Chapter 1.2, Problem 83E

(a)

To determine

The $x$ -intercept of the graph of the equation $y=ax+b,a\ne 0$, where $a$ and $b$ are real numbers.

The $x$ -intercept of the graph of the equation $y=ax+b,a\ne 0$, where $a$ and $b$ are real numbers is $\left(\frac{-b}{a},0\right)$.

Explanation:

Consider the equation $y=ax+b,a\ne 0$.

For finding the $x$ -intercept, put $y=0$.

$\begin{array}{c}0=ax+b\\ 0-b=ax-b\\ \frac{-b}{a}=\frac{ax}{a}\\ \frac{-b}{a}=x\end{array}$

Hence, the $x$ -intercept of the graph of the equation $y=ax+b,a\ne 0$, where $a$ and $b$ are real numbers is $\left(\frac{-b}{a},0\right)$.

(b)

To determine

The $y$ -intercept of the graph of the equation $y=ax+b,a\ne 0$, where $a$ and $b$ are real numbers.

The $y$ -intercept of the graph of the equation $y=ax+b,a\ne 0$, where $a$ and $b$ are real numbers is $\left(0,b\right)$.

Explanation:

Consider the equation $y=ax+b,a\ne 0$.

To find the $y$ -intercept, put $x=0$.

$\begin{array}{l}y=a\left(0\right)+b\\ y=b\end{array}$

Hence, the $y$ -intercept of the graph of the equation $y=ax+b,a\ne 0$, where $a$ and $b$ are real numbers is $\left(0,b\right)$.

(c)

To determine

To calculate: The $x$ and $y$ -intercept of the graph of the equation $y=5x+10$ using part (a) and (b).

The $x$ and $y$ -intercept of the graph of the equation $y=5x+10$ are $\left(-2,0\right)\text{ and }\left(\text{0,10}\right)$ respectively.

Calculation:

Consider the equation $y=5x+10$.

As the $x$ - intercept is given by $\frac{-b}{a}$,

The $x$ -intercept of the graph of the equation $y=5x+10$ is

$\begin{array}{c}\frac{-b}{a}=\frac{-10}{5}\\ =-2\end{array}$

As the $y$ -intercept is given by $b$,

So $b=10$.

Hence, the $x$ and $y$ -intercept of the graph of the equation $y=5x+10$ are $\left(-2,0\right)\text{ and }\left(\text{0,10}\right)$ respectively.