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

Dec 22, 2015

The vertex is at:$\left(4 , - 11\right)$

#### Explanation:

y=(x−3)^2−2x−4 => expand to simplify:
$y = {x}^{2} - 6 x + 9 - 2 x - 4$ => simplify add/subtract like terms:
$y = {x}^{2} - 8 x + 5$ => quadratic function in standard/general form of :
$f \left(x\right) = y = a {x}^{2} + b x + c$=>where the x and y coordinates of the vertex are:
$\left(x , y\right) = \left[- \frac{b}{2 a} , f \left(- \frac{b}{2 a}\right)\right]$
so in this case:
$f \left(x\right) = y = {x}^{2} - 8 x + 5$=> where:$a = 1 , b = - 8 , c = 5$, then:
$x = - \left(- \frac{8}{2}\right) = 4$, and:
$f \left(4\right) = {4}^{2} - 8 \cdot 4 + 5 = - 11$
hence the vertex is at:
$\left(4 , - 11\right)$