# How do you write the quadratic in vertex form given vertex is (-2,6) and y intercpet is 12 ?

May 30, 2015

Quadrati equation $y = a {x}^{2} + b x + c$.

c = 12, find a and b.

x of vertex: $\left(- \frac{b}{2} a\right) = - 2 \to b = 4 a$

y of vertex: #f(-2) = 6 = 4a - 2b + 12 --> 4a - 8a + 12 = 6 ->

-> $a = \frac{6}{4} = \frac{3}{2}$

Equation $y = \frac{3}{2} \left({x}^{2}\right) + 6 x + 12$

Check:

x of vertex: $\left(- \frac{b}{2 a}\right) = - \frac{6}{3} = - 2$ OK
$f \left(- 2\right) = \left(\frac{3}{2}\right) 4 - 12 + 12 = 6$ OK