A parabola opening up or down has vertex (0, – 1) and passes through (6, – 4). Write its equation in vertex form.

Answer 1

a parabola has vertex (0, -1) and passes through (6,-4). It opens downward. (0, -1) is the maximum point on the parabola. the general form of the equation is y= a(x-h)^2 + k, where (h,k) is the vertex = (0, -1) and a<0

You know it opens downward because (6,-4) is down below the vertex. -4 < -1

when it opens downward a<0. a is the coefficient of the x^2 term

y = a(x-0)^2 - 1 = ax^2 -1. plug in the other point to solve for a

-4 = a(6)^2 -1 = 36a -1

36a = -4+1 = -3

a = -3/36 = -1/12

y = f(x) = (-1/12)x^2 - 1 is the equation of the parabola

in vertex form f(x) = (-1/12)(x - 0)^2 -1

or f(x) = -x^2/12 -1

or 12y = -x^2 -12 removing fractions