that the degree is 5, the leading term is 4x5, the leading coeο¬cient is 4 and the constant term is β5. 2. The form given in Deο¬nition 3.2 has the highest power of x ο¬rst. To that end, we re-write g(x) = 12x + x3 = x3 + 12x, and see that the degree of g is 3, the leading term is x3, the leading coeο¬cient is 1 and the constant term is 0. 3. We need to rewrite the formula for h so that it resembles the form given in Deο¬nition 3.2: 5 x, the leading 5. The degree of h is 1, the leading term is β 1 h(x) = 4βx 5 = 4 coeο¬cient is β and the constant term is 4 5. 4. It may seem that we have some work ahead of us to get p in the form of Deο¬nition 3.2. However, it is possible to glean the information requested about p without multiplying out the entire expression (2x β 1)3(x β 2)(3x + 2). The leading term of p will be the term which has the highest power of x. The way to get this |
term is to multiply the terms with the highest power of x from each factor together - in other words, the leading term of p(x) is the product of the leading terms of the factors of p(x). Hence, the leading term of p is (2x)3(x)(3x) = 24x5. This means that the degree of p is 5 and the leading coeο¬cient is 24. As for the constant term, we can perform a similar trick. The constant term is obtained by multiplying the constant terms from each of the factors (β1)3(β2)(2) = 4. Our next example shows how polynomials of higher degree arise βnaturallyβ4 in even the most basic geometric applications. 3Technically, 00 is an indeterminant form, which is a special case of being undeο¬ned. The authors realize this is beyond pedantry, but we wouldnβt mention it if we didnβt feel it was neccessary. 4this is a dangerous word... 238 Polynomial Functions Example 3.1.3. A box with no top is to be fashioned from a 10 inch Γ 12 inch piece of cardboard by cutting out congruent squares from each corner of the cardboard and then folding the resulting tabs. Let x denote the length of the side of the square which is removed from each corner. 12 in x x x x x x x x 10 in height depth width 1. Find the volume V of the box as a function of x. Include an appropriate applied domain. 2. Use a graphing calculator to graph y = V (x) on the domain you found in part 1 and approximate the dimensions of the box with maximum volume to two decimal places. What is the maximum volume? Solution. 1. From Geometry, we know that Volume = width Γ height Γ depth. The key is to ο¬nd each of these quantities in terms of x. From the ο¬gure, we see that the height of the box is x itself. The cardboard piece is initially 10 inches wide. Removing squares with a side length of x inches from each corner leaves 10 β 2x inches for the width.5 As for the depth, the cardboard is initially 12 inches long, so after cutting out x inches from each side, we would have 12 β 2x inches remaining. As a function6 of x, the volume is V (x) = x(10 β 2 |
x)(12 β 2x) = 4x3 β 44x2 + 120x To ο¬nd a suitable applied domain, we note that to make a box at all we need x > 0. Also the shorter of the two dimensions of the cardboard is 10 inches, and since we are removing 2x inches from this dimension, we also require 10 β 2x > 0 or x < 5. Hence, our applied domain is 0 < x < 5. 2. Using a graphing calculator, we see that the graph of y = V (x) has a relative maximum. For 0 < x < 5, this is also the absolute maximum. Using the βMaximumβ feature of the calculator, we get x β 1.81, y β 96.77. This yields a height of x β 1.81 inches, a width of 10 β 2x β 6.38 inches, and a depth of 12 β 2x β 8.38 inches. The y-coordinate is the maximum volume, which is approximately 96.77 cubic inches (also written in3). 5Thereβs no harm in taking an extra step here and making sure this makes sense. If we chopped out a 1 inch square from each side, then the width would be 8 inches, so chopping out x inches would leave 10 β 2x inches. 6When we write V (x), it is in the context of function notation, not the volume V times the quantity x. 3.1 Graphs of Polynomials 239 In order to solve Example 3.1.3, we made good use of the graph of the polynomial y = V (x), so we ought to turn our attention to graphs of polynomials in general. Below are the graphs of y = x2, y = x4 and y = x6, side-by-side. We have omitted the axes to allow you to see that as the exponent increases, the βbottomβ becomes βο¬atterβ and the βsidesβ become βsteeper.β If you take the the time to graph these functions by hand,7 you will see why. y = x2 y = x4 y = x6 All of these functions are even, (Do you remember how to show this?) and it is exactly because the exponent is even.8 This symmetry is important, but we want to explore a diο¬erent yet |
equally important feature of these functions which we can be seen graphically β their end behavior. The end behavior of a function is a way to describe what is happening to the function values (the y-values) as the x-values approach the βendsβ of the x-axis.9 That is, what happens to y as x becomes small without bound10 (written x β ββ) and, on the ο¬ip side, as x becomes large without bound11 (written x β β). For example, given f (x) = x2, as x β ββ, we imagine substituting x = β100, x = β1000, etc., into f to get f (β100) = 10000, f (β1000) = 1000000, and so on. Thus the function values are becoming larger and larger positive numbers (without bound). To describe this behavior, we write: as x β ββ, f (x) β β. If we study the behavior of f as x β β, we see that in this case, too, f (x) β β. (We told you that the symmetry was important!) The same can be said for any function of the form f (x) = xn where n is an even natural number. If we generalize just a bit to include vertical scalings and reο¬ections across the x-axis,12 we have 7Make sure you choose some x-values between β1 and 1. 8Herein lies one of the possible origins of the term βevenβ when applied to functions. 9Of course, there are no ends to the x-axis. 10We think of x as becoming a very large (in the sense of its absolute value) negative number far to the left of zero. 11We think of x as moving far to the right of zero and becoming a very large positive number. 12See Theorems 1.4 and 1.5 in Section 1.7. 240 Polynomial Functions End Behavior of functions f (x) = axn, n even. Suppose f (x) = axn where a = 0 is a real number and n is an even natural number. The end behavior of the graph of y = f (x) matches one of the following: for a > 0, as x β ββ, f (x) β β and as x β β, f (x) β β |
for a < 0, as x β ββ, f (x) β ββ and as x β β, f (x) β ββ Graphically: a > 0 a < 0 We now turn our attention to functions of the form f (x) = xn where n β₯ 3 is an odd natural number. (We ignore the case when n = 1, since the graph of f (x) = x is a line and doesnβt ο¬t the general pattern of higher-degree odd polynomials.) Below we have graphed y = x3, y = x5, and y = x7. The βο¬atteningβ and βsteepeningβ that we saw with the even powers presents itself here as well, and, it should come as no surprise that all of these functions are odd.13 The end behavior of these functions is all the same, with f (x) β ββ as x β ββ and f (x) β β as x β β. y = x3 y = x5 y = x7 As with the even degreed functions we studied earlier, we can generalize their end behavior. End Behavior of functions f (x) = axn, n odd. Suppose f (x) = axn where a = 0 is a real number and n β₯ 3 is an odd natural number. The end behavior of the graph of y = f (x) matches one of the following: for a > 0, as x β ββ, f (x) β ββ and as x β β, f (x) β β for a < 0, as x β ββ, f (x) β β and as x β β, f (x) β ββ Graphically: a > 0 a < 0 13And are, perhaps, the inspiration for the moniker βodd functionβ. 3.1 Graphs of Polynomials 241 Despite having diο¬erent end behavior, all functions of the form f (x) = axn for natural numbers n share two properties which help distinguish them from other animals in the algebra zoo: they are continuous and smooth. While these concepts are formally deο¬ned using Calculus,14 informally, graphs of continuous functions have no βbreaksβ or βholesβ in them, and the graphs of smooth |
functions have no βsharp turnsβ. It turns out that these traits are preserved when functions are added together, so general polynomial functions inherit these qualities. Below we ο¬nd the graph of a function which is neither smooth nor continuous, and to its right we have a graph of a polynomial, for comparison. The function whose graph appears on the left fails to be continuous where it has a βbreakβ or βholeβ in the graph; everywhere else, the function is continuous. The function is continuous at the βcornerβ and the βcuspβ, but we consider these βsharp turnsβ, so these are places where the function fails to be smooth. Apart from these four places, the function is smooth and continuous. Polynomial functions are smooth and continuous everywhere, as exhibited in the graph on the right. βholeβ βcornerβ βcuspβ βbreakβ Pathologies not found on graphs of polynomials The graph of a polynomial The notion of smoothness is what tells us graphically that, for example, f (x) = |x|, whose graph is the characteristic ββ¨β shape, cannot be a polynomial. The notion of continuity is what allowed us to construct the sign diagram for quadratic inequalities as we did in Section 2.4. This last result is formalized in the following theorem. Theorem 3.1. The Intermediate Value Theorem (Zero Version): Suppose f is a continuous function on an interval containing x = a and x = b with a < b. If f (a) and f (b) have diο¬erent signs, then f has at least one zero between x = a and x = b; that is, for at least one real number c such that a < c < b, we have f (c) = 0. The Intermediate Value Theorem is extremely profound; it gets to the heart of what it means to be a real number, and is one of the most often used and under appreciated theorems in Mathematics. With that being said, most students see the result as common sense since it says, geometrically, that the graph of a polynomial function cannot be above the x-axis at one point and below the x-axis at another point without crossing the x-axis somewhere in between. The following example |
uses the Intermediate Value Theorem to establish a fact that that most students take for granted. Many students, and sadly some instructors, will ο¬nd it silly. 14In fact, if you take Calculus, youβll ο¬nd that smooth functions are automatically continuous, so that saying βpolynomials are continuous and smoothβ is redundant. 242 Polynomial Functions Example 3.1.4. Use the Intermediate Value Theorem to establish that Solution. Consider the polynomial function f (x) = x2 β 2. Then f (1) = β1 and f (3) = 7. Since f (1) and f (3) have diο¬erent signs, the Intermediate Value Theorem guarantees us a real number c between 1 and 3 with f (c) = 0. If c2 β 2 = 0 then c = Β± 2. Since c is between 1 and 3, c is positive, so c = 2 is a real number. β β 2. β Our primary use of the Intermediate Value Theorem is in the construction of sign diagrams, as in Section 2.4, since it guarantees us that polynomial functions are always positive (+) or always negative (β) on intervals which do not contain any of its zeros. The general algorithm for polynomials is given below. Steps for Constructing a Sign Diagram for a Polynomial Function Suppose f is a polynomial function. 1. Find the zeros of f and place them on the number line with the number 0 above them. 2. Choose a real number, called a test value, in each of the intervals determined in step 1. 3. Determine the sign of f (x) for each test value in step 2, and write that sign above the corresponding interval. Example 3.1.5. Construct a sign diagram for f (x) = x3(x β 3)2(x + 2) x2 + 1. Use it to give a rough sketch of the graph of y = f (x). Solution. First, we ο¬nd the zeros of f by solving x3(x β 3)2(x + 2) x2 + 1 = 0. We get x = 0, x = 3 and x = β2. (The equation x2 + 1 = 0 produces no real solutions.) These three points divide the real number line into four intervals: (ββ, β2 |
), (β2, 0), (0, 3) and (3, β). We select the test values x = β3, x = β1, x = 1 and x = 4. We ο¬nd f (β3) is (+), f (β1) is (β) and f (1) is (+) as is f (4). Wherever f is (+), its graph is above the x-axis; wherever f is (β), its graph is below the x-axis. The x-intercepts of the graph of f are (β2, 0), (0, 0) and (3, 0). Knowing f is smooth and continuous allows us to sketch its graph. y (+) 0 (β) β2 0 (+) 0 (+) 0 3 β3 β1 1 4 x A sketch of y = f (x) A couple of notes about the Example 3.1.5 are in order. First, note that we purposefully did not label the y-axis in the sketch of the graph of y = f (x). This is because the sign diagram gives us the zeros and the relative position of the graph - it doesnβt give us any information as to how high or low the graph strays from the x-axis. Furthermore, as we have mentioned earlier in the text, without Calculus, the values of the relative maximum and minimum can only be found approximately using a calculator. If we took the time to ο¬nd the leading term of f, we would ο¬nd it to be x8. Looking 3.1 Graphs of Polynomials 243 at the end behavior of f, we notice that it matches the end behavior of y = x8. This is no accident, as we ο¬nd out in the next theorem. Theorem 3.2. End Behavior for Polynomial Functions: The end behavior of a polynomial f (x) = anxn +anβ1xnβ1 +...+a2x2 +a1x+a0 with an = 0 matches the end behavior of y = anxn. To see why Theorem 3.2 is true, letβs ο¬rst look at a speciο¬c example. Consider f (x) = 4x3 β x + 5. If we wish to examine end behavior, we look to see the behavior of f as x β Β±β. Since weβre |
concerned with xβs far down the x-axis, we are far away from x = 0 so can rewrite f (x) for these values of x as f (x) = 4x3 1 β 1 4x2 + 5 4x3 4x2 and 5 1 As x becomes unbounded (in either direction), the terms 0, as the table below indicates. 4x3 become closer and closer to x β1000 β100 β10 10 100 1000 1 4x2 5 4x3 0.00000025 β0.00000000125 β0.00000125 β0.00125 0.00125 0.00000125 0.00000000125 0.000025 0.0025 0.0025 0.000025 0.00000025 In other words, as x β Β±β, f (x) β 4x3 (1 β 0 + 0) = 4x3, which is the leading term of f. The formal proof of Theorem 3.2 works in much the same way. Factoring out the leading term leaves f (x) = anxn 1 + anβ1 anx +... + a2 anxnβ2 + a1 anxnβ1 + a0 anxn As x β Β±β, any term with an x in the denominator becomes closer and closer to 0, and we have f (x) β anxn. Geometrically, Theorem 3.2 says that if we graph y = f (x) using a graphing calculator, and continue to βzoom outβ, the graph of it and its leading term become indistinguishable. Below are the graphs of y = 4x3 β x + 5 (the thicker line) and y = 4x3 (the thinner line) in two diο¬erent windows. A view βcloseβ to the origin. A βzoomed outβ view. 244 Polynomial Functions Letβs return to the function in Example 3.1.5, f (x) = x3(xβ3)2(x+2) x2 + 1, whose sign diagram and graph are reproduced below for reference. Theorem 3.2 tells us that the end behavior is the same as that of its leading term x8. This tells us that the graph of y = f (x) starts and ends above the x-axis. In other words, f (x) is |
(+) as x β Β±β, and as a result, we no longer need to evaluate f at the test values x = β3 and x = 4. Is there a way to eliminate the need to evaluate f at the other test values? What we would really need to know is how the function behaves near its zeros does it cross through the x-axis at these points, as it does at x = β2 and x = 0, or does it simply touch and rebound like it does at x = 3. From the sign diagram, the graph of f will cross the x-axis whenever the signs on either side of the zero switch (like they do at x = β2 and x = 0); it will touch when the signs are the same on either side of the zero (as is the case with x = 3). What we need to determine is the reason behind whether or not the sign change occurs. y (+) 0 (β) β2 0 (+) 0 (+) 0 3 β3 β1 1 4 x A sketch of y = f (x) Fortunately, f was given to us in factored form: f (x) = x3(x β 3)2(x + 2). When we attempt to determine the sign of f (β4), we are attempting to ο¬nd the sign of the number (β4)3(β7)2(β2), which works out to be (β)(+)(β) which is (+). If we move to the other side of x = β2, and ο¬nd the sign of f (β1), we are determining the sign of (β1)3(β4)2(+1), which is (β)(+)(+) which gives us the (β). Notice that signs of the ο¬rst two factors in both expressions are the same in f (β4) and f (β1). The only factor which switches sign is the third factor, (x + 2), precisely the factor which gave us the zero x = β2. If we move to the other side of 0 and look closely at f (1), we get the sign pattern (+1)3(β2)2(+3) or (+)(+)(+) and we note that, once again, going from f (β1) to f (1), the only factor which changed sign was the ο¬rst factor, x3, which corresponds to the zero x = 0. Finally, to ο¬nd |
f (4), we substitute to get (+4)3(+2)2(+5) which is (+)(+)(+) or (+). The sign didnβt change for the middle factor (x β 3)2. Even though this is the factor which corresponds to the zero x = 3, the fact that the quantity is squared kept the sign of the middle factor the same on either side of 3. If we look back at the exponents on the factors (x + 2) and x3, we see that they are both odd, so as we substitute values to the left and right of the corresponding zeros, the signs of the corresponding factors change which results in the sign of the function value changing. This is the key to the behavior of the function near the zeros. We need a deο¬nition and then a theorem. Deο¬nition 3.3. Suppose f is a polynomial function and m is a natural number. If (x β c)m is a factor of f (x) but (x β c)m+1 is not, then we say x = c is a zero of multiplicity m. Hence, rewriting f (x) = x3(x β 3)2(x + 2) as f (x) = (x β 0)3(x β 3)2(x β (β2))1, we see that x = 0 is a zero of multiplicity 3, x = 3 is a zero of multiplicity 2 and x = β2 is a zero of multiplicity 1. 3.1 Graphs of Polynomials 245 Theorem 3.3. The Role of Multiplicity: Suppose f is a polynomial function and x = c is a zero of multiplicity m. If m is even, the graph of y = f (x) touches and rebounds from the x-axis at (c, 0). If m is odd, the graph of y = f (x) crosses through the x-axis at (c, 0). Our last example shows how end behavior and multiplicity allow us to sketch a decent graph without appealing to a sign diagram. Example 3.1.6. Sketch the graph of f (x) = β3(2x β 1)(x + 1)2 using end behavior and the multiplicity of its zeros. 2 and x = β1 as zeros. To ο¬nd the multiplicity of x = 1 |
Solution. The end behavior of the graph of f will match that of its leading term. To ο¬nd the leading term, we multiply by the leading terms of each factor to get (β3)(2x)(x)2 = β6x3. This tells us that the graph will start above the x-axis, in Quadrant II, and ο¬nish below the x-axis, in Quadrant IV. Next, we ο¬nd the zeros of f. Fortunately for us, f is factored.15 Setting each factor equal to zero gives is x = 1 2 we note that it corresponds to the factor (2x β 1). This isnβt strictly in the form required in Deο¬nition 3.3. If we factor out the 2, however, we get (2x β 1) = 2 x β 1, and we see that the multiplicity of x = 1 2 2 is 1. Since 1 is an odd number, we know from Theorem 3.3 that the graph of f will cross through the x-axis at 1 2, 0. Since the zero x = β1 corresponds to the factor (x + 1)2 = (x β (β1))2, we ο¬nd its multiplicity to be 2 which is an even number. As such, the graph of f will touch and rebound from the x-axis at (β1, 0). Though weβre not asked to, we can ο¬nd the y-intercept by ο¬nding f (0) = β3(2(0) β 1)(0 + 1)2 = 3. Thus (0, 3) is an additional point on the graph. Putting this together gives us the graph below. y x 15Obtaining the factored form of a polynomial is the main focus of the next few sections. 246 3.1.1 Exercises Polynomial Functions In Exercises 1 - 10, ο¬nd the degree, the leading term, the leading coeο¬cient, the constant term and the end behavior of the given polynomial. 1. f (x) = 4 β x β 3x2 3. q(r) = 1 β 16r4 β 5. f (x) = 3x17 + 22.5x10 β Οx7 + 1 3 2. g(x) = 3x5 β 2 |
x2 + x + 1 4. Z(b) = 42b β b3 6. s(t) = β4.9t2 + v0t + s0 7. P (x) = (x β 1)(x β 2)(x β 3)(x β 4) 8. p(t) = βt2(3 β 5t)(t2 + t + 4) 9. f (x) = β2x3(x + 1)(x + 2)2 10. G(t) = 4(t β 2)2 t + 1 2 In Exercises 11 - 20, ο¬nd the real zeros of the given polynomial and their corresponding multiplicities. Use this information along with a sign chart to provide a rough sketch of the graph of the polynomial. Compare your answer with the result from a graphing utility. 11. a(x) = x(x + 2)2 12. g(x) = x(x + 2)3 13. f (x) = β2(x β 2)2(x + 1) 14. g(x) = (2x + 1)2(x β 3) 15. F (x) = x3(x + 2)2 16. P (x) = (x β 1)(x β 2)(x β 3)(x β 4) 17. Q(x) = (x + 5)2(x β 3)4 18. h(x) = x2(x β 2)2(x + 2)2 19. H(t) = (3 β t)(t2 + 1) 20. Z(b) = b(42 β b2) In Exercises 21 - 26, given the pair of functions f and g, sketch the graph of y = g(x) by starting with the graph of y = f (x) and using transformations. Track at least three points of your choice through the transformations. State the domain and range of g. 21. f (x) = x3, g(x) = (x + 2)3 + 1 22. f (x) = x4, g(x) = (x + 2)4 + 1 23. f (x) = x4, g(x) = 2 β 3(x β 1)4 24. f (x) = x5, g(x) |
= βx5 β 3 25. f (x) = x5, g(x) = (x + 1)5 + 10 26. f (x) = x6, g(x) = 8 β x6 27. Use the Intermediate Value Theorem to prove that f (x) = x3 β 9x + 5 has a real zero in each of the following intervals: [β4, β3], [0, 1] and [2, 3]. 28. Rework Example 3.1.3 assuming the box is to be made from an 8.5 inch by 11 inch sheet of paper. Using scissors and tape, construct the box. Are you surprised?16 16Consider decorating the box and presenting it to your instructor. If done well enough, maybe your instructor will issue you some bonus points. Or maybe not. 3.1 Graphs of Polynomials 247 In Exercises 29 - 31, suppose the revenue R, in thousands of dollars, from producing and selling x hundred LCD TVs is given by R(x) = β5x3 + 35x2 + 155x for 0 β€ x β€ 10.07. 29. Use a graphing utility to graph y = R(x) and determine the number of TVs which should be sold to maximize revenue. What is the maximum revenue? 30. Assume that the cost, in thousands of dollars, to produce x hundred LCD TVs is given by C(x) = 200x + 25 for x β₯ 0. Find and simplify an expression for the proο¬t function P (x). (Remember: Proο¬t = Revenue - Cost.) 31. Use a graphing utility to graph y = P (x) and determine the number of TVs which should be sold to maximize proο¬t. What is the maximum proο¬t? 32. While developing their newest game, Sasquatch Attack!, the makers of the PortaBoy (from Example 2.1.5) revised their cost function and now use C(x) =.03x3 β 4.5x2 + 225x + 250, for x β₯ 0. As before, C(x) is the cost to make x PortaBoy Game Systems. Market research indicates that the demand function p(x) = β1.5x + 250 remains unchanged. Use a graphing utility to ο¬nd the production level x that maximizes the proο¬t made |
by producing and selling x PortaBoy game systems. 33. According to US Postal regulations, a rectangular shipping box must satisfy the inequality βLength + Girth β€ 130 inchesβ for Parcel Post and βLength + Girth β€ 108 inchesβ for other services. Letβs assume we have a closed rectangular box with a square face of side length x as drawn below. The length is the longest side and is clearly labeled. The girth is the distance around the box in the other two dimensions so in our case it is the sum of the four sides of the square, 4x. (a) Assuming that weβll be mailing a box via Parcel Post where Length + Girth = 130 inches, express the length of the box in terms of x and then express the volume V of the box in terms of x. (b) Find the dimensions of the box of maximum volume that can be shipped via Parcel Post. (c) Repeat parts 33a and 33b if the box is shipped using βother servicesβ. x x length 248 Polynomial Functions 34. We now revisit the data set from Exercise 6b in Section 2.5. In that exercise, you were given a chart of the number of hours of daylight they get on the 21st of each month in Fairbanks, Alaska based on the 2009 sunrise and sunset data found on the U.S. Naval Observatory website. We let x = 1 represent January 21, 2009, x = 2 represent February 21, 2009, and so on. The chart is given again for reference. Month Number Hours of Daylight 1 2 3 4 5 6 7 8 9 10 11 12 5.8 9.3 12.4 15.9 19.4 21.8 19.4 15.6 12.4 9.1 5.6 3.3 Find cubic (third degree) and quartic (fourth degree) polynomials which model this data and comment on the goodness of ο¬t for each. What can we say about using either model to make predictions about the year 2020? (Hint: Think about the end behavior of polynomials.) Use the models to see how many hours of daylight they got on your birthday and then check the website to see how accurate the models are. Knowing that Sasquatch are largely nocturnal, what days of the year according to your models are going to allow for at least 14 hours of darkness for ο¬eld research on |
the elusive creatures? 35. An electric circuit is built with a variable resistor installed. For each of the following resistance values (measured in kilo-ohms, kβ¦), the corresponding power to the load (measured in milliwatts, mW ) is given in the table below. 17 Resistance: (kβ¦) Power: (mW ) 1.012 1.063 2.199 1.496 3.275 1.610 4.676 1.613 6.805 1.505 9.975 1.314 (a) Make a scatter diagram of the data using the Resistance as the independent variable and Power as the dependent variable. (b) Use your calculator to ο¬nd quadratic (2nd degree), cubic (3rd degree) and quartic (4th degree) regression models for the data and judge the reasonableness of each. (c) For each of the models found above, ο¬nd the predicted maximum power that can be delivered to the load. What is the corresponding resistance value? (d) Discuss with your classmates the limitations of these models - in particular, discuss the end behavior of each. 36. Show that the end behavior of a linear function f (x) = mx + b is as it should be according to the results weβve established in the section for polynomials of odd degree.18 (That is, show that the graph of a linear function is βup on one side and down on the otherβ just like the graph of y = anxn for odd numbers n.) 17The authors wish to thank Don Anthan and Ken White of Lakeland Community College for devising this problem and generating the accompanying data set. 18Remember, to be a linear function, m = 0. 3.1 Graphs of Polynomials 249 37. There is one subtlety about the role of multiplicity that we need to discuss further; speciο¬cally we need to see βhowβ the graph crosses the x-axis at a zero of odd multiplicity. In the section, we deliberately excluded the function f (x) = x from the discussion of the end behavior of f (x) = xn for odd numbers n and we said at the time that it was due to the fact that f (x) = x didnβt ο¬t the pattern we were trying to establish. You just showed in the previous exercise that |
the end behavior of a linear function behaves like every other polynomial of odd degree, so what doesnβt f (x) = x do that g(x) = x3 does? Itβs the βο¬atteningβ for values of x near zero. It is this local behavior that will distinguish between a zero of multiplicity 1 and one of higher odd multiplicity. Look again closely at the graphs of a(x) = x(x + 2)2 and F (x) = x3(x + 2)2 from Exercise 3.1.1. Discuss with your classmates how the graphs are fundamentally diο¬erent at the origin. It might help to use a graphing calculator to zoom in on the origin to see the diο¬erent crossing behavior. Also compare the behavior of a(x) = x(x + 2)2 to that of g(x) = x(x + 2)3 near the point (β2, 0). What do you predict will happen at the zeros of f (x) = (x β 1)(x β 2)2(x β 3)3(x β 4)4(x β 5)5? 38. Here are a few other questions for you to discuss with your classmates. (a) How many local extrema could a polynomial of degree n have? How few local extrema can it have? (b) Could a polynomial have two local maxima but no local minima? (c) If a polynomial has two local maxima and two local minima, can it be of odd degree? Can it be of even degree? (d) Can a polynomial have local extrema without having any real zeros? (e) Why must every polynomial of odd degree have at least one real zero? (f) Can a polynomial have two distinct real zeros and no local extrema? (g) Can an x-intercept yield a local extrema? Can it yield an absolute extrema? (h) If the y-intercept yields an absolute minimum, what can we say about the degree of the polynomial and the sign of the leading coeο¬cient? 250 3.1.2 Answers 1. f (x) = 4 β x β 3x2 Degree 2 Leading term β3x2 Leading coeο¬ |
cient β3 Constant term 4 As x β ββ, f (x) β ββ As x β β, f (x) β ββ 3. q(r) = 1 β 16r4 Degree 4 Leading term β16r4 Leading coeο¬cient β16 Constant term 1 As r β ββ, q(r) β ββ As r β β, q(r) β ββ β 5. f (x) = 3x17 + 22.5x10 β Οx7 + 1 3 β 3x17 β Degree 17 Leading term Leading coeο¬cient Constant term 1 3 As x β ββ, f (x) β ββ As x β β, f (x) β β 3 Polynomial Functions 2. g(x) = 3x5 β 2x2 + x + 1 Degree 5 Leading term 3x5 Leading coeο¬cient 3 Constant term 1 As x β ββ, g(x) β ββ As x β β, g(x) β β 4. Z(b) = 42b β b3 Degree 3 Leading term βb3 Leading coeο¬cient β1 Constant term 0 As b β ββ, Z(b) β β As b β β, Z(b) β ββ 6. s(t) = β4.9t2 + v0t + s0 Degree 2 Leading term β4.9t2 Leading coeο¬cient β4.9 Constant term s0 As t β ββ, s(t) β ββ As t β β, s(t) β ββ 7. P (x) = (x β 1)(x β 2)(x β 3)(x β 4) 8. p(t) = βt2(3 β 5t)(t2 + t + 4) Degree 4 Leading term x4 Leading coeο¬cient 1 Constant term 24 As x β ββ, P (x) β β As x β β, P (x) β β Degree 5 Leading term 5t5 Leading coeο¬cient 5 Constant term 0 As t β ββ, p(t) β ββ As t β β, p(t) β οΏ½ |
οΏ½ 3.1 Graphs of Polynomials 251 9. f (x) = β2x3(x + 1)(x + 2)2 Degree 6 Leading term β2x6 Leading coeο¬cient β2 Constant term 0 As x β ββ, f (x) β ββ As x β β, f (x) β ββ 10. G(t) = 4(t β 2)2 t + 1 2 Degree 3 Leading term 4t3 Leading coeο¬cient 4 Constant term 8 As t β ββ, G(t) β ββ As t β β, G(t) β β 11. a(x) = x(x + 2)2 x = 0 multiplicity 1 x = β2 multiplicity 2 12. g(x) = x(x + 2)3 x = 0 multiplicity 1 x = β2 multiplicity 3 y y β2 β1 x β2 β1 x 13. f (x) = β2(x β 2)2(x + 1) x = 2 multiplicity 2 x = β1 multiplicity 1 14. g(x) = (2x + 1)2(x β 3) x = β 1 x = 3 multiplicity 1 2 multiplicity 2 y y β2 β1 1 2 x β1 1 2 3 x 252 Polynomial Functions 15. F (x) = x3(x + 2)2 x = 0 multiplicity 3 x = β2 multiplicity 2 y β2 β1 x 17. Q(x) = (x + 5)2(x β 3)4 x = β5 multiplicity 2 x = 3 multiplicity 4 y 16. P (x) = (x β 1)(x β 2)(x β 3)(x β 4) x = 1 multiplicity 1 x = 2 multiplicity 1 x = 3 multiplicity 1 x = 4 multiplicity 1 y 1 2 3 4 x 18. f (x) = x2(x β 2)2(x + 2)2 x = β2 multiplicity 2 x = 0 multiplicity 2 x = 2 multiplicity 2 y β5β4β3β2β1 1 2 3 4 5 x β2 β1 1 2 x 19. H(t) = (3 β t) t2 + 1 x = 3 multiplicity 1 y 1 2 3 t 20. |
Z(b) = b(42 β b2) β 42 multiplicity 1 b = β b = 0 multiplicity 1 b = 42 multiplicity 1 β y β6β5β4β3β2β1 1 2 3 4 5 6 b 3.1 Graphs of Polynomials 253 21. g(x) = (x + 2)3 + 1 domain: (ββ, β) range: (ββ, β) 22. g(x) = (x + 2)4 + 1 domain: (ββ, β) range: [1, β) y x 12 11 10 1 β2 β3 β4 β5 β6 β7 β8 β9 β10 β4 β3 β2 β1 23. g(x) = 2 β 3(x β 1)4 domain: (ββ, β) range: (ββ, 2] y 1 2 x 2 1 β1 β2 β3 β4 β5 β6 β7 β8 β9 β10 β11 β12 β13 y 21 20 19 18 17 16 15 14 13 12 11 10 4 β3 β2 β1 x 24. g(x) = βx5 β 3 domain: (ββ, β) range: (ββ, β) y 10 1 β2 β3 β4 β5 β6 β7 β8 β9 β10 β1 1 x 254 Polynomial Functions 25. g(x) = (x + 1)5 + 10 domain: (ββ, β) range: (ββ, β) 26. g(x) = 8 β x6 domain: (ββ, β) range: (ββ, 8] y 21 20 19 18 17 16 15 14 13 12 11 10 4 β3 β2 β1 x y 10 1 β2 β3 β4 β5 β6 β7 β8 β9 β10 β1 1 x 27. We have f (β4) = β23, f (β3) = 5, f (0) = 5, f (1) = β3, f (2) = β5 and f (3) = 5 so the Intermediate Value Theorem tells us that f (x) = x3 β 9x + 5 has real zeros in the intervals [β4, β3], [0, 1] and [2, 3]. 28. |
V (x) = x(8.5 β 2x)(11 β 2x) = 4x3 β 39x2 + 93.5x, 0 < x < 4.25. Volume is maximized when x β 1.58, so the dimensions of the box with maximum volume are: height β 1.58 inches, width β 5.34 inches, and depth β 7.84 inches. The maximum volume is β 66.15 cubic inches. 29. The calculator gives the location of the absolute maximum (rounded to three decimal places) as x β 6.305 and y β 1115.417. Since x represents the number of TVs sold in hundreds, x = 6.305 corresponds to 630.5 TVs. Since we canβt sell half of a TV, we compare R(6.30) β 1115.415 and R(6.31) β 1115.416, so selling 631 TVs results in a (slightly) higher revenue. Since y represents the revenue in thousands of dollars, the maximum revenue is $1,115,416. 30. P (x) = R(x) β C(x) = β5x3 + 35x2 β 45x β 25, 0 β€ x β€ 10.07. 31. The calculator gives the location of the absolute maximum (rounded to three decimal places) as x β 3.897 and y β 35.255. Since x represents the number of TVs sold in hundreds, x = 3.897 corresponds to 389.7 TVs. Since we canβt sell 0.7 of a TV, we compare P (3.89) β 35.254 and P (3.90) β 35.255, so selling 390 TVs results in a (slightly) higher revenue. Since y represents the revenue in thousands of dollars, the maximum revenue is $35,255. 32. Making and selling 71 PortaBoys yields a maximized proο¬t of $5910.67. 3.1 Graphs of Polynomials 255 33. (a) Our ultimate goal is to maximize the volume, so weβll start with the maximum Length + Girth of 130. This means the length is 130 β 4x. The volume of a rectangular box is always length Γ width Γ height so we get V (x) = x2(130 β 4x) = β |
4x3 + 130x2. (b) Graphing y = V (x) on [0, 33] Γ [0, 21000] shows a maximum at (21.67, 20342.59) so the dimensions of the box with maximum volume are 21.67in. Γ 21.67in. Γ 43.32in. for a volume of 20342.59in.3. (c) If we start with Length + Girth = 108 then the length is 108 β 4x and the volume is V (x) = β4x3 + 108x2. Graphing y = V (x) on [0, 27] Γ [0, 11700] shows a maximum at (18.00, 11664.00) so the dimensions of the box with maximum volume are 18.00in. Γ 18.00in. Γ 36in. for a volume of 11664.00in.3. (Calculus will conο¬rm that the measurements which maximize the volume are exactly 18in. by 18in. by 36in., however, as Iβm sure you are aware by now, we treat all calculator results as approximations and list them as such.) 34. The cubic regression model is p3(x) = 0.0226x3 β 0.9508x2 + 8.615x β 3.446. It has R2 = 0.93765 which isnβt bad. The graph of y = p3(x) in the viewing window [β1, 13] Γ [0, 24] along with the scatter plot is shown below on the left. Notice that p3 hits the x-axis at about x = 12.45 making this a bad model for future predictions. To use the model to approximate the number of hours of sunlight on your birthday, youβll have to ο¬gure out what decimal value of x is close enough to your birthday and then plug it into the model. My (Jeο¬βs) birthday is July 31 which is 10 days after July 21 (x = 7). Assuming 30 days in a month, I think x = 7.33 should work for my birthday and p3(7.33) β 17.5. The website says there will be about 18.25 hours of daylight that day. To have 14 hours of darkness we need 10 hours of daylight. We see that p |
3(1.96) β 10 and p3(10.05) β 10 so it seems reasonable to say that weβll have at least 14 hours of darkness from December 21, 2008 (x = 0) to February 21, 2009 (x = 2) and then again from October 21,2009 (x = 10) to December 21, 2009 (x = 12). The quartic regression model is p4(x) = 0.0144x4 β0.3507x3 +2.259x2 β1.571x+5.513. It has R2 = 0.98594 which is good. The graph of y = p4(x) in the viewing window [β1, 15] Γ [0, 35] along with the scatter plot is shown below on the right. Notice that p4(15) is above 24 making this a bad model as well for future predictions. However, p4(7.33) β 18.71 making it much better at predicting the hours of daylight on July 31 (my birthday). This model says weβll have at least 14 hours of darkness from December 21, 2008 (x = 0) to about March 1, 2009 (x = 2.30) and then again from October 10, 2009 (x = 9.667) to December 21, 2009 (x = 12). y = p3(x) y = p4(x) 256 Polynomial Functions 35. (a) The scatter plot is shown below with each of the three regression models. (b) The quadratic model is P2(x) = β0.02x2 + 0.241x + 0.956 with R2 = 0.77708. The cubic model is P3(x) = 0.005x3 β 0.103x2 + 0.602x + 0.573 with R2 = 0.98153. The quartic model is P4(x) = β0.000969x4 + 0.0253x3 β 0.240x2 + 0.944x + 0.330 with R2 = 0.99929. (c) The maximums predicted by the three models are P2(5.737) β 1.648, P3(4.232) β 1.657 and P4(3.784) β 1.630, respectively. y = P2 |
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