Exercise 37 Page 162

We will determine the $x\text{-}$ and $y\text{-}$intercepts one at a time. Let's start with the $x\text{-}$intercept.

Finding the $x\text{-}$intercept

Think of the point where the graph of an equation crosses the $x\text{-}$axis. The $y$-value of that $(x,y)$ coordinate pair is $0,$ and the $x$-value is the $x\text{-}$intercept. To find the $x\text{-}$intercept of the equation, we should substitute $0$ for $y$ and solve for $x.$

$y=x-1$

Substitute

$y=0$

$0=x-1$

AddEqn

$\text{LHS}+1=\text{RHS}+1$

$1=x$

RearrangeEqn

Rearrange equation

$x=1$

An $x\text{-}$intercept of $1$ means that the graph passes through the $x\text{-}$axis at the point $\left(1,0\right).$

Finding the $y\text{-}$intercept

We could use the same concept to find the $y\text{-}$intercept, but take a close look at the given equation.

$$\begin{array}{c}y=x-1\iff y=x+(\text{-}1)\end{array}$$

The given equation is in slope-intercept form, $y=mx+b.$ For an equation in this form, we know that the constant $b$ represents the $y\text{-}$intercept. Therefore, the $y\text{-}$intercept is $\text{-}1,$ and the graph passes through the $y\text{-}$axis at the point $(0,\text{-}1).$