Chapter 3.3, Problem 45PPS

To determine

The rate change for the linear function

Answer to Problem 45PPS

The slope is $\frac{2}{3}$ .

Explanation of Solution

Given information:

  

A table in which there are x-intercepts and y-intercepts. According to graph the points are $\left(\frac{5}{2},-\frac{3}{2}\right)\text{and }\left(\frac{-1}{2},\frac{1}{2}\right)$

Concept Used:

The slope of a line connecting the points $\left(x_1,y_1\right),\text{ and }\left(x_2,y_2\right)$ is given by the formula

  $\text{Slope }\left(m\right)=\frac{y_2-y_1}{x_2-x_1}$

Calculation:

In order to find the slope of the line passing through the points $\left(\frac{5}{2},-\frac{3}{2}\right)\text{and }\left(\frac{-1}{2},\frac{1}{2}\right)$ , use the slope formula and substitute the values $\left(x_1,y_1\right)=\left(\frac{5}{2},-\frac{3}{2}\right)\text{and }\left(x_2,y_2\right)=\left(\frac{-1}{2},\frac{1}{2}\right)$ and simplify further as shown below,

  $\begin{array}{c}\text{Slope }\left(m\right)=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{\frac{1}{2}-\left(-\frac{3}{2}\right)}{\left(\frac{-1}{2}\right)-\frac{5}{2}}=\frac{-\frac{4}{2}}{\frac{-6}{2}}\\ =\frac{-2}{-3}\\ =\frac{2}{3}\end{array}$

Thus, the slope is $\frac{2}{3}$ .