# What is the vertex of y= 2(x - 4)^2 - x^2+4x-1 ?

Dec 16, 2015

vertex$= \left(6 , - 5\right)$

#### Explanation:

Start by expanding the brackets, then simplifying the terms:

$y = 2 {\left(x - 4\right)}^{2} - {x}^{2} + 4 x - 1$
$y = 2 \left(x - 4\right) \left(x - 4\right) - {x}^{2} + 4 x - 1$
$y = 2 \left({x}^{2} - 8 x + 16\right) - {x}^{2} + 4 x - 1$
$y = 2 {x}^{2} - 16 + 32 - {x}^{2} + 4 x - 1$
$y = {x}^{2} - 12 x + 31$

Take the simplified equation and rewrite it in vertex form:

$y = {x}^{2} - 12 x + 31$
$y = \left({x}^{2} - 12 x\right) + 31$
$y = \left({x}^{2} - 12 x + {\left(\frac{12}{2}\right)}^{2} - {\left(\frac{12}{2}\right)}^{2}\right) + 31$
$y = \left({x}^{2} - 12 x + {\left(6\right)}^{2} - {\left(6\right)}^{2}\right) + 31$
$y = \left({x}^{2} - 12 x + 36 - 36\right) + 31$
$y = \left({x}^{2} - 12 x + 36\right) + 31 - \left(36 \cdot 1\right)$
$y = {\left(x - 6\right)}^{2} + 31 - 36$
$y = {\left(x - 6\right)}^{2} - 5$

Recall that the general equation of a quadratic equation written in vertex form is:

$y = a {\left(x - h\right)}^{2} + k$

where:
$h =$x-coordinate of the vertex
$k =$y-coordinate of the vertex

So in this case, the vertex is $\left(6 , - 5\right)$.