What is the vertex form of #y= 36x^2+132x+121 #?

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Answer 1 · 2018-07-11T18:38:01

#(x+11/6)^2=1/36y#

Given equation:

#y=36x^2+132x+121#

#y=36(x^2+11/3x)+121#

#y=36(x^2+11/3x+(11/6)^2)-36(11/6)^2+121#

#y=36(x+11/6)^2#

#(x+11/6)^2=1/36y#

The above is the vertex form of parabola with vertex at

#(x+11/6=0, y=0)\equiv(-11/6, 0)#