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

Dec 23, 2015

The vertex is $\left(\frac{5}{\sqrt{2}} , - 30\right)$

#### Explanation:

Expand and simplify the expression first
$y = {x}^{2} + 5 \left({x}^{2} - 6 x + 9\right)$
$y = 6 {x}^{2} - 30 x + 45$
$y = 3 \left(2 {x}^{2} - 10 x + 15\right)$
The use completing the square to get vertex form
$y = 3 \left({\left(\sqrt{2} x - 5\right)}^{2} - 25 + 15\right)$
$y = 3 {\left(\sqrt{2} x - 5\right)}^{2} - 30$
The vertex is $\left(\frac{5}{\sqrt{2}} , - 30\right)$