# How do you write the quadratic in vertex and standard form given the vertex ( -1, 0) and passes through ( -4, -72)?

May 21, 2015

The standard form is : y = ax^2 + bx + c. Find a, b, and c.
x of vertex: $- \frac{b}{2} a = - 1 \to b = 2 a$
y of vertex: f(-1) = 0 = a - b + c = a - 2a + c = 0 -> a = c

The parabola passes at point (-4, -72):

$- 72 = a {\left(- 4\right)}^{2} - 4 b + c = 16 a - 8 a + a = 7 a \to a = - 8 \to b = - 16 \mathmr{and} c = - 8$

Standard form:$y = - 8 {x}^{2} - 16 x - 8$

Check:
x of vertex: -b/2a = 16/-16 = -1. OK
y of vertex: f(-1) = -8 + 16 - 8 = 0 . OK