# How do you find the quadratic function with vertex (-2,5) and point (0,9)?

Jan 7, 2018

$y = {x}^{2} + 4 x + 9$

#### Explanation:

Vertex form of a quadratic equation is $y = a {\left(x - h\right)}^{2} + k$ where $\left(h , k\right)$ is the vertex

In this case, h is -2 and k is 5.

Thus the equation becomes $y = a {\left(x + 2\right)}^{2} + 5$

In order to find out a, we must plug in the x- and y-values of the point we want the equation to pass through.

The equation becomes:
$9 = a {\left(0 + 2\right)}^{2} + 5$

$\implies 9 = a \cdot {2}^{2} + 5$

$\implies 4 a = 4$

$\implies a = 1$

Finally the equation is $y = {\left(x + 2\right)}^{2} + 5$

Another way to put it is:
$y = {x}^{2} + 4 x + 9$