What is the vertex form of #y=4x^2 + 12x + 9 #?

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Answer 1 · 2016-05-17T15:38:17

#y=4(x+1.5)^2#

Given -

#y=4x^2+12x+9#

Find the #x, y #co-ordinates of the vertex

#x=(-b)/(2a)=(-12)/(2xx4)=(-12)/8=-1.5#

At #x=1.5; y= 4(-1.5)^2+12(-1.5)+9#

#y=4(2.25)-18+9#
#y=9-18+9=0#

Equation in vertex form -

#y=a(x-h)^2+k#
#h # = x- coordinate
#k #= y- coordinate
#a#= coefficient of #x^2#
Then -

#y=4(x-(-1.5))^2+0#
#y=4(x+1.5)^2#