Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 8 - Review Exercises - Page 575: 42

Answer

Vertex: $\left(2,3\right)$ Axis of Symmetry: $x=2$ Domain: set of all real numbers Range: $\{y|y\le3\}$ Graph of $f(x)=-2x^2+8x-5$

Work Step by Step

To find the properties of the given function, $ f(x)=-2x^2+8x-5 ,$ convert to the form $f(x)=a(x-h)^2+k$. Grouping the $x$-variables together and making the coefficient of $x^2$ equal to $1$, the given equation is equivalent to \begin{align*} f(x)&=(-2x^2+8x)-5 \\ f(x)&=-2(x^2-4x)-5 .\end{align*} Completing the square of the right-side expression by adding $\left(\dfrac{b}{2}\right)^2,$ the equation above is equivalent to \begin{align*} f(x)&=-2\left(x^2-4x+\left(\dfrac{-4}{2}\right)^2\right)+\left[-5-(-2)\left(\dfrac{-4}{2}\right)^2\right] \\\\ f(x)&=-2\left(x^2-4x+4\right)+\left[-5+8\right] \\ f(x)&=-2\left(x-2\right)^2+3 .\end{align*}(Note that $a\left(\dfrac{b}{2}\right)^2\Rightarrow (-2)\left(\dfrac{-4}{2}\right)^2 $ should be subtracted as well to cancel out the term that was added to complete the square.) Since the vertex of the function $f(x)=a(x-h)^2+k$ is given by $(h,k)$, then the vertex of the quadratic function, $ f(x)=-2\left(x-2\right)^2+3 $, is $ \left(2,3\right) $. The axis of symmetry of the function $f(x)=a(x-h)^2+k$ is given by $x=h$. With $h= 2 $ then the axis of symmetry is $ x=2 $. To graph the parabola, find points that are on the parabla. This can be done by substituting values of $x$ and solving the corresponding value of $y$. Let $y=f(x).$ Then $ y=-2x^2+8x-5 $. Substituting values of $x$ and solving $y$ results to \begin{array}{l|r} \text{If }x=0: & \text{If }x=1: \\\\ y=-2(0)^2+8(0)-5 & y=-2(1)^2+8(1)-5 \\ y=-2(0)+8(0)-5 & y=-2(1)+8(1)-5 \\ y=0+0-5 & y=-2+8-5 \\ y=-5 & y=1 .\end{array} Hence, the points $ (0,-5) $ and $ (1,1) $ are on the parabola. Reflecting these points about the axis of symmetry, the points $ (3,1) $ and $ (4,-5) $ are also on the parabola. Using the points $\{ (0,-5), (1,1), \left(2,3\right), (3,1), (4,-5) \}$ the graph of the parabola is determined (see graph above). Using the graph, the domain (values of $x$ used in the graph) is the set of all real numbers. The range (values of $y$ used in the graph) is $ \{y|y\le3\} $.
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