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

2 Answers
Mar 22, 2018

Answer:

#"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]}

Mar 22, 2018

Answer:

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.