# What is the vertex of y=(x-2)^2+16x-1 ?

Feb 7, 2018

(-6, 33)

#### Explanation:

The graph $y = {\left(x - 2\right)}^{2} + 16 x - 1$ can be expanded.

$y = {x}^{2} - 4 x + 4 + 16 x - 1$ is the new equation.

Combining like terms, we get

$y = {x}^{2} + 12 x + 3$.

We can change this into $y = a \left(x - h\right) + k$ form.

$y = {\left(x + 6\right)}^{2} - 33$.

The vertex must be $\left(- 6 , - 33\right)$.

To check, here is our graph: graph{y=x^2+12x+3 [-37.2, 66.8, -34.4, 17.64]}

Yay!