# How do you write the equation ins vertex form for the quadratic whose vertex is at (-2, 5) and has a point at (2,-2)?

Jul 28, 2016

$y = - \frac{7}{16} {\left(x + 2\right)}^{2} + 5$

#### Explanation:

The equation of a parabola in $\textcolor{b l u e}{\text{vertex form}}$ is

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where (h ,k) are the coordinates of the vertex and a, is a constant.

here (h ,k) = (-2 ,5)

$\Rightarrow y = a {\left(x + 2\right)}^{2} + 5 \text{ is the partial equation}$

Since the parabola passes through (2 ,-2) then the coordinates of this point make the equation true.
Substitute x = 2 and y = -2 into the partial equation to find a.

$\Rightarrow a {\left(2 + 2\right)}^{2} + 5 = - 2 \Rightarrow 16 a = - 7 \Rightarrow a = - \frac{7}{16}$

$\Rightarrow y = - \frac{7}{16} {\left(x + 2\right)}^{2} + 5 \text{ is the equation in vertex form}$
graph{-7/16(x+2)^2+5 [-10, 10, -5, 5]}