How to graph a parabola #y=3/2x^2 - 3x-5/2#?

1 Answer
Jun 21, 2015

Complete the square to find the vertex, axis of symmetry, etc.

Explanation:

#3/2(x-1)^2 = 3/2(x^2-2x+1) = 3/2x^2-3x+3/2#

So

#y = 3/2x^2-3x-5/2#

#= 3/2(x-1)^2-3/2-5/2#

#= 3/2(x-1)^2-4#

This is in vertex form,

allowing us to tell that the vertex is at #(1, -4)#

and the axis of symmetry is #x=1#.

The intersection with the #y# axis is at #(0, 3/2)#

The intersections with the #x# axis are where #y=0#, so

#0 = 3/2(x-1)^2 - 4#

Add #4# to both sides to get:

#3/2(x-1)^2 = 4#

Multiply both sides by #2/3# to get:

#(x-1)^2 = 2/3*4#

Hence #x = 1+-2sqrt(2/3) ~= 1+-1.633#

So the intersections with the #y# axis are approximately:

#(-0.633,0)# and #(2.633, 0)#

graph{3/2x^2-3x-5/2 [-10, 10, -5, 5]}