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

Jul 9, 2017

$\left(- 3 , 1\right)$

#### Explanation:

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