# How do you write a quadratic equation with y-intercept of -4 and vertex at (3, -7)?

May 3, 2017

$y = \frac{1}{3} {x}^{2} - 2 x - 4$

#### Explanation:

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

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

$\text{here } \left(h , k\right) = \left(3 , - 7\right)$

$\Rightarrow y = a {\left(x - 3\right)}^{2} - 7$

$\text{to find a, use y-intercept of - 4} \rightarrow \left(0 , - 4\right)$

$- 4 = 9 a - 7$

$\Rightarrow a = \frac{1}{3}$

$\Rightarrow y = \frac{1}{3} {\left(x - 3\right)}^{2} - 7 \leftarrow \textcolor{red}{\text{ in vertex form}}$

$\text{distributing and simplifying gives}$

$y = \frac{1}{3} \left({x}^{2} - 6 x + 9\right) - 7$

$\Rightarrow y = \frac{1}{3} {x}^{2} - 2 x - 4 \leftarrow \textcolor{red}{\text{ in standard form}}$