Chapter 1.1, Problem 47E

To determine

To find: An expression for the function whose graph is the line segment joining the points $\left(1,-3\right)$ and $\left(5,7\right)$.

Answer to Problem 47E

The function of a line joining the points $\left(1,-3\right)$ and $\left(5,7\right)$ is $\underset{\_}{f\left(x\right)=\frac{5}{2}x-\frac{11}{2}\text{ where }1\le x\le 5}$.

Explanation of Solution

Given:

The graph is the line segment joining the points $\left(1,-3\right)\text{ and }\left(5,7\right)$.

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 $\left(1,-3\right)\text{ and }\left(5,7\right)$.

Here, $x_1=1,y_1=-3,x_2=5,y_2=7$.

$\begin{array}{c}m=\frac{7-\left(-3\right)}{5-1}\\ =\frac{7+3}{4}\\ =\frac{10}{4}\\ =\frac{5}{2}\end{array}$

Use formula (2) and obtain the equation of a line with slope $m=\frac{5}{2}$ and the point $\left(1,-3\right)$.

$\begin{array}{l}y-\left(-3\right)=\frac{5}{2}\left(x-1\right)\\ y+3=\frac{5}{2}x-\frac{5}{2}\\ y=\frac{5}{2}x-\frac{5}{2}-3\\ y=\frac{5}{2}x-\frac{11}{2}\end{array}$

Therefore the function is $f\left(x\right)=\frac{5}{2}x-\frac{11}{2}\text{ where }1\le x\le 5$.