Question

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

Answer 1

`"x-intercept "=-5," y-intercept "=5`

Explanation:

`"to find the intercepts, that is where the graph crosses the"`
`"x and y axes"`

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

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

`x=0rArry=0+5=5larrcolor(red)"y-intercept"`

`y=0rArrx+5=0rArrx=-5larrcolor(red)"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]}

Answer 2

**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) = `color(blue)(1)` and

y-intercept = `color(blue)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 `color(blue)(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.