# How do you write the quadratic in vertex form given vertex is (3,-6). and y intercept is 2?

May 8, 2015

The general vertex form of a quadratic is
$y = m {\left(x - a\right)}^{2} + b$
where the vertex is at $\left(a , b\right)$

We are told that the vertex is at $\left(3 , - 6\right)$
so this becomes
$y = m {\left(x - 3\right)}^{2} - 6$
$= m {x}^{2} - 6 m x + 9 m - 6$

We are also told that when $x = 0$ then $y = 2$ (that's what it means to say the y-intercept is $2$).

So $m {\left(0\right)}^{2} - 6 m \left(0\right) + 9 m - 6 = 2$

$9 m = 8$

$m = \frac{8}{9}$

And the equation of the quadratic in vertex form is
$y = \frac{8}{9} {\left(x - 3\right)}^{2} - 6$