# How do you find the x and y intercepts for y = x + 5 ?

##### 2 Answers
Mar 22, 2018

$\text{x-intercept "=-5," y-intercept } = 5$

#### Explanation:

$\text{to find the intercepts, that is where the graph crosses the}$
$\text{x and y axes}$

• " let x = 0, in the equation for y-intercept"

• " let y = 0, in the equation for x-intercept"

$x = 0 \Rightarrow y = 0 + 5 = 5 \leftarrow \textcolor{red}{\text{y-intercept}}$

$y = 0 \Rightarrow x + 5 = 0 \Rightarrow x = - 5 \leftarrow \textcolor{red}{\text{x-intercept}}$
graph{(y-x-5)((x-0)^2+(y-5)^2-0.04)((x+5)^2+(y-0)^2-0.04)=0 [-10, 10, -5, 5]}

Mar 22, 2018

x-intercept color(blue)(=(-5,0)

y-intercept color(blue)(=(0,5)

#### Explanation:

Given:

Step 1

color(blue)(y=mx+b is the Slope-Intercept Form of the linear equation of a line, where color(blue)(m is the Slope and color(blue)(b is the y-intercept.

The linear equation color(red)(y=x+5 is in Slope-Intercept Form.

Hence,

Slope (m) = $\textcolor{b l u e}{1}$ and

y-intercept = $\textcolor{b l u e}{5}$

To plot this point on the graph, we write it as color(blue)((0,5)

Step 2

To find the x-intercept, set color(red)(y=0.

Hence, the given equation color(red)(y=x+5 can be written as

color(red)(0=x+5

rArr color(red)(x+5=0

Subtract $\textcolor{b l u e}{5}$ from both sides of the equation.

rArr color(red)(x+5-color(blue)(5)=0-color(blue)(5)

rArr color(red)(x+cancel(5)-color(blue)(cancel(5))=0-color(blue)(5)

rArr color(blue)(x=-5

To plot this point on the graph, we write it as color(blue)((-5,0)

Hence,

x-intercept color(blue)(=(-5,0)

y-intercept color(blue)(=(0,5)

Please refer to the graph below to verify the solutions visually:

graph{y=x+5 [-20, 20, -10, 10]}

Hope it helps.