# What is the vertex form of y=x^2 + 8x - 9?

Dec 30, 2015

$y = {\left(x + 4\right)}^{2} - 25$ The vertex form is obtained by completing the square. Please go through the explanation on how we can get the solution.

#### Explanation:

To find the vertex form of the quadratic function we need to make the function into $y = a {\left(x - h\right)}^{2} + k$ where $\left(h , k\right)$ is the vertex.

$y = {x}^{2} + 8 x - 9$

Knowledge about completing squares would help.

$y + 9 = {x}^{2} + 8 x$ Firstly move the constant to the other side, this can be done by adding 9 to both the sides.

We need to make ${x}^{2} + 8 x$ into a whole square.

*In case you are confused with what a whole square is, it is nothing but expressing the given expression in the form ${\left(x - h\right)}^{2}$

On expanding ${\left(x - h\right)}^{2}$ would be ${x}^{2} - 2 h x + {h}^{2}$

Now let us see that the coefficient of $x$ is $2 h$ and we need ${h}^{2}$ so we divide coefficient of $x$ by $2$ and square the result to get ${h}^{2}$ That would help completing the square. The steps to complete the square for our question is given below.*

Divide the coefficient of $x$ by $2$ and square the result and add it to both the sides.

$y + 9 + {\left(\frac{8}{2}\right)}^{2} = {x}^{2} + 8 x + {\left(\frac{8}{2}\right)}^{2}$
$y + 9 + 16 = {x}^{2} + 8 x + 16$

$y + 25 = {\left(x + 4\right)}^{2}$

$y = {\left(x + 4\right)}^{2} - 25$ Vertex form