Chapter 1.1, Problem 48E

To determine

To find: An expression for the function whose graph is the line segment.

Answer to Problem 48E

The function of a line joining the two points $\left(-5,10\right)\text{ and }\left(7,-10\right)$ is $f\left(x\right)=\frac{5}{3}x+\frac{5}{3}\text{ where }-5\le x\le 7$.

Explanation of Solution

Given:

The graph is the line segment joining the points $\left(-5,10\right)\text{ and }\left(7,-10\right)$.

Formula used:

The slope (m) of a line segment joining the points $\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)$ is,

$m=\frac{y_2-y_1}{x_2-x_1}$ (1)

The equation of a line with slope (m) and a point $\left(x_1,y_1\right)$ is,

$y-y_1=m\left(x-x_1\right)$ (2)

Calculation:

Use formula (1) and obtain the slope of a line segment joining the points $\left(-5,10\right)\text{ and }\left(7,-10\right)$.

Here, $x_1=-5,y_1=10,x_2=7,y_2=-10$.

$\begin{array}{c}m=\frac{-10-\left(10\right)}{7-\left(-5\right)}\\ =-\frac{20}{12}\\ =-\frac{5}{3}\end{array}$

Use formula (2) and obtain the equation of line with slope $m=-\frac{5}{3}$ and the point (−5, 10).

$\begin{array}{c}y-10=-\frac{5}{3}\left(x-\left(-5\right)\right)\\ =-\frac{5}{3}x-\frac{25}{3}\end{array}$

Rearrange the terms and simplify.

$\begin{array}{c}y=-\frac{5}{3}x-\frac{25}{3}+10\\ =\frac{5}{3}x+\frac{5}{3}\end{array}$

Therefore the function is $f\left(x\right)=\frac{5}{3}x+\frac{5}{3}\text{ where }-5\le x\le 7$.