Question

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

Answer 1

The standard form is : y = ax^2 + bx + c. Find a, b, and c.
x of vertex: `-b/2a = -1 -> b = 2a`
y of vertex: f(-1) = 0 = a - b + c = a - 2a + c = 0 -> a = c

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

`-72 = a(-4)^2 - 4b + c = 16a - 8a + a = 7a -> a = -8 -> b = -16 and c = -8`

Standard form:`y = -8x^2 - 16x - 8`

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