# How do you write the vertex form equation of the parabola y = x^2 + 16x + 71?

May 10, 2018

Vertex form of equation is $y = {\left(x + 8\right)}^{2} + 7$

#### Explanation:

$y = {x}^{2} + 16 x + 71$ or

$y = \left({x}^{2} + 16 x + 64\right) + 7$ or

$y = {\left(x + 8\right)}^{2} + 7$ Comparing with vertex form of

equation f(x) = a(x-h)^2+k ; (h,k) being vertex we find

here $h = - 8 , k = 7 \therefore$ Vertex is at $\left(- 8 , 7\right)$ and vertex form of

equation is $y = {\left(x + 8\right)}^{2} + 7$

graph{x^2+16 x+71 [-40, 40, -20, 20]} [Ans]

May 10, 2018

$y = {\left(x + 8\right)}^{2} + 7$

The vertex is at $\left(- 8 , 7\right)$

#### Explanation:

Vertex form is obtained by the process of completing the square,

$y = {x}^{2} + 16 x + 71$

$y = {x}^{2} + 16 x \textcolor{b l u e}{+ 64 - 64} + 71 \text{ } \leftarrow \textcolor{b l u e}{+ {\left(\frac{b}{2}\right)}^{2} - {\left(\frac{b}{2}\right)}^{2}}$

$y = \left({x}^{2} + 16 x + 64\right) + \left(- 64 + 71\right)$

$y = {\left(x + 8\right)}^{2} + 7 \text{ } \leftarrow$ vertex form.

The vertex is at $\left(- 8 , 7\right)$ graph{y =x^2+16x+71 [-71, 89, 7.1, 87.1]}