# How do you write standard form of quadratic equation of vertex (-1,5) and yint =3?

Jun 27, 2016

$y = - 2 {\left(x + 1\right)}^{2} + 5$

#### Explanation:

The vertex form for a quadratic is
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{m} {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$
for a parabola with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$
$\textcolor{g r e e n}{m}$ indicates if the parabola opens upward (if positive) or downward (if negative) and the "spread" of the parabola.

Since we want a parabola with vertex at $\left(\textcolor{red}{- 1} , \textcolor{b l u e}{5}\right)$
It must have the form:
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{m} {\left(x + 1\right)}^{2} + 5$

So what remains is to determine the value of $\textcolor{g r e e n}{m}$

W are told that $\text{yint} = 3$ (I assume this means the y-intercept)

Remembering that the y-intercept is the value of $y$ when $x = 0$
by replacing $y$ with $3$ and $x$ with $0$ we have:
$\textcolor{w h i t e}{\text{XXX}} 3 = \textcolor{g r e e n}{m} {\left(0 + 1\right)}^{2} + 5$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow 3 = m + 5$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow m = - 2$

So the complete form for the equation of this parabola is
$\textcolor{w h i t e}{\text{XXX}} y = - 2 {\left(x + 1\right)}^{2} + 5$

Here is a graph of $y = - 2 {\left(x + 1\right)}^{2} + 5$ to help verify our result:
graph{(-2)*(x+1)^2+5 [-6.735, 7.31, -0.513, 6.507]}