Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.2 - Quadratic Functions - Exercise Set - Page 332: 86

Answer

The graph is shown below.
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Work Step by Step

Let us consider the parabola defined by the provided quadratic equation $y=-4{{x}^{2}}+20x+160$. The coefficient of ${{x}^{2}}$ is $-4$. Therefore, the value of $a$ is negative and the parabola would be opening downwards. Compare the equation with the standard form of the quadratic equation, which is $f\left( x \right)=a{{x}^{2}}+bx+c$. So, $\begin{align} & a=-4 \\ & b=20 \\ & c=160 \end{align}$ The general form of the parabola vertex is $\left( -\frac{b}{2a},f\left( -\frac{b}{2a} \right) \right)$. Putting in the values of $a$ and $b$ to find the $x\text{-}$ coordinate of the vertex, we get: $\begin{align} & x=-\frac{b}{2a} \\ & =-\frac{20}{2\left( -4 \right)} \\ & =-\frac{20}{\left( -8 \right)} \\ & =2.5 \end{align}$ Putting in the value of $x$ in the equation to find the $y\text{-}$ coordinate of the vertex, we get: $\begin{align} & f\left( x \right)=-4{{x}^{2}}+20x+160 \\ & f\left( 2.5 \right)=-4{{\left( 2.5 \right)}^{2}}+20\left( 2.5 \right)+160 \\ & =-25+210 \\ & =185 \end{align}$ So, the vertex of the parabola is $\left( -\frac{b}{2a},f\left( -\frac{b}{2a} \right) \right)=\left( 2.5,185 \right)$. Use a graphing utility to plot the graph as follows: Step 1: Write the function. Step 2: Set the window $x:\left( -24,40,4 \right)$ and $y:\left( -210,300,30 \right)$. Step 3: Plot the graph.
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