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

1 Answer
Jul 8, 2015

Answer:

This is an inverted V shape, with sides of slope #+-2#, vertex at #(-3/2, 3)#, intersection with the #y# axis at #(0, 0)# and intersections with the #x# axis at #(0, 0)# and #(-3, 0)#.

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

Explanation:

When #x <= -3/2#, #2x+3 <= 0# so #abs(2x+3) = -2x-3#

and #f(x) = 3-abs(2x+3) = 3 - (-2x-3) = 2x+6#

So this part of the curve is a line of slope #2# which intersects the #x# axis where #2x+6 = 0#, that is at #(-3, 0)#

When #2x+3 = 0#, #x = -3/2# and #f(x) = 2x+6 = 3#.

So the point #(-3/2, 3)# is the minimum of #abs(2x+3)# and the maximum of #f(x)#, thus the vertex.

When #x >= -3/2#, #2x+3 >= 0# so #abs(2x+3) = 2x+3#

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

So this part of the curve is a line of slope #-2# which intersects the #x# and #y# axis at #(0, 0)#.