Chapter 3, Problem 14PT

To determine

To find:Slope intercept form for the linear function.

Answer to Problem 14PT

  $y=\frac{-2}{5}x+\frac{11}{5}$

Explanation of Solution

From the given Graph we can say that the line passed through the points

  $\left(x_1,y_1\right)=\left(3,1\right)$

  $\left(x_2,y_2\right)=\left(-2,3\right)$

Using two-point formula

  $y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)\left(x-x_1\right)$

On substituting values

We get

  $\begin{array}{l}y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)\left(x-x_1\right)\\ \Rightarrow y-1=\left(\frac{3-1}{-2-3}\right)\left(x-3\right)\\ \Rightarrow y-1=\left(\frac{2}{-5}\right)\left(x-3\right)\\ \Rightarrow y-1=\frac{-2}{5}x+\frac{2}{5}\left(3\right)\\ \Rightarrow y-1=\frac{-2}{5}x+\frac{6}{5}\\ \Rightarrow y=\frac{-2}{5}x+\frac{6}{5}+1\\ \Rightarrow y=\frac{-2}{5}x+\frac{6+5}{5}\\ \Rightarrow y=\frac{-2}{5}x+\frac{11}{5}\end{array}$

Therefore

Slope intercept form of given linear function is $y=\frac{-2}{5}x+\frac{11}{5}$