What is the vertex of #y=8(3x+7)^2+5#?

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Answer 1 · 2017-10-20T17:45:48

#(-7/3, 5)=(-2.bar(3), 5)#

First get this into vertex form:

#y=a(b(x-h))^2+k# where #(h,k)# is the vertex

by factoring out the #3# in the parentheses:

#y=8(3(x+7/3))^2+5#

Then factor out a negative #1#:

#y=8(3(x-1(-7/3)))^2+5#

So it's now in vertex form:

#y=8(3(x-(-7/3)))^2+5# where #h=-7/3# and #k=5#

So our vertex is #(-7/3, 5)=(-2.bar(3), 5)#