# How do you write an equation with Vertex (-3, 12), point (-1, 0) on quadratic?

Aug 19, 2017

$y = - 3 {\left(x + 3\right)}^{2} + 12$

#### Explanation:

$\text{the equation of a parabola in "color(blue)"vertex form}$ is.

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

$\text{here } \left(h , k\right) = \left(- 3 , 12\right)$

$\Rightarrow y = a {\left(x + 3\right)}^{2} + 12$

$\text{to find a substitute "(-1,0)" into the equation}$

$0 = 4 a + 12 \Rightarrow a = - 3$

$\Rightarrow y = - 3 {\left(x + 3\right)}^{2} + 12 \leftarrow \textcolor{red}{\text{ in vertex form}}$