What is the vertex of # y= 4(x+2)^2-2x^2-4x+3#?

1 Answer
matt h. · Shwetank Mauria
Jul 9, 2017

#(-3,1)#

Explanation:

Firstly, expand the squared brackets:
#y=4(x^2+4x+4)-2x^2-4x+3#
Then, expand the brackets:
#y=4x^2+16x+16-2x^2-4x+3#
Collect like terms:
#y=2x^2+12x+19#
Use formula for x-turning point: (#-b/{2a}#)
thus, x=#-3#
Plug -3 back into the original formula for y coordinate:
#4(-3+2)^2-2(-3)^2-4(-3)+3=4-18+12+3=1#
therefore the vertex is: #(-3,1)#

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