Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Example 2.3.5. Graph \( f\left( x\right) = \left| {{x}^{2} - x - 6}\right| \) . | Solution. Using the definition of absolute value, Definition 2.4, we have\n\n\[ f\left( x\right) = \left\{ \begin{array}{rrr} - \left( {{x}^{2} - x - 6}\right) , & \text{ if } & {x}^{2} - x - 6 < 0 \\ {x}^{2} - x - 6, & \text{ if } & {x}^{2} - x - 6 \geq 0 \end{array}\right. \]\n\nThe trouble is that we have yet to dev... | Yes |
1. Solve \( f\left( x\right) = g\left( x\right) \) . | 1. To solve \( f\left( x\right) = g\left( x\right) \), we replace \( f\left( x\right) \) with \( {2x} - 1 \) and \( g\left( x\right) \) with 5 to get \( {2x} - 1 = 5 \) . Solving for \( x \), we get \( x = 3 \) . | Yes |
1. Solve \( f\left( x\right) = g\left( x\right) \) . | 1. To solve \( f\left( x\right) = g\left( x\right) \), we look for where the graphs of \( f \) and \( g \) intersect. These appear to be at the points \( \left( {-1,2}\right) \) and \( \left( {1,2}\right) \), so our solutions to \( f\left( x\right) = g\left( x\right) \) are \( x = - 1 \) and \( x = 1 \) . | Yes |
Theorem 2.4. Inequalities Involving the Absolute Value: Let \( c \) be a real number.\n\n- For \( c > 0,\left| x\right| < c \) is equivalent to \( - c < x < c \) .\n\n- For \( c > 0,\left| x\right| \leq c \) is equivalent to \( - c \leq x \leq c \) .\n\n- For \( c \leq 0,\left| x\right| < c \) has no solution, and for ... | 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 \( \left( {0, c}... | Yes |
1. \( \left| {x - 1}\right| \geq 3 \) | From Theorem 2.4, \( \left| {x - 1}\right| \geq 3 \) is equivalent to \( x - 1 \leq - 3 \) or \( x - 1 \geq 3 \) . Solving, we get \( x \leq - 2 \) or \( x \geq 4 \), which, in interval notation is \( \left( {-\infty , - 2\rbrack \cup \lbrack 4,\infty }\right) \) . Graphically, we have\n\n\n\nWe can tell both from the correlation coefficient as well as the graph that the regression line is a good fit to the data. | No |
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, \( x = - \frac{b}{2a} \approx - \frac{9.464}{2\left( {-{0.321}}\right) } \approx {14.741} \) . This corresponds to roughly 2: 45 PM. To find the temperature, we substitute \( x = {14.741} \) into \( y... | Yes |
Example 3.1.1. Determine if the following functions are polynomials. Explain your reasoning. | 1. We note directly that the domain of \( g\left( x\right) = \frac{{x}^{3} + 4}{x} \) is \( x \neq 0 \) . By definition, a polynomial has all real numbers as its domain. Hence, \( g \) can’t be a polynomial.\n\n2. Even though \( p\left( x\right) = \frac{{x}^{3} + {4x}}{x} \) simplifies to \( p\left( x\right) = {x}^{2} ... | Yes |
Example 3.1.2. Find the degree, leading term, leading coefficient and constant term of the following polynomial functions.\n\n1. \( f\left( x\right) = 4{x}^{5} - 3{x}^{2} + {2x} - 5 \) 2. \( g\left( x\right) = {12x} + {x}^{3} \)\n\n3. \( h\left( x\right) = \frac{4 - x}{5} \) 4. \( p\left( x\right) = {\left( 2x - 1\righ... | ## Solution.\n\n1. There are no surprises with \( f\left( x\right) = 4{x}^{5} - 3{x}^{2} + {2x} - 5 \) . It is written in the form of Definition 3.2, and we see that the degree is 5, the leading term is \( 4{x}^{5} \), the leading coefficient is 4 and the constant term is -5 .\n\n2. The form given in Definition 3.2 has... | Yes |
1. Find the volume \( V \) of the box as a function of \( x \) . Include an appropriate applied domain. | ## Solution.\n\n1. From Geometry, we know that Volume \( = \) width \( \times \) height \( \times \) depth. The key is to find each of these quantities in terms of \( x \) . From the figure, we see that the height of the box is \( x \) itself. The cardboard piece is initially 10 inches wide. Removing squares with a sid... | Yes |
Use the Intermediate Value Theorem to establish that \( \sqrt{2} \) is a real number. | Consider the polynomial function \( f\left( x\right) = {x}^{2} - 2 \) . Then \( f\left( 1\right) = - 1 \) and \( f\left( 3\right) = 7 \) . Since \( f\left( 1\right) \) and \( f\left( 3\right) \) have different signs, the Intermediate Value Theorem guarantees us a real number \( c \) between 1 and 3 with \( f\left( c\ri... | Yes |
Construct a sign diagram for \( f\left( x\right) = {x}^{3}{\left( x - 3\right) }^{2}\left( {x + 2}\right) \left( {{x}^{2} + 1}\right) \) . Use it to give a rough sketch of the graph of \( y = f\left( x\right) \) . | Solution. First, we find the zeros of \( f \) by solving \( {x}^{3}{\left( x - 3\right) }^{2}\left( {x + 2}\right) \left( {{x}^{2} + 1}\right) = 0 \) . We get \( x = 0 \) , \( x = 3 \) and \( x = - 2 \) . (The equation \( {x}^{2} + 1 = 0 \) produces no real solutions.) These three points divide the real number line int... | Yes |
Theorem 3.2. End Behavior for Polynomial Functions: The end behavior of a polynomial \( f\left( x\right) = {a}_{n}{x}^{n} + {a}_{n - 1}{x}^{n - 1} + \ldots + {a}_{2}{x}^{2} + {a}_{1}x + {a}_{0} \) with \( {a}_{n} \neq 0 \) matches the end behavior of \( y = {a}_{n}{x}^{n} \) . | To see why Theorem 3.2 is true, let’s first look at a specific example. Consider \( f\left( x\right) = 4{x}^{3} - x + 5 \) . If we wish to examine end behavior, we look to see the behavior of \( f \) as \( x \rightarrow \pm \infty \) . Since we’re concerned with \( x \) ’s far down the \( x \) -axis, we are far away fr... | Yes |
Sketch the graph of \( f\left( x\right) = - 3\left( {{2x} - 1}\right) {\left( x + 1\right) }^{2} \) using end behavior and the multiplicity of its zeros. | Solution. The end behavior of the graph of \( f \) will match that of its leading term. To find the leading term, we multiply by the leading terms of each factor to get \( \left( {-3}\right) \left( {2x}\right) {\left( x\right) }^{2} = - 6{x}^{3} \) . This tells us that the graph will start above the \( x \) -axis, in Q... | Yes |
Theorem 3.5. The Remainder Theorem: Suppose \( p \) is a polynomial of degree at least 1 and \( c \) is a real number. When \( p\left( x\right) \) is divided by \( x - c \) the remainder is \( p\left( c\right) \) . | The proof of Theorem 3.5 is a direct consequence of Theorem 3.4. When a polynomial is divided by \( x - c \), the remainder is either 0 or has degree less than the degree of \( x - c \) . Since \( x - c \) is degree 1 , the degree of the remainder must be 0 , which means the remainder is a constant. Hence, in either ca... | Yes |
Theorem 3.6. The Factor Theorem: Suppose \( p \) is a nonzero polynomial. The real number \( c \) is a zero of \( p \) if and only if \( \left( {x - c}\right) \) is a factor of \( p\left( x\right) \) . | The proof of The Factor Theorem is a consequence of what we already know. If \( \left( {x - c}\right) \) is a factor of \( p\left( x\right) \), this means \( p\left( x\right) = \left( {x - c}\right) q\left( x\right) \) for some polynomial \( q \) . Hence, \( p\left( c\right) = \left( {c - c}\right) q\left( c\right) = 0... | Yes |
Use synthetic division to perform the following polynomial divisions. Find the quotient and the remainder polynomials, then write the dividend, quotient and remainder in the form given in Theorem 3.4. | 1. When setting up the synthetic division tableau, we need to enter 0 for the coefficient of \( x \) in the dividend. Doing so gives\n\n\n\nSince the dividend was a third degree polynomial, the quotient is a quadrati... | Yes |
1. Find \( p\left( {-2}\right) \) using The Remainder Theorem. Check your answer by substitution. | 1. The Remainder Theorem states \( p\left( {-2}\right) \) is the remainder when \( p\left( x\right) \) is divided by \( x - \left( {-2}\right) \) . We set up our synthetic division tableau below. We are careful to record the coefficient of \( {x}^{2} \) as 0 , and proceed as above.\n\n\n\nFrom the first divi... | Yes |
Theorem 3.7. Suppose \( f \) is a polynomial of degree \( n \geq 1 \) . Then \( f \) has at most \( n \) real zeros, counting multiplicities. | Theorem 3.7 is a consequence of the Factor Theorem and polynomial multiplication. Every zero \( c \) of \( f \) gives us a factor of the form \( \left( {x - c}\right) \) for \( f\left( x\right) \) . Since \( f \) has degree \( n \), there can be at most \( n \) of these factors. | Yes |
Theorem 3.8. Cauchy’s Bound: Suppose \( f\left( x\right) = {a}_{n}{x}^{n} + {a}_{n - 1}{x}^{n - 1} + \ldots + {a}_{1}x + {a}_{0} \) is a polynomial of degree \( n \) with \( n \geq 1 \) . Let \( M \) be the largest of the numbers: \( \frac{\left| {a}_{0}\right| }{\left| {a}_{n}\right| },\frac{\left| {a}_{1}\right| }{\l... | The proof of this fact is not easily explained within the confines of this text. This paper contains the result and gives references to its proof. | No |
Example 3.3.1. Let \( f\left( x\right) = 2{x}^{4} + 4{x}^{3} - {x}^{2} - {6x} - 3 \) . Determine an interval which contains all of the real zeros of \( f \) . | Solution. To find the \( M \) stated in Cauchy’s Bound, we take the absolute value of the leading coefficient, in this case \( \left| 2\right| = 2 \) and divide it into the largest (in absolute value) of the remaining coefficients, in this case \( \left| {-6}\right| = 6 \) . This yields \( M = 3 \) so it is guaranteed ... | Yes |
Theorem 3.9. Rational Zeros Theorem: Suppose \( f\left( x\right) = {a}_{n}{x}^{n} + {a}_{n - 1}{x}^{n - 1} + \ldots + {a}_{1}x + {a}_{0} \) is a polynomial of degree \( n \) with \( n \geq 1 \), and \( {a}_{0},{a}_{1},\ldots {a}_{n} \) are integers. If \( r \) is a rational zero of \( f \), then \( r \) is of the form ... | The Rational Zeros Theorem gives us a list of numbers to try in our synthetic division and that is a lot nicer than simply guessing. If none of the numbers in the list are zeros, then either the polynomial has no real zeros at all, or all of the real zeros are irrational numbers. To see why the Rational Zeros Theorem w... | Yes |
Let \( f\left( x\right) = 2{x}^{4} + 4{x}^{3} - {x}^{2} - {6x} - 3 \) . Use the Rational Zeros Theorem to list all of the possible rational zeros of \( f \) . | Solution. To generate a complete list of rational zeros, we need to take each of the factors of constant term, \( {a}_{0} = - 3 \), and divide them by each of the factors of the leading coefficient \( {a}_{4} = 2 \) . The factors of \( - 3 \) are \( \pm 1 \) and \( \pm 3 \) . Since the Rational Zeros Theorem tacks on a... | Yes |
Let \( f\left( x\right) = 2{x}^{4} + 4{x}^{3} - {x}^{2} - {6x} - 3 \) .\n\n1. Graph \( y = f\left( x\right) \) on the calculator using the interval obtained in Example 3.3.1 as a guide.\n\n2. Use the graph to shorten the list of possible rational zeros obtained in Example 3.3.2.\n\n3. Use synthetic division to find the... | ## Solution.\n\n1. In Example 3.3.1, we determined all of the real zeros of \( f \) lie in the interval \( \left\lbrack {-4,4}\right\rbrack \) . We set our window accordingly and get\n\n\n\n2. In Example 3.3.2, we le... | Yes |
Use Cauchy’s Bound to determine an interval in which all of the real zeros of \( f \) lie. | Applying Cauchy’s Bound, we find \( M = {12} \), so all of the real zeros lie in the interval \( \left\lbrack {-{13},{13}}\right\rbrack \) . | Yes |
Let \( f\left( x\right) = 2{x}^{4} + 4{x}^{3} - {x}^{2} - {6x} - 3 \) . Use Descartes’ Rule of Signs to determine the possible number and location of the real zeros of \( f \) . | Solution. As noted above, the variations of sign of \( f\left( x\right) \) is 1 . This means, counting multiplicities, \( f \) has exactly 1 positive real zero. Since \( f\left( {-x}\right) = 2{\left( -x\right) }^{4} + 4{\left( -x\right) }^{3} - {\left( -x\right) }^{2} - 6\left( {-x}\right) - 3 = \) \( 2{x}^{4} - 4{x}^... | Yes |
Theorem 3.11. Upper and Lower Bounds: Suppose \( f \) is a polynomial of degree \( n \geq 1 \) . - If \( c > 0 \) is synthetically divided into \( f \) and all of the numbers in the final line of the division tableau have the same signs, then \( c \) is an upper bound for the real zeros of \( f \) . That is, there are ... | The Upper and Lower Bounds Theorem works because of Theorem 3.4. For the upper bound part of the theorem, suppose \( c > 0 \) is divided into \( f \) and the resulting line in the division tableau contains, for example, all nonnegative numbers. This means \( f\left( x\right) = \left( {x - c}\right) q\left( x\right) + r... | Yes |
1. Find all of the real zeros of \( f \) and their multiplicities. | We know from Cauchy’s Bound that all of the real zeros lie in the interval \( \left\lbrack {-4,4}\right\rbrack \) and that our possible rational zeros are \( \pm \frac{1}{2}, \pm 1, \pm \frac{3}{2} \) and \( \pm 3 \) . Descartes’ Rule of Signs guarantees us at least one negative real zero and exactly one positive real ... | Yes |
Example 3.3.8. Suppose the profit \( P \), in thousands of dollars, from producing and selling \( x \) hundred LCD TVs is given by \( P\left( x\right) = - 5{x}^{3} + {35}{x}^{2} - {45x} - {25},0 \leq x \leq {10.07} \) . How many TVs should be produced to make a profit? Check your answer using a graphing utility. | Solution. To ’make a profit’ means to solve \( P\left( x\right) = - 5{x}^{3} + {35}{x}^{2} - {45x} - {25} > 0 \), which we do analytically using a sign diagram. To simplify things, we first factor out the -5 common to all the coefficients to get \( - 5\left( {{x}^{3} - 7{x}^{2} + {9x} - 5}\right) > 0 \), so we can just... | Yes |
Perform the indicated operations. Write your answer in the form \( {}^{5}a + {bi} \) . | 1. As mentioned earlier, we treat expressions involving \( i \) as we would any other radical. We combine like terms to get \( \left( {1 - {2i}}\right) - \left( {3 + {4i}}\right) = 1 - {2i} - 3 - {4i} = - 2 - {6i} \) . | No |
Theorem 3.12. Properties of the Complex Conjugate: Let \( z \) and \( w \) be complex numbers.\n\n- \( \overline{\bar{z}} = z \)\n\n- \( \bar{z} + \bar{w} = \overline{z + w} \)\n\n- \( \bar{z}\bar{w} = \overline{zw} \)\n\n- \( {\left( \bar{z}\right) }^{n} = \overline{{z}^{n}} \), for any natural number \( n \)\n\n- \( ... | Essentially, Theorem 3.12 says that complex conjugation works well with addition, multiplication and powers. The proof of these properties can best be achieved by writing out \( z = a + {bi} \) and \( w = c + {di} \) for real numbers \( a, b, c \) and \( d \) . Next, we compute the left and right hand sides of each equ... | No |
1. Find all of the complex zeros of \( f \) and state their multiplicities. | Since \( f \) is a fifth degree polynomial, we know that we need to perform at least three successful divisions to get the quotient down to a quadratic function. At that point, we can find the remaining zeros using the Quadratic Formula, if necessary. Using the techniques developed in Section 3.3, we get\n\n | No |
Find a polynomial \( p \) of lowest degree that has integer coefficients and satisfies all of the following criteria:\n\n- the graph of \( y = p\\left( x\\right) \) touches (but doesn’t cross) the \( x \) -axis at \( \\left( {\\frac{1}{3},0}\\right) \)\n\n- \( x = {3i} \) is a zero of \( p \).\n\n- as \( x \\rightarrow... | Solution. To solve this problem, we will need a good understanding of the relationship between the \( x \) -intercepts of the graph of a function and the zeros of a function, the Factor Theorem, the role of multiplicity, complex conjugates, the Complex Factorization Theorem, and end behavior of polynomial functions. (I... | Yes |
Find the domain of the following rational functions. Write them in the form \( \frac{p\left( x\right) }{q\left( x\right) } \) for polynomial functions \( p \) and \( q \) and simplify. | 1. To find the domain of \( f \), we proceed as we did in Section 1.4: we find the zeros of the denominator and exclude them from the domain. Setting \( x + 1 = 0 \) results in \( x = - 1 \) . Hence, our domain is \( \left( {-\infty , - 1}\right) \cup \left( {-1,\infty }\right) \) . The expression \( f\left( x\right) \... | No |
Theorem 4.1. Location of Vertical Asymptotes and Holes: \( {}^{a} \) Suppose \( r \) is a rational function which can be written as \( r\left( x\right) = \frac{p\left( x\right) }{q\left( x\right) } \) where \( p \) and \( q \) have no common zeros. \( {}^{b} \) Let \( c \) be a real number which is not in the domain of... | - If \( q\left( c\right) \neq 0 \), then the graph of \( y = r\left( x\right) \) has a hole at \( \left( {c,\frac{p\left( c\right) }{q\left( c\right) }}\right) \).\n\n- If \( q\left( c\right) = 0 \), then the line \( x = c \) is a vertical asymptote of the graph of \( y = r\left( x\right) \) . | Yes |
Find the vertical asymptotes of, and/or holes in, the graphs of the following rational functions. Verify your answers using a graphing calculator, and describe the behavior of the graph near them using proper notation. | 1. To use Theorem 4.1, we first find all of the real numbers which aren’t in the domain of \( f \) . To do so, we solve \( {x}^{2} - 3 = 0 \) and get \( x = \pm \sqrt{3} \) . Since the expression \( f\left( x\right) \) is in lowest terms, there is no cancellation possible, and we conclude that the lines \( x = - \sqrt{... | Yes |
1. Find and interpret \( P\left( 0\right) \) . | Substituting \( t = 0 \) gives \( P\left( 0\right) = \frac{100}{{\left( 5 - 0\right) }^{2}} = 4 \), which means 4000 bacteria are initially introduced into the environment. | Yes |
List the horizontal asymptotes, if any, of the graphs of the following functions. Verify your answers using a graphing calculator, and describe the behavior of the graph near them using proper notation. | 1. The numerator of \( f\left( x\right) \) is \( {5x} \), which has degree 1 . The denominator of \( f\left( x\right) \) is \( {x}^{2} + 1 \), which has degree 2. Applying Theorem 4.2, \( y = 0 \) is the horizontal asymptote. Sure enough, we see from the graph that as \( x \rightarrow - \infty, f\left( x\right) \righta... | Yes |
Example 4.1.5. The number of students \( N \) at local college who have had the flu \( t \) months after the semester begins can be modeled by the formula \( N\left( t\right) = {500} - \frac{450}{1 + {3t}} \) for \( t \geq 0 \). | ## Solution.\n\n1. \( N\left( 0\right) = {500} - \frac{450}{1 + 3\left( 0\right) } = {50} \) . This means that at the beginning of the semester,50 students\n\nhave had the flu.\n\n2. We set \( N\left( t\right) = {300} \) to get \( {500} - \frac{450}{1 + {3t}} = {300} \) and solve. Isolating the fraction gives \( \frac{... | Yes |
Theorem 4.3. Determination of Slant Asymptotes: Suppose \( r \) is a rational function and \( r\left( x\right) = \frac{p\left( x\right) }{q\left( x\right) } \), where the degree of \( p \) is exactly one more than the degree of \( q \) . Then the graph of \( y = r\left( x\right) \) has the slant asymptote \( y = L\left... | In the same way that Theorem 4.2 gives us an easy way to see if the graph of a rational function \( r\left( x\right) = \frac{p\left( x\right) }{q\left( x\right) } \) has a horizontal asymptote by comparing the degrees of the numerator and denominator, Theorem 4.3 gives us an easy way to check for slant asymptotes. Unli... | Yes |
Find the slant asymptotes of the graphs of the following functions if they exist. Verify your answers using a graphing calculator and describe the behavior of the graph near them using proper notation. | 1. The degree of the numerator is 2 and the degree of the denominator is 1 , so Theorem 4.3 guarantees us a slant asymptote. To find it, we divide \( 1 - x = - x + 1 \) into \( {x}^{2} - {4x} + 2 \) and get a quotient of \( - x + 3 \), so our slant asymptote is \( y = - x + 3 \) . We confirm this graphically, and we se... | Yes |
Sketch the graph of \( r\left( x\right) = \frac{{x}^{4} + 1}{{x}^{2} + 1} \) . | Solution.\n\n1. The denominator \( {x}^{2} + 1 \) is never zero so the domain is \( \left( {-\infty ,\infty }\right) \).\n\n2. With no real zeros in the denominator, \( {x}^{2} + 1 \) is an irreducible quadratic. Our only hope of reducing \( r\left( x\right) \) is if \( {x}^{2} + 1 \) is a factor of \( {x}^{4} + 1 \) .... | No |
1. Solve \( \frac{{x}^{3} - {2x} + 1}{x - 1} = \frac{1}{2}x - 1 \) . | To solve the equation, we clear denominators\n\n\[ \frac{{x}^{3} - {2x} + 1}{x - 1} = \frac{1}{2}x - 1 \]\n\n\[ \left( \frac{{x}^{3} - {2x} + 1}{x - 1}\right) \cdot 2\left( {x - 1}\right) = \left( {\frac{1}{2}x - 1}\right) \cdot 2\left( {x - 1}\right) \]\n\n\[ 2{x}^{3} - {4x} + 2 = {x}^{2} - {3x} + 2 \]\n\n\[ 2{x}^{3} ... | Yes |
Working alone, Taylor can weed the garden in 4 hours. If Carl helps, they can weed the garden in 3 hours. How long would it take for Carl to weed the garden on his own? | Solution. The key relationship between work and time which we use in this problem is:\namount of work done \( = \) rate of work \( \cdot \) time spent working\n\nWe are told that, working alone, Taylor can weed the garden in 4 hours. In Taylor's case then:\namount of work Taylor does \( = \) rate of Taylor working \( \... | Yes |
1. Find an expression for the average cost function \( \bar{C}\left( x\right) \) . | 1. From \( \bar{C}\left( x\right) = \frac{C\left( x\right) }{x} \), we obtain \( \bar{C}\left( x\right) = \frac{{80x} + {150}}{x} \). The domain of \( C \) is \( x \geq 0 \), but since \( x = 0 \) causes problems for \( \bar{C}\left( x\right) \), we get our domain to be \( x > 0 \), or \( \left( {0,\infty }\right) \) . | Yes |
1. Express the height \( h \) in centimeters as a function of the width \( x \) and state the applied domain. | 1. We are told that the volume of the box is 1000 cubic centimeters and that \( x \) represents the width, in centimeters. From geometry, we know Volume \( = \) width \( \times \) height \( \times \) depth. Since the base of the box is a square, the width and the depth are both \( x \) centimeters. Using \( h \) for th... | Yes |
1. Hooke’s Law: The force \( F \) exerted on a spring is directly proportional the extension \( x \) of the spring. | 1. Applying the definition of direct variation, we get \( F = {kx} \) for some constant \( k \) . | Yes |
1. Use your calculator to generate a scatter diagram for these data using \( V \) as the independent variable and \( P \) as the dependent variable. Does it appear from the graph that \( P \) is inversely proportional to \( V \) ? Explain. | 1. If \( P \) really does vary inversely with \( V \), then \( P = \frac{k}{V} \) for some constant \( k \) . From the data plot, the points do seem to lie along a curve like \( y = \frac{k}{x} \) . | No |
1. \( \left( {g \circ f}\right) \left( 1\right) \) | Using Definition 5.1, \( \left( {g \circ f}\right) \left( 1\right) = g\left( {f\left( 1\right) }\right) \) . We find \( f\left( 1\right) = - 3 \), so\n\n\[ \left( {g \circ f}\right) \left( 1\right) = g\left( {f\left( 1\right) }\right) = g\left( {-3}\right) = 2 \] | Yes |
Theorem 5.1. Properties of Function Composition: Suppose \( f, g \), and \( h \) are functions.\n\n- \( h \circ \left( {g \circ f}\right) = \left( {h \circ g}\right) \circ f \), provided the composite functions are defined.\n\n- If \( I \) is defined as \( I\left( x\right) = x \) for all real numbers \( x \), then \( I... | By repeated applications of Definition 5.1, we find \( \left( {h \circ \left( {g \circ f}\right) }\right) \left( x\right) = h\left( {\left( {g \circ f}\right) \left( x\right) }\right) = h\left( {g\left( {f\left( x\right) }\right) }\right) \) . Similarly, \( \left( {\left( {h \circ g}\right) \circ f}\right) \left( x\rig... | No |
The surface area \( S \) of a sphere is a function of its radius \( r \) and is given by the formula \( S\left( r\right) = {4\pi }{r}^{2} \) . Suppose the sphere is being inflated so that the radius of the sphere is increasing according to the formula \( r\left( t\right) = 3{t}^{2} \), where \( t \) is measured in seco... | Solution. If we look at the functions \( S\left( r\right) \) and \( r\left( t\right) \) individually, we see the former gives the surface area of a sphere of a given radius while the latter gives the radius at a given time. So, given a specific time, \( t \), we could find the radius at that time, \( r\left( t\right) \... | Yes |
1. \( F\left( x\right) = \left| {{3x} - 1}\right| \) | Our goal is to express the function \( F \) as \( F = g \circ f \) for functions \( g \) and \( f \) . From Definition 5.1, we know \( F\left( x\right) = g\left( {f\left( x\right) }\right) \), and we can think of \( f\left( x\right) \) as being the ’inside’ function and \( g \) as being the ’outside’ function. Looking ... | Yes |
Theorem 5.2. Properties of Inverse Functions: Suppose \( f \) and \( g \) are inverse functions.\n\n- The range \( {}^{a} \) of \( f \) is the domain of \( g \) and the domain of \( f \) is the range of \( g \)\n\n- \( f\left( a\right) = b \) if and only if \( g\left( b\right) = a \)\n\n- \( \left( {a, b}\right) \) is ... | Theorem 5.2 is a consequence of Definition 5.2 and the Fundamental Graphing Principle for Functions. We note the third property in Theorem 5.2 tells us that the graphs of inverse functions are reflections about the line \( y = x \) . For a proof of this, see Example 1.1.7 in Section 1.1 and Exercise 72 in Section 2.1. | No |
Determine if the following functions are one-to-one in two ways: (a) analytically using Definition 5.3 and (b) graphically using the Horizontal Line Test.\n\n1. \( f\left( x\right) = \frac{1 - {2x}}{5} \) 2. \( g\left( x\right) = \frac{2x}{1 - x} \)\n\n3. \( h\left( x\right) = {x}^{2} - {2x} + 4 \) 4. \( F = \{ \left( ... | 1. (a) To determine if \( f \) is one-to-one analytically, we assume \( f\left( c\right) = f\left( d\right) \) and attempt to deduce that \( c = d \) .\n\n\[ f\left( c\right) = f\left( d\right) \]\n\n\[ \frac{1 - {2c}}{5} = \frac{1 - {2d}}{5} \]\n\n\[ 1 - {2c} = 1 - {2d} \]\n\n\[ - {2c} = - {2d} \]\n\n\[ c = d\checkmar... | Yes |
Find the inverse of the following one-to-one functions. Check your answers analytically using function composition and graphically.\n\n1. \( f\left( x\right) = \frac{1 - {2x}}{5} \) | Solution.\n\n1. As we mentioned earlier, it is possible to think our way through the inverse of \( f \) by recording the steps we apply to \( x \) and the order in which we apply them and then reversing those steps in the reverse order. We encourage the reader to do this. We, on the other hand, will practice the algori... | Yes |
1. \( j\left( x\right) = {x}^{2} - {2x} + 4, x \leq 1 \) | The function \( j \) is a restriction of the function \( h \) from Example 5.2.1. Since the domain of \( j \) is restricted to \( x \leq 1 \), we are selecting only the ’left half’ of the parabola. We see that the graph of \( j \) passes the Horizontal Line Test and thus \( j \) is invertible.\n\n![aac1f2b5-88f4-48cc-b... | Yes |
1. Explain why \( p \) is one-to-one and find a formula for \( {p}^{-1}\left( x\right) \). State the restricted domain. | 1. We leave to the reader to show the graph of \( p\left( x\right) = - {1.5x} + {250},0 \leq x \leq {166} \), is a line segment from \( \left( {0,{250}}\right) \) to \( \left( {{166},1}\right) \), and as such passes the Horizontal Line Test. Hence, \( p \) is one-to-one. We find the expression for \( {p}^{-1}\left( x\r... | No |
Theorem 5.6. Properties of Radicals: Let \( x \) and \( y \) be real numbers and \( m \) and \( n \) be natural numbers. If \( \sqrt[n]{x},\sqrt[n]{y} \) are real numbers, then\n\n- Product Rule: \( \sqrt[n]{xy} = \sqrt[n]{x}\sqrt[n]{y} \)\n\n- Powers of Radicals: \( \sqrt[n]{{x}^{m}} = {\left( \sqrt[n]{x}\right) }^{m}... | The proof of Theorem 5.6 is based on the definition of the principal roots and properties of exponents. To establish the product rule, consider the following. If \( n \) is odd, then by definition \( \sqrt[n]{xy} \) is the unique real number such that \( {\left( \sqrt[n]{xy}\right) }^{n} = {xy} \) . Given that \( {\lef... | No |
For the following functions, state their domains and create sign diagrams. Check your answer graphically using your calculator. | 1. As far as domain is concerned, \( f\left( x\right) \) has no denominators and no even roots, which means its domain is \( \left( {-\infty ,\infty }\right) \) . To create the sign diagram, we find the zeros of \( f \) .\n\n\[ f\left( x\right) = 0 \]\n\n\[ {3x}\sqrt[3]{2 - x} = 0 \]\n\n\[ {3x} = 0\text{ or }\sqrt[3]{2... | Yes |
1. Express the total cost \( C \) of connecting the Junction Box to the Outpost as a function of \( x \) , the number of miles the cable is run along Route 117 before heading off road directly towards the Outpost. Determine a reasonable applied domain for the problem. | 1. The cost is broken into two parts: the cost to run cable along Route 117 at \( \$ {15} \) per mile, and the cost to run it off road at \( \$ {20} \) per mile. Since \( x \) represents the miles of cable run along Route 117, the cost for that portion is \( {15x} \) . From the diagram, we see that the number of miles ... | Yes |
1. Find and interpret \( V\left( 0\right) \) . | To find \( V\left( 0\right) \), we replace \( x \) with 0 to obtain \( V\left( 0\right) = {25}{\left( \frac{4}{5}\right) }^{0} = {25} \) . Since \( x \) represents the age of the car in years, \( x = 0 \) corresponds to the car being brand new. Since \( V\left( x\right) \) is measured in thousands of dollars, \( V\left... | Yes |
1. Find and interpret \( T\left( 0\right) \) . | To find \( T\left( 0\right) \), we replace every occurrence of the independent variable \( t \) with 0 to obtain \( T\left( 0\right) = {70} + {90}{e}^{-{0.1}\left( 0\right) } = {160} \) . This means that the coffee was served at \( {160}^{ \circ }\mathrm{F} \) . | Yes |
Theorem 6.2. Properties of Logarithmic Functions: Suppose \( f\left( x\right) = {\log }_{b}\left( x\right) \) . | - The domain of \( f \) is \( \left( {0,\infty }\right) \) and the range of \( f \) is \( \left( {-\infty ,\infty }\right) \) .\n\n- \( \left( {1,0}\right) \) is on the graph of \( f \) and \( x = 0 \) is a vertical asymptote of the graph of \( f \) .\n\n- \( f \) is one-to-one, continuous and smooth\n\n- \( {b}^{a} = ... | Yes |
Simplify the following.\n\n1. \( {\log }_{3}\left( {81}\right) \) 2. \( {\log }_{2}\left( \frac{1}{8}\right) \) 3. \( {\log }_{\sqrt{5}}\left( {25}\right) \) 4. \( \ln \left( \sqrt[3]{{e}^{2}}\right) \)\n\n5. \( \log \left( {0.001}\right) \) 6. \( {2}^{{\log }_{2}\left( 8\right) } \) 7. \( {117}^{-{\log }_{117}\left( 6... | Solution.\n\n1. The number \( {\log }_{3}\left( {81}\right) \) is the exponent we put on 3 to get 81 . As such, we want to write 81 as a power of 3 . We find \( {81} = {3}^{4} \), so that \( {\log }_{3}\left( {81}\right) = 4 \) .\n\n2. To find \( {\log }_{2}\left( \frac{1}{8}\right) \), we need rewrite \( \frac{1}{8} \... | Yes |
Find the domain of the following functions. Check your answers graphically using the calculator.\n\n1. \( f\left( x\right) = 2\log \left( {3 - x}\right) - 1 \) | ## Solution.\n\n1. We set \( 3 - x > 0 \) to obtain \( x < 3 \), or \( \left( {-\infty ,3}\right) \) . The graph from the calculator below verifies this. Note that we could have graphed \( f \) using transformations. Taking a cue from Theorem 1.7, we rewrite \( f\left( x\right) = 2{\log }_{10}\left( {-x + 3}\right) - 1... | Yes |
1. Graph \( f \) using transformations and state the domain and range of \( f \) . | If we identify \( g\left( x\right) = {2}^{x} \), we see \( f\left( x\right) = g\left( {x - 1}\right) - 3 \) . We pick the points \( \left( {-1,\frac{1}{2}}\right) ,\left( {0,1}\right) \) and \( \left( {1,2}\right) \) on the graph of \( g \) along with the horizontal asymptote \( y = 0 \) to track through the transforma... | Yes |
Theorem 6.6. (Algebraic Properties of Logarithmic Functions) Let \( g\left( x\right) = {\log }_{b}\left( x\right) \) be a logarithmic function \( \left( {b > 0, b \neq 1}\right) \) and let \( u > 0 \) and \( w > 0 \) be real numbers.\n\n- Product Rule: \( g\left( {uw}\right) = g\left( u\right) + g\left( w\right) \) . I... | Consider the product rule: \( {\log }_{b}\left( {uw}\right) = {\log }_{b}\left( u\right) + {\log }_{b}\left( w\right) \) . Let \( a = {\log }_{b}\left( {uw}\right), c = {\log }_{b}\left( u\right) \), and \( d = {\log }_{b}\left( w\right) \) . Then, by definition, \( {b}^{a} = {uw},{b}^{c} = u \) and \( {b}^{d} = w \) .... | Yes |
Expand the following using the properties of logarithms and simplify. Assume when necessary that all quantities represent positive real numbers.\n\n1. \( {\log }_{2}\left( \frac{8}{x}\right) \) | To expand \( {\log }_{2}\left( \frac{8}{x}\right) \), we use the Quotient Rule identifying \( u = 8 \) and \( w = x \) and simplify.\n\n\[ {\log }_{2}\left( \frac{8}{x}\right) = {\log }_{2}\left( 8\right) - {\log }_{2}\left( x\right) \;\text{ Quotient Rule } \]\n\n\[ = 3 - {\log }_{2}\left( x\right) \;\text{Since}{2}^{... | Yes |
Use the properties of logarithms to write the following as a single logarithm. | 1. The difference of logarithms requires the Quotient Rule: \( {\log }_{3}\left( {x - 1}\right) - {\log }_{3}\left( {x + 1}\right) = {\log }_{3}\left( \frac{x - 1}{x + 1}\right) \) . | Yes |
Theorem 6.7. (Change of Base Formulas) Let \( a, b > 0, a, b \neq 1 \) .\n\n- \( {a}^{x} = {b}^{x{\log }_{b}\left( a\right) } \) for all real numbers \( x \) .\n\n- \( {\log }_{a}\left( x\right) = \frac{{\log }_{b}\left( x\right) }{{\log }_{b}\left( a\right) } \) for all real numbers \( x > 0 \) . | The proofs of the Change of Base formulas are a result of the other properties studied in this section. If we start with \( {b}^{x{\log }_{b}\left( a\right) } \) and use the Power Rule in the exponent to rewrite \( x{\log }_{b}\left( a\right) \) as \( {\log }_{b}\left( {a}^{x}\right) \) and then apply one of the Invers... | Yes |
Example 6.2.3. Use an appropriate change of base formula to convert the following expressions to ones with the indicated base. Verify your answers using a calculator, as appropriate.\n\n1. \( {3}^{2} \) to base 10 2. \( {2}^{x} \) to base \( e \) | ## Solution.\n\n1. We apply the Change of Base formula with \( a = 3 \) and \( b = {10} \) to obtain \( {3}^{2} = {10}^{2\log \left( 3\right) } \) . Typing the latter in the calculator produces an answer of 9 as required.\n\n2. Here, \( a = 2 \) and \( b = e \) so we have \( {2}^{x} = {e}^{x\ln \left( 2\right) } \) . T... | Yes |
1. \( {2}^{3x} = {16}^{1 - x} \) | Since 16 is a power of 2, we can rewrite \( {2}^{3x} = {16}^{1 - x} \) as \( {2}^{3x} = {\left( {2}^{4}\right) }^{1 - x} \) . Using properties of exponents, we get \( {2}^{3x} = {2}^{4\left( {1 - x}\right) } \) . Using the one-to-one property of exponential functions, we get \( {3x} = 4\left( {1 - x}\right) \) which gi... | Yes |
1. \( {2}^{{x}^{2} - {3x}} - {16} \geq 0 \) | Since we already have 0 on one side of the inequality, we set \( r\left( x\right) = {2}^{{x}^{2} - {3x}} - {16} \) . The domain of \( r \) is all real numbers, so in order to construct our sign diagram, we need to find the zeros of \( r \) . Setting \( r\left( x\right) = 0 \) gives \( {2}^{{x}^{2} - {3x}} - {16} = 0 \)... | Yes |
Recall from Example 6.1.2 that the temperature of coffee \( T \) (in degrees Fahrenheit) \( t \) minutes after it is served can be modeled by \( T\left( t\right) = {70} + {90}{e}^{-{0.1t}} \) . When will the coffee be warmer than \( {100}^{ \circ }\mathrm{F} \) ? | Solution. We need to find when \( T\left( t\right) > {100} \), or in other words, we need to solve the inequality \( {70} + {90}{e}^{-{0.1t}} > {100} \) . Getting 0 on one side of the inequality, we have \( {90}{e}^{-{0.1t}} - {30} > 0 \), and we set \( r\left( t\right) = {90}{e}^{-{0.1t}} - {30} \) . The domain of \( ... | Yes |
The function \( f\left( x\right) = \frac{5{e}^{x}}{{e}^{x} + 1} \) is one-to-one. Find a formula for \( {f}^{-1}\left( x\right) \) and check your answer graphically using your calculator. | Solution. We start by writing \( y = f\left( x\right) \), and interchange the roles of \( x \) and \( y \) . To solve for \( y \), we first clear denominators and then isolate the exponential function.\n\n\[ y = \frac{5{e}^{x}}{{e}^{x} + 1} \]\n\n\[ x = \frac{5{e}^{y}}{{e}^{y} + 1}\;\text{ Switch }x\text{ and }y \]\n\n... | No |
1. \( {\log }_{117}\left( {1 - {3x}}\right) = {\log }_{117}\left( {{x}^{2} - 3}\right) \) | Since we have the same base on both sides of the equation \( {\log }_{117}\left( {1 - {3x}}\right) = {\log }_{117}\left( {{x}^{2} - 3}\right) \) , we equate what’s inside the logs to get \( 1 - {3x} = {x}^{2} - 3 \) . Solving \( {x}^{2} + {3x} - 4 = 0 \) gives \( x = - 4 \) and \( x = 1 \) . To check these answers usin... | Yes |
1. \( \frac{1}{\ln \left( x\right) + 1} \leq 1 \) | We start solving \( \frac{1}{\ln \left( x\right) + 1} \leq 1 \) by getting 0 on one side of the inequality: \( \frac{1}{\ln \left( x\right) + 1} - 1 \leq 0 \) . Getting a common denominator yields \( \frac{1}{\ln \left( x\right) + 1} - \frac{\ln \left( x\right) + 1}{\ln \left( x\right) + 1} \leq 0 \) which reduces to \... | Yes |
In order to successfully breed Ippizuti fish the \( \mathrm{{pH}} \) of a freshwater tank must be at least 7.8 but can be no more than 8.5. Determine the corresponding range of hydrogen ion concentration, and check your answer using a calculator. | Solution. Recall from Exercise 77 in Section 6.1 that \( \mathrm{{pH}} = - \log \left\lbrack {\mathrm{H}}^{ + }\right\rbrack \) where \( \left\lbrack {\mathrm{H}}^{ + }\right\rbrack \) is the hydrogen ion concentration in moles per liter. We require \( {7.8} \leq - \log \left\lbrack {\mathrm{H}}^{ + }\right\rbrack \leq... | Yes |
The function \( f\left( x\right) = \frac{\log \left( x\right) }{1 - \log \left( x\right) } \) is one-to-one. Find a formula for \( {f}^{-1}\left( x\right) \) and check your answer graphically using your calculator. | Solution. We first write \( y = f\left( x\right) \) then interchange the \( x \) and \( y \) and solve for \( y \) .\n\n\[ y = f\left( x\right) \]\n\n\[ y = \frac{\log \left( x\right) }{1 - \log \left( x\right) } \]\n\n\[ x = \frac{\log \left( y\right) }{1 - \log \left( y\right) } \]\nInterchange \( x \) and \( y \) .\... | Yes |
1. Express the amount \( A \) in the account as a function of the term of the investment \( t \) in years. | 1. Substituting \( P = {2000}, r = {0.07125} \), and \( n = {12} \) (since interest is compounded monthly) into Equation 6.2 yields \( A\left( t\right) = {2000}{\left( 1 + \frac{0.07125}{12}\right) }^{12t} = {2000}{\left( {1.0059375}\right) }^{12t} \) . | Yes |
In order to perform arthrosclerosis research, epithelial cells are harvested from discarded umbilical tissue and grown in the laboratory. A technician observes that a culture of twelve thousand cells grows to five million cells in one week. Assuming that the cells follow The Law of Uninhibited Growth, find a formula fo... | Solution. We begin with \( N\left( t\right) = {N}_{0}{e}^{kt} \) . Since \( N \) is to give the number of cells in thousands, we have \( {N}_{0} = {12} \), so \( N\left( t\right) = {12}{e}^{kt} \) . In order to complete the formula, we need to determine the growth rate \( k \) . We know that after one week, the number ... | Yes |
Iodine-131 is a commonly used radioactive isotope used to help detect how well the thyroid is functioning. Suppose the decay of Iodine-131 follows the model given in Equation 6.5, and that the half-life \( {}^{10} \) of Iodine-131 is approximately 8 days. If 5 grams of Iodine-131 is present initially, find a function w... | Solution. Since we start with 5 grams initially, Equation 6.5 gives \( A\left( t\right) = 5{e}^{kt} \) . Since the half-life is 8 days, it takes 8 days for half of the Iodine-131 to decay, leaving half of it behind. Hence, \( A\left( 8\right) = {2.5} \) which means \( 5{e}^{8k} = {2.5} \) . Solving, we get \( k = \frac... | Yes |
1. Assuming the temperature of the roast follows Newton's Law of Warming, find a formula for the temperature of the roast \( T \) as a function of its time in the oven, \( t \), in hours. | 1. The initial temperature of the roast is \( {40}^{ \circ }\mathrm{F} \), so \( {T}_{0} = {40} \) . The environment in which we are placing the roast is the \( {350}^{ \circ }\mathrm{F} \) oven, so \( {T}_{a} = {350} \) . Newton’s Law of Warming tells us \( T\left( t\right) = {350} + \left( {{40} - {350}}\right) {e}^{... | Yes |
The number of people \( N \), in hundreds, at a local community college who have heard the rumor 'Carl is afraid of Virginia Woolf' can be modeled using the logistic equation\n\n\[ N\left( t\right) = \frac{84}{1 + {2799}{e}^{-t}} \]\n\nwhere \( t \geq 0 \) is the number of days after April 1,2009.\n\n1. Find and interp... | ## Solution.\n\n1. We find \( N\left( 0\right) = \frac{84}{1 + {2799}{e}^{0}} = \frac{84}{2800} = \frac{3}{100} \). Since \( N\left( t\right) \) measures the number of people who have heard the rumor in hundreds, \( N\left( 0\right) \) corresponds to 3 people. Since \( t = 0 \) corresponds to April 1,2009, we may concl... | Yes |
1. If a 7 character case-sensitive \( {}^{16} \) password is comprised of letters and numbers only, find the associated information entropy. | 1. There are 26 letters in the alphabet, 52 if upper and lower case letters are counted as different. There are 10 digits (0 through 9) for a total of \( N = {62} \) symbols. Since the password is to be 7 characters long, \( L = 7 \) . Thus, \( H = 7{\log }_{2}\left( {62}\right) = \frac{7\ln \left( {62}\right) }{\ln \l... | Yes |
Find the partial pressure of carbon dioxide in arterial blood if the \( \mathrm{{pH}} \) is 7.4. | Solution. We set \( \mathrm{{pH}} = {7.4} \) and get \( {7.4} = {6.1} + \log \left( \frac{800}{x}\right) \), or \( \log \left( \frac{800}{x}\right) = {1.3} \) . Solving, we find \( x = \frac{800}{{10}^{1 \cdot 3}} \approx {40.09} \) . Hence, the partial pressure of carbon dioxide in the blood is about 40 torr. | Yes |
Write the standard equation of the circle with center \( \left( {-2,3}\right) \) and radius 5. | Here, \( \left( {h, k}\right) = \left( {-2,3}\right) \) and \( r = 5 \), so we get\n\n\[ \n{\left( x - \left( -2\right) \right) }^{2} + {\left( y - 3\right) }^{2} = {\left( 5\right) }^{2} \n\]\n\n\[ \n{\left( x + 2\right) }^{2} + {\left( y - 3\right) }^{2} = {25} \n\] | Yes |
Graph \( {\left( x + 2\right) }^{2} + {\left( y - 1\right) }^{2} = 4 \) . Find the center and radius. | Solution. From the standard form of a circle, Equation 7.1, we have that \( x + 2 \) is \( x - h \), so \( h = - 2 \) and \( y - 1 \) is \( y - k \) so \( k = 1 \) . This tells us that our center is \( \left( {-2,1}\right) \) . Furthermore, \( {r}^{2} = 4 \), so \( r = 2 \) . Thus we have a circle centered at \( \left(... | Yes |
Complete the square to find the center and radius of \( 3{x}^{2} - {6x} + 3{y}^{2} + {4y} - 4 = 0 \) . | \[ 3{x}^{2} - {6x} + 3{y}^{2} + {4y} - 4 = 0 \] \[ 3{x}^{2} - {6x} + 3{y}^{2} + {4y} = 4 \] add 4 to both sides \[ 3\left( {{x}^{2} - {2x}}\right) + 3\left( {{y}^{2} + \frac{4}{3}y}\right) = 4 \] factor out leading coefficients \[ 3\left( {{x}^{2} - {2x} + \underline{1}}\right) + 3\left( {{y}^{2} + \frac{4}{3}y + \unde... | Yes |
Write the standard equation of the circle which has \( \left( {-1,3}\right) \) and \( \left( {2,4}\right) \) as the endpoints of a diameter. | Solution. We recall that a diameter of a circle is a line segment containing the center and two points on the circle. Plotting the given data yields\n\n\n\nSince the given points are endpoints of a diameter, we know ... | Yes |
Find the points on the unit circle with \( y \) -coordinate \( \frac{\sqrt{3}}{2} \) . | Solution. We replace \( y \) with \( \frac{\sqrt{3}}{2} \) in the equation \( {x}^{2} + {y}^{2} = 1 \) to get\n\n\[ \n{x}^{2} + {y}^{2} = 1 \n\] \n\n\[ \n{x}^{2} + {\left( \frac{\sqrt{3}}{2}\right) }^{2} = 1 \n\] \n\n\[ \n\frac{3}{4} + {x}^{2} = 1 \n\] \n\n\[ \n{x}^{2} = \frac{1}{4} \n\] \n\n\[ \nx = \pm \sqrt{\frac{1}... | Yes |
Graph \( {\left( x + 1\right) }^{2} = - 8\left( {y - 3}\right) \) . Find the vertex, focus, and directrix. | Solution. We recognize this as the form given in Equation 7.2. Here, \( x - h \) is \( x + 1 \) so \( h = - 1 \) , and \( y - k \) is \( y - 3 \) so \( k = 3 \) . Hence, the vertex is \( \left( {-1,3}\right) \) . We also see that \( {4p} = - 8 \) so \( p = - 2 \) . Since \( p < 0 \), the focus will be below the vertex ... | Yes |
Find the standard form of the parabola with focus \( \\left( {2,1}\\right) \) and directrix \( y = - 4 \) . | Solution. Sketching the data yields,\n\n\n\n---\n\n\\({}^{2}\\) No, I’m not making this up.\n\n\\({}^{3}\\) Consider this an exercise to show what follows.\n\n---\n\n\n\nFrom the diagram, we see the parabola opens up... | Yes |
Graph \( {\left( y - 2\right) }^{2} = {12}\left( {x + 1}\right) \) . Find the vertex, focus, and directrix. | Solution. We recognize this as the form given in Equation 7.3. Here, \( x - h \) is \( x + 1 \) so \( h = - 1 \) , and \( y - k \) is \( y - 2 \) so \( k = 2 \) . Hence, the vertex is \( \left( {-1,2}\right) \) . We also see that \( {4p} = {12} \) so \( p = 3 \) . Since \( p > 0 \), the focus will be the right of the v... | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.