Precalculus (6th Edition) Blitzer

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

Chapter 7 - Review Exercises - Page 877: 14

Answer

The required quadratic equation is $ y=3{{x}^{2}}-4x+5$.

Work Step by Step

We know that the graph is passing through the point $\left( 1,4 \right);$ the equation becomes: $ a+b+c=4$ Now, the graph is passing through the point $\left( 3,20 \right);$ the equation becomes: $9a+3b+c=20$ Then the graph is passing through the point $\left( -2,25 \right);$ the equation becomes: $4a-2b+c=25$ Subtract equation $ a+b+c=4$ from equation $9a+3b+c=20$, given that $\begin{align} & 8a+2b=16 \\ & b=8-4a \end{align}$ Subtract equation $ a+b+c=4$ from equation $4a-2b+c=25$ to get: $\begin{align} & 3a-3b=21 \\ & a-b=7 \end{align}$ Substitute the value of b from equation $ b=8-4a $ in equation $ a-b=7$ as given below: $\begin{align} & a-\left( 8-4a \right)=7 \\ & a-8+4a=7 \\ & 5a=15 \\ & a=3 \end{align}$ From equation $ b=8-4a $, $\begin{align} & b=8-4\times 3 \\ & =-4 \end{align}$ From equation $9a+3b+c=20$, $\begin{align} & c=4-a-b \\ & =4-3+4 \\ & =5 \end{align}$ Therefore, $ a=3,b=-4,\text{ and }c=5$; put these values in the quadratic equation $ y=a{{x}^{2}}+bx+c $ as given below: $ y=3{{x}^{2}}-4x+5$ Hence, the quadratic equation is $ y=3{{x}^{2}}-4x+5$.
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