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

1 Answer
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)