# What is the vertex form of y=x^2+4x+16?

Jan 9, 2016

$y = {\left(x + 2\right)}^{2} + 12$

#### Explanation:

The standard form of a quadratic equation is:

$y = a {x}^{2} + b x + c$

The vertex form is : $y = {\left(x - h\right)}^{2} + k$ where (h , k ) are the coordinates of the vertex.

For the given function $a = 1$ , $b = 4$, and $c = 16$.

The x-coordinate of the vertex (h) $= - \frac{b}{2 a} = - \frac{4}{2} = - 2$

and the corresponding y-coordinate is found by substituting x =- 2 into the equation :

$\Rightarrow y = {\left(- 2\right)}^{2} + 4 \left(- 2\right) + 16 = 4 - 8 + 16 = 12$

the coordinates of the vertex are (- 2 , 12 ) = (h , k )

the vertex form of $y = {x}^{2} + 4 x + 16$ is then :

$y = {\left(x + 2\right)}^{2} + 12$

check:

${\left(x + 2\right)}^{2} + 12 = {x}^{2} + 4 x + 16$