Precalculus (6th Edition) Blitzer

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

Chapter 10 - Cumulative Review Exercises - Page 1128: 36


The linear function in slope-intercept form is $ y=\frac{x}{4}+\frac{3}{2}$.

Work Step by Step

Let us consider the coordinates $\left( -2,1 \right)$ and $\left( 6,3 \right)$ as $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ respectively. And find the equation of the function by taking $ f\left( x \right)$ as $ y $: $\begin{align} & \frac{y-{{y}_{1}}}{x-{{x}_{1}}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \\ & \frac{y-1}{x-\left( -2 \right)}=\frac{3-1}{6-\left( -2 \right)} \\ & \frac{y-1}{x+2}=\frac{2}{8} \\ & \frac{y-1}{x+2}=\frac{1}{4} \end{align}$ Then, simplify it further, to get $\begin{align} & 4\left( y-1 \right)=x+2 \\ & 4y-4=x+2 \\ & 4y=x+2+4 \\ & y=\frac{x+6}{4} \end{align}$ And breaking the terms into individual terms, $\begin{align} & y=\frac{x}{4}+\frac{6}{4} \\ & y=\frac{x}{4}+\frac{3}{2} \end{align}$ Thus, the slope-intercept form is $ y=\frac{x}{4}+\frac{3}{2}$.
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