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: 88

Answer

The graph is shown below:
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Work Step by Step

Let us consider the parabola defined by the quadratic equation $f\left( x \right)=0.01{{x}^{2}}+0.6x+100$. The coefficient of ${{x}^{2}}$ is $0.01$. Therefore, the value of $a$ is positive and the parabola would be opening upwards. Compare the equation with the standard form of the quadratic equation $f\left( x \right)=a{{x}^{2}}+bx+c$. So, $\begin{align} & a=0.01 \\ & b=0.6 \\ & c=100 \end{align}$ 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: $\begin{align} & x=-\frac{b}{2a} \\ & =-\frac{0.6}{2\left( 0.01 \right)} \\ & =-\frac{600}{20} \\ & =-30 \end{align}$ Putting in the value of $x$ in the equation to find the $y\text{-}$ coordinate of the vertex, we get: $\begin{align} & f\left( x \right)=0.01{{x}^{2}}+0.6x+100 \\ & f\left( 30 \right)=0.01{{\left( -30 \right)}^{2}}+0.6\left( -30 \right)+100 \\ & =9-18+100 \\ & =91 \end{align}$ Thus, the vertex of the parabola is $\left( -\frac{b}{2a},f\left( -\frac{b}{2a} \right) \right)=\left( -30,91 \right)$. Use a graphing utility to plot the graph: Step 1: Write the function. Step 2: Set the window $x:\left( -250,250,60 \right)$ and $y:\left( 0,400,30 \right)$. Step 3: Plot the graph.
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