How do you find the intercepts for #y=-3x+3#?

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Answer 1 · 2015-06-14T22:40:02

The intercept with the #y# axis is where #x=0#, hence #y = -3*0 + 3 = 3#. That is #(0, 3)#

The intercept with the #x# axis is where #y=0#, hence #0 = -3x + 3#, hence #x = 1#. That is #(1, 0)#

To find where #y = -3x + 3# intercepts the #y# axis, substitute #x=0# into the equation to find:

#y = (-3*0) + 3 = 0 + 3 = 3#.

That is #(0, 3)#

To find where #y = -3x + 3# intercepts the #x# axis, substitute #y=0# into the equation to find:

#0 = -3x+3#

Add #3x to both sides to get:

#3x=3#

Divide both sides by #3# to get:

#x = 1#.

That is #(1, 0)#

graph{-3x+3 [-9.5, 10.5, -3, 7]}