# How do you find the vertex of  (x+2)^2=-8(y-1)?

Mar 14, 2016

Vertex is at (-2,1)

#### Explanation:

$8 \left(y - 1\right) = - {\left(x + 2\right)}^{2} \mathmr{and} y - 1 = - \frac{1}{8} \cdot {\left(x + 2\right)}^{2} \mathmr{and} y = - \frac{1}{8} \cdot {\left(x + 2\right)}^{2} + 1$ So vertex is at $\left(- 2 , 1\right)$ graph{-1/8*(x+2)^2+1 [-10, 10, -5, 5]}[Answer]

Mar 14, 2016

Vertex is $\left(- 2 , 1\right)$

#### Explanation:

This is a quadratic equation, with a vertex at the top.

${\left(x + 2\right)}^{2} = - 8 \left(y - 1\right)$ --> Expand the left hand side
${x}^{2} + 4 x + 4 = - 8 \left(y - 1\right)$ --> Divide both sides by -1/8
$- \frac{1}{8} {x}^{2} - \frac{1}{2} x - \frac{1}{2} = y - 1$ --> Add one to both sides
$- \frac{1}{8} {x}^{2} - \frac{1}{2} x - \frac{1}{2} + 1 = y$ --> Simplify
$y = - \frac{1}{8} {x}^{2} - \frac{1}{2} x + \frac{1}{2}$

Apply the equation for the $x$-value of the vertex of a parabola. For a quadratic of the form $y = a {x}^{2} + b x + c$, the $x$-value of the vertex is

$x = - \frac{b}{2 a}$

For $b = - \frac{1}{2}$ and $a = - \frac{1}{8}$, we have vertex $x = - 2$. Plugging $x = - 2$ into $y = - \frac{1}{8} {\left(- 2\right)}^{2} - \frac{1}{2} \left(- 2\right) + \frac{1}{2}$ gives $y = 1$. So the vertex is $\left(- 2 , 1\right)$.

Graphically,

graph{-1/8x^2-.5x+.5 [-10, 10, -5, 5]}