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

Feb 27, 2016

Put the equation into vertex form to find that the vertex is at $\left(- 8 , - 65\right)$

#### Explanation:

The vertex form of a quadratic equation is

$y = a {\left(x - h\right)}^{2} + k$

and the vertex of that graph is $\left(h , k\right)$

To obtain the vertex form, we use a process called completing the square. Doing so in this case is as follows:

$y = {x}^{2} + 16 x - 1$

$= {x}^{2} + 16 x + 64 - 65$

$= {\left(x + 8\right)}^{2} - 65$

$= {\left(x - \left(- 8\right)\right)}^{2} - 65$

Thus the vertex is at $\left(- 8 , - 65\right)$