# What is the vertex of  y= (x-3)^2-5x^2+6x-9?

Oct 16, 2016

the vertex is $\left(0 , 0\right)$

#### Explanation:

Developing the equation
$y = {\left(x - 3\right)}^{2} - 5 {x}^{2} + 6 x - 9 = {x}^{2} - 6 x + 9 - 5 {x}^{2} + 6 x - 9$
Simplifying $y = - 4 {x}^{2}$
So it's a parabola and the vertex is at (0,0)
Here is the graph of the function
graph{(x-3)^2-5x^2+6x-9 [-10, 10, -5, 5]}