# What is the vertex form of y=x^2/2+4x+8 ?

The vertex form is

${\left(x - - 4\right)}^{2} = 2 \left(y - 0\right) \text{ }$with vertex at $\left(h , k\right) = \left(- 4 , 0\right)$

#### Explanation:

The given equation is

$y = \frac{1}{2} {x}^{2} + 4 x + 8$

$y = \frac{1}{2} \left({x}^{2} + 8 x\right) + 8$

$y = \frac{1}{2} \left({x}^{2} + 8 x + 16 - 16\right) + 8$

$y = \frac{1}{2} \left({\left(x + 4\right)}^{2} - 16\right) + 8$

$y = \frac{1}{2} {\left(x + 4\right)}^{2} - 8 + 8$

$y = \frac{1}{2} {\left(x + 4\right)}^{2}$

$2 \left(y - 0\right) = {\left(x + 4\right)}^{2}$

${\left(x + 4\right)}^{2} = 2 \left(y - 0\right)$

The vertex form is

${\left(x - - 4\right)}^{2} = 2 \left(y - 0\right) \text{ }$with vertex at $\left(h , k\right) = \left(- 4 , 0\right)$

God bless...I hope the explanation is useful.