Exercise 60 Page 10

To determine the $x\text{-}$ and $y\text{-}$intercepts of a line we need to substitute $0$ for one variable, solve, then repeat for the other variable.

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 $x.$

$y=x+2$

Substitute

$y=0$

$0=x+2$

SubEqn

$\text{LHS}-2=\text{RHS}-2$

$\text{-}2=x$

RearrangeEqn

Rearrange equation

$x=\text{-}2$

An $x\text{-}$intercept of $\text{-}2$ means that the graph passes through the $x\text{-}$axis at the point $(\text{-}2,0).$

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.

$$\begin{array}{c}y=x+2\end{array}$$

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 $2,$ and the graph passes through the $y\text{-}$axis at the point $(0,2).$