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In any right triangle, whereaandbare the lengths of the legs, andcis the length of the hypotenuse,a2+b2=c2.
https://openstax.org/books/intermediate-algebra-2e/pages/9-key-concepts
Projectile motionThe height in feet,h, of an object shot upwards into the air with initial velocity,v0, aftertseconds is given by the formulah= −16t2+v0t.
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The height in feet,h, of an object shot upwards into the air with initial velocity,v0, aftertseconds is given by the formulah= −16t2+v0t.
https://openstax.org/books/intermediate-algebra-2e/pages/9-key-concepts
Parabola OrientationFor the graph of the quadratic functionf(x)=ax2+bx+c,f(x)=ax2+bx+c,ifa> 0, the parabola opens upward.a< 0, the parabola opens downward.
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For the graph of the quadratic functionf(x)=ax2+bx+c,f(x)=ax2+bx+c,ifa> 0, the parabola opens upward.a< 0, the parabola opens downward.
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a> 0, the parabola opens upward.
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a< 0, the parabola opens downward.
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Axis of Symmetry and Vertex of a Parabola The graph of the functionf(x)=ax2+bx+cf(x)=ax2+bx+cis a parabola where:the axis of symmetry is the vertical linex=−b2a.x=−b2a.the vertex is a point on the axis of symmetry, so itsx-coordinate is−b2a.−b2a.they-coordinate of the vertex is found by substitutingx=−b2ax=âˆ...
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the axis of symmetry is the vertical linex=−b2a.x=−b2a.
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the vertex is a point on the axis of symmetry, so itsx-coordinate is−b2a.−b2a.
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they-coordinate of the vertex is found by substitutingx=−b2ax=−b2ainto the quadratic equation.
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Find the Intercepts of a ParabolaTo find the intercepts of a parabola whose function isf(x)=ax2+bx+c:f(x)=ax2+bx+c:y-interceptx-interceptsLetx=0and solve forf(x).Letf(x)=0and solve forx.y-interceptx-interceptsLetx=0and solve forf(x).Letf(x)=0and solve forx.
https://openstax.org/books/intermediate-algebra-2e/pages/9-key-concepts
To find the intercepts of a parabola whose function isf(x)=ax2+bx+c:f(x)=ax2+bx+c:y-interceptx-interceptsLetx=0and solve forf(x).Letf(x)=0and solve forx.y-interceptx-interceptsLetx=0and solve forf(x).Letf(x)=0and solve forx.
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How to graph a quadratic function using properties.Step 1.Determine whether the parabola opens upward or downward.Step 2.Find the equation of the axis of symmetry.Step 3.Find the vertex.Step 4.Find they-intercept. Find the point symmetric to they-intercept across the axis of symmetry.Step 5.Find thex-intercepts. Find a...
https://openstax.org/books/intermediate-algebra-2e/pages/9-key-concepts
Step 1.Determine whether the parabola opens upward or downward.
https://openstax.org/books/intermediate-algebra-2e/pages/9-key-concepts
Step 2.Find the equation of the axis of symmetry.
https://openstax.org/books/intermediate-algebra-2e/pages/9-key-concepts
Step 3.Find the vertex.
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Step 4.Find they-intercept. Find the point symmetric to they-intercept across the axis of symmetry.
https://openstax.org/books/intermediate-algebra-2e/pages/9-key-concepts
Step 5.Find thex-intercepts. Find additional points if needed.
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Step 6.Graph the parabola.
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Minimum or Maximum Values of a Quadratic EquationThey-coordinate of the vertex of the graph of a quadratic equation is theminimumvalue of the quadratic equation if the parabola opensupward.maximumvalue of the quadratic equation if the parabola opensdownward.
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They-coordinate of the vertex of the graph of a quadratic equation is the
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minimumvalue of the quadratic equation if the parabola opensupward.
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maximumvalue of the quadratic equation if the parabola opensdownward.
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Graph a Quadratic Function of the formf(x)=x2+kf(x)=x2+kUsing a Vertical ShiftThe graph off(x)=x2+kf(x)=x2+kshifts the graph off(x)=x2f(x)=x2vertically k units.Ifk> 0, shift the parabola vertically upkunits.Ifk< 0, shift the parabola vertically down|k||k|units.
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The graph off(x)=x2+kf(x)=x2+kshifts the graph off(x)=x2f(x)=x2vertically k units.Ifk> 0, shift the parabola vertically upkunits.Ifk< 0, shift the parabola vertically down|k||k|units.
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Ifk> 0, shift the parabola vertically upkunits.
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Ifk< 0, shift the parabola vertically down|k||k|units.
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Graph a Quadratic Function of the formf(x)=(x−h)2f(x)=(x−h)2Using a Horizontal ShiftThe graph off(x)=(x−h)2f(x)=(x−h)2shifts the graph off(x)=x2f(x)=x2horizontally h units.Ifh> 0, shift the parabola horizontally lefthunits.Ifh< 0, shift the parabola horizontally right|h||h|units.
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The graph off(x)=(x−h)2f(x)=(x−h)2shifts the graph off(x)=x2f(x)=x2horizontally h units.Ifh> 0, shift the parabola horizontally lefthunits.Ifh< 0, shift the parabola horizontally right|h||h|units.
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Ifh> 0, shift the parabola horizontally lefthunits.
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Ifh< 0, shift the parabola horizontally right|h||h|units.
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Graph of a Quadratic Function of the formf(x)=ax2f(x)=ax2The coefficientain the functionf(x)=ax2f(x)=ax2affects the graph off(x)=x2f(x)=x2by stretching or compressing it.If0<|a|<1,0<|a|<1,then the graph off(x)=ax2f(x)=ax2will be “wider” than the graph off(x)=x2.f(x)=x2.If|a|>1,|a|>1,then the graph off(x)=ax2f(x)=ax...
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The coefficientain the functionf(x)=ax2f(x)=ax2affects the graph off(x)=x2f(x)=x2by stretching or compressing it.If0<|a|<1,0<|a|<1,then the graph off(x)=ax2f(x)=ax2will be “wider” than the graph off(x)=x2.f(x)=x2.If|a|>1,|a|>1,then the graph off(x)=ax2f(x)=ax2will be “skinnier” than the graph off(x)=x2.f(x)=x2.
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How to graph a quadratic function using transformationsStep 1.Rewrite the function inf(x)=a(x−h)2+kf(x)=a(x−h)2+kform by completing the square.Step 2.Graph the function using transformations.
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Step 1.Rewrite the function inf(x)=a(x−h)2+kf(x)=a(x−h)2+kform by completing the square.
https://openstax.org/books/intermediate-algebra-2e/pages/9-key-concepts
Step 2.Graph the function using transformations.
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Graph a quadratic function in the vertex formf(x)=a(x−h)2+kf(x)=a(x−h)2+kusing propertiesStep 1.Rewrite the function inf(x)=a(x−h)2+kf(x)=a(x−h)2+kform.Step 2.Determine whether the parabola opens upward,a> 0, or downward, a < 0.Step 3.Find the axis of symmetry,x=h.Step 4.Find the vertex, (h,k).Step 5.Find they-...
https://openstax.org/books/intermediate-algebra-2e/pages/9-key-concepts
Step 1.Rewrite the function inf(x)=a(x−h)2+kf(x)=a(x−h)2+kform.
https://openstax.org/books/intermediate-algebra-2e/pages/9-key-concepts
Step 2.Determine whether the parabola opens upward,a> 0, or downward, a < 0.
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Step 3.Find the axis of symmetry,x=h.
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Step 4.Find the vertex, (h,k).
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Step 5.Find they-intercept. Find the point symmetric to they-intercept across the axis of symmetry.
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Step 6.Find thex-intercepts, if possible.
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Step 7.Graph the parabola.
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Solve a Quadratic Inequality GraphicallyStep 1.Write the quadratic inequality in standard form.Step 2.Graph the functionf(x)=ax2+bx+cf(x)=ax2+bx+cusing properties or transformations.Step 3.Determine the solution from the graph.
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Step 1.Write the quadratic inequality in standard form.
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Step 2.Graph the functionf(x)=ax2+bx+cf(x)=ax2+bx+cusing properties or transformations.
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Step 3.Determine the solution from the graph.
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How to Solve a Quadratic Inequality AlgebraicallyStep 1.Write the quadratic inequality in standard form.Step 2.Determine the critical points -- the solutions to the related quadratic equation.Step 3.Use the critical points to divide the number line into intervals.Step 4.Above the number line show the sign of each quadr...
https://openstax.org/books/intermediate-algebra-2e/pages/9-key-concepts
Step 1.Write the quadratic inequality in standard form.
https://openstax.org/books/intermediate-algebra-2e/pages/9-key-concepts
Step 2.Determine the critical points -- the solutions to the related quadratic equation.
https://openstax.org/books/intermediate-algebra-2e/pages/9-key-concepts
Step 3.Use the critical points to divide the number line into intervals.
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Step 4.Above the number line show the sign of each quadratic expression using test points from each interval substituted into the original inequality.
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Step 5.Determine the intervals where the inequality is correct. Write the solution in interval notation.
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discriminant : In the Quadratic Formula,x=−b±b2−4ac2a,x=−b±b2−4ac2a,the quantityb2− 4acis called the discriminant.
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quadratic function : A quadratic function, wherea,b, andcare real numbers andaâ‰0,aâ‰0,is a function of the formf(x)=ax2+bx+c.f(x)=ax2+bx+c.
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quadratic inequality : A quadratic inequality is an inequality that contains a quadratic expression.
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x x -axis, the line y = 0 y = 0
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x x -axis, the line y = 0 y = 0
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log a 1 = 0 log a 1 = 0
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ln 1 = 0 ln 1 = 0
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log a a = 1 log a a = 1
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ln e = 1 ln e = 1
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a log a x = x log a a x = x a log a x = x log a a x = x
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e ln x = x ln e x = x e ln x = x ln e x = x
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log a ( M · N ) = log a M + log a N log a ( M · N ) = log a M + log a N
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ln ( M · N ) = ln M + ln N ln ( M · N ) = ln M + ln N
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log a M N = log a M − log a N log a M N = log a M − log a N
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ln M N = ln M − ln N ln M N = ln M − ln N
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log a M p = p log a M log a M p = p log a M
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ln M p = p ln M ln M p = p ln M
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asymptote : A line which a graph of a function approaches closely but never touches.
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common logarithmic function : The functionf(x)=logxf(x)=logxis the common logarithmic function with base10,10,wherex>0.x>0.y=logxis equivalent tox=10yy=logxis equivalent tox=10y
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exponential function : An exponential function, wherea>0a>0andaâ‰1,aâ‰1,is a function of the formf(x)=ax.f(x)=ax.
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logarithmic function : The functionf(x)=logaxf(x)=logaxis the logarithmic function with basea,a,wherea>0,a>0,x>0,x>0,andaâ‰1.aâ‰1.y=logaxis equivalent tox=ayy=logaxis equivalent tox=ay
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natural base : The numbereis defined as the value of(1+1n)n,(1+1n)n,asngets larger and larger. We say, asnincreases without bound,e≈2.718281827...e≈2.718281827...
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natural exponential function : The natural exponential function is an exponential function whose base ise:f(x)=ex.f(x)=ex.The domain is(−∞,∞)(−∞,∞)and the range is(0,∞).(0,∞).
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natural logarithmic function : The functionf(x)=lnxf(x)=lnxis the natural logarithmic function with basee,e,wherex>0.x>0.y=lnxis equivalent tox=eyy=lnxis equivalent tox=ey
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one-to-one function : A function is one-to-one if each value in the range has exactly one element in the domain. For each ordered pair in the function, eachy-value is matched with only onex-value.
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x = − b 2 a x = − b 2 a
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x = h x = h
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Substitute x = − b 2 a x = − b 2 a and solve for y .
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Let x = 0 x = 0
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Let x = 0 x = 0
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Let y = 0 y = 0
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Let y = 0 y = 0
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y = − b 2 a y = − b 2 a
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y = k y = k
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Substitute y = − b 2 a y = − b 2 a and solve for x .
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Let x = 0 x = 0
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Let x = 0 x = 0
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Let y = 0 y = 0
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Let y = 0 y = 0
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y = b a x , y = b a x , y = − b a x y = − b a x
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y = a b x , y = a b x , y = − a b x y = − a b x
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Use a units left/right of center b units above/below the center
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Use a units above/below the center b units left/right of center
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x = 3 y 2 − 2 y + 1 x = 3 y 2 − 2 y + 1
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x 2 - x 2 - and y 2 - y 2 - terms must have the same coefficients and they must be the same sign as the constant after the = sign
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