# What is the the vertex of x = (y -3)^2 - 5y-5 ?

Jul 5, 2018

$\frac{11}{2} , - \frac{105}{4}$

#### Explanation:

Let $f \left(y\right) = {\left(y - 3\right)}^{2} - 5 y - 5$
then we get by using

${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$

$f \left(y\right) = {y}^{2} - 6 y + 9 - 5 y - 5$
combining like terms

$f \left(y\right) = {y}^{2} - 11 y + 4$

we compute the coordinates of the vertex:_

$f ' \left(y\right) = 2 y - 11$
so

$f ' \left(y\right) = 0$ if $y = \frac{11}{2}$

and
$f \left(\frac{11}{2}\right) = - \frac{105}{4}$