Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 2 Linear Equations and Functions - 2.7 Use Absolute Value Functions and Transformations - 2.7 Exercises - Skill Practice - Page 128: 35

Answer

$h=1$ and $k=4-3a$

Work Step by Step

We are given the function: $$y=a|x-h|+k.\tag1$$ Substitute the values of $(x,y)$ of the two given points in Eq. $(1)$: $$\begin{align*} \begin{cases} a|-2-h|+k&=4\\ a|4-h|+k&=4 \end{cases} \end{align*}$$ Subtract the two equations side by side: $$\begin{align*} a|-2-h|+k-a|4-h|-k&=4-4\\ a(|h+2|-|4-h|)&=0. \end{align*}$$ As $a\not=0$, we have: $$|h+2|=|4-h|.\tag2$$ We solve Eq. $(2)$ for $h$ using the graphical method. We find the solution: $$h=1.$$ From the system of equations we get: $$a|4-1|+k=4$$ $$3a+k=4$$ Eq. $(1)$ can be written: $$y=a|x-1|+4-3a.$$ So $h=1$ and $k=4-3a$.
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