What is the vertex form of #3y=8x^2 + 17x - 13?

1 Answer
Nov 25, 2015

The vertex form is #y=8/3(x+17/16)^2-235/32#.

Explanation:

First, let's rewrite the equation so the numbers are all on one side:

#3y=8x^2+17x-13#
#y=(8x^2)/3+(17x)/3-13/3#

To find the vertex form of the equation, we must complete the square:

#y=(8x^2)/3+(17x)/3-13/3#

#y=8/3(x^2+17/8x)-13/3#

#y=8/3(x^2+17/8x+(17/8-:2)^2-(17/8-:2)^2)-13/3#

#y=8/3(x^2+17/8x+(17/8*1/2)^2-(17/8*1/2)^2)-13/3#

#y=8/3(x^2+17/8x+(17/16)^2-(17/16)^2)-13/3#

#y=8/3(x^2+17/8x+(289/256)-(289/256))-13/3#

#y=8/3(x^2+17/8x+(289/256))-13/3-(289/256*8/3)#

#y=8/3(x+17/16)^2-13/3-289/96#

#y=8/3(x+17/16)^2-235/32#