How do you graph #y = abs(3x+2) #?

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Answer 1 · 2015-06-20T07:53:09

Basically, it's a V-shape with slope #+-3# and vertex #(-2/3, 0)#

The vertex is where #abs(3x+2) = 0# at #(-2/3, 0)#.

When #x < -2/3# the inner expression #3x + 2 < 0# so

#abs(3x+2) = -(3x-2)#

and the equation becomes #y = -3x + 2#

This is a straight line of slope #-3#.

When #x > 2/3# the inner expression #3x + 2 > 0# so

#abs(3x+2) = 3x - 2#

and the equation becomes #y = 3x - 2#

This is a straight line of slope #3#

So the original equation represents a V shape with slope #+-3# and vertex #(-2/3, 0)#

graph{abs(3x+2) [-10, 10, -5, 5]}