Question

What is the equation of the parabola that has a vertex at `(6, 3)` and passes through point `(3, -9)`?

Answer 1

y = `-4/3 x^2 + 16x -45`

Explanation:

start by writing the equation in vertex form since coords of vertex are given.

vertex form is : y =`a(x - h )^2 + k", (h,k) being coords of vertex "`

hence partial equation is : y =`a(x - 6 )^2 + 3`

To find a , substitute (3 , -9) into the equation

thus : `a(3 - 6 )^2 + 3 = -9 → 9a = - 12 → a = - 4/3`

`rArr y = -4/3 (x - 6 )^2 + 3 " is the equation "`

distribute bracket and the equation in standard form is

`y = -4/3 x^2 + 16x - 45`