How do you write #f(x)= 2x^2+6x # into vertex form?

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Answer 1 · 2017-05-04T04:55:58

#y=2(x+1.5)^2-4.5#

Given -

#y=2x^2+6x#

First, find the vertex

Find #x# coordinate of the vertex

#x=(-b)/(2a)=(-6)/(2xx2)=-6/4=-1.5#
Find #y# coordinat of the vertex
#y=2(-1.5)^2+6(-1.5)#
#y=2(2.25)-9#

#y=4.5-9=-4.5#

Vertex #(-1.5, -4.5)#

The vertex form of the parabola is -

#y=a(x-h)^2-k#

Where -

#a=2# the coefficient of #x^2#
#h=-1.5# this is #x#coordinate of the vertex
#k=-4.5# this is #y# coordinate of the vertex

Substitute these values.

#y=2(x-(-1.5))^2+(-4.5)#
#y=2(x+1.5)^2-4.5#
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