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

1 Answer
Jim G.
Mar 15, 2016

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 #

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