# How do you write a quadratic function in vertex form whose has vertex (-3,-2) and passes through point (1,14)?

Mar 2, 2017

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

#### Explanation:

Any quadratic function with vertex at $\left(\textcolor{red}{- 3} , \textcolor{b l u e}{- 2}\right)$
can be written in the form:
$\textcolor{w h i t e}{\text{XXX")y=color(green)m(x-color(red)(""(-3)))^2+color(blue)(} \left(- 2\right)}$ for some constant $\textcolor{g r e e n}{m}$
or (simplified)
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{m} {\left(x + 3\right)}^{2} - 2$

If $\left(\textcolor{m a \ge n t a}{x} , \textcolor{b r o w n}{y}\right) = \left(\textcolor{m a \ge n t a}{1} , \textcolor{b r o w n}{14}\right)$ is a solution for some specific quadratic with this form:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b r o w n}{14} = \textcolor{g r e e n}{m} {\left(\textcolor{m a \ge n t a}{1} + 3\right)}^{2} - 2$

$\rightarrow \textcolor{w h i t e}{\text{XXX}} 14 = 16 \textcolor{g r e e n}{m} - 2$

$\rightarrow \textcolor{w h i t e}{\text{XXX}} 16 \textcolor{g r e e n}{m} = 16$

$\rightarrow \textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{m} = 1$

$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{1} {\left(x + 3\right)}^{2} - 2$
As verification here is the graph of $y = {\left(x + 3\right)}^{2} - 2$ 