What equation is #y=(x+3)^2 + (x+4)^2# rewritten in vertex form?

1 Answer
Jun 2, 2018

#y=2(x+7/2)^2+1/2#

Explanation:

This is a bit of a sneaky question. It isn't immediately obvious that this is a parabola, but "vertex form" is a form of equation specifically for one. It is a parabola, a closer look reveals, which is fortunate... It's the same thing as "completing the square" - we want the equation in the form #a(x-h)^2+k#.

To get there from here, we first multiply out the two brackets, then collect up terms, then divide through to make the #x^2# coefficient 1:
#1/2y=x^2+7x+25/2#

Then we find a square bracket that gives us the correct #x# coefficient. Note that in general
#(x+n)^2=x^2+2n+n^2#
So we choose #n# to be half our existing #x# coefficient, i.e. #7/2#. Then we need to subtract off the extra #n^2=49/4# that we've introduced. So
#1/2y=(x+7/2)^2-49/4+25/2=(x+7/2)^2+1/4#

Multiply back through to get #y#:
#y=2(x+7/2)^2+1/2#