# What is the vertex of y= 2(x - 4)^2 - 8x+3 ?

Apr 26, 2017

The vertex is $\left(6 , - 27\right)$

#### Explanation:

Given: $y = 2 {\left(x - 4\right)}^{2} - 8 x + 3$

Expand the square:

$y = 2 \left({x}^{2} - 8 x + 16\right) - 8 x + 3$

Distribute the 2:

$y = 2 {x}^{2} - 16 x + 32 - 8 x + 3$

Combine like terms:

$y = 2 {x}^{2} - 24 x + 35$

The x coordinate of the vertex, h, can be computed using the following equation:

$h = - \frac{b}{2 a}$ where $b = - 24$ and $a = 2$

h = -(-24)/(2(2)

$h = 6$

The y coordinate of the vertex, k, can be computed by evaluating the function at the value of h, (6):

$k = 2 {\left(6 - 4\right)}^{2} - 8 \left(6\right) + 3$

$k = - 37$

The vertex is $\left(6 , - 27\right)$