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

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Answer 1 · 2018-05-16T18:03:59

vertex: (2, 5)

#y = 3(x-2)^2+5#

this is a parabola because of one variable squared and the other one is not so now write it in the standard form of parabolas which it is = to

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Vertical :

#(x-h)^2=4p(y-k)#

Horizontal:

#(y-k)^2=4p(x-h)^2#

vertex = (h, k)

this #y=3(x-2)^2+5# equation is vertical since #x# is squared

subtract 5 from both sides:

#y-5=3(x-2)^2#

divide both sides by 3:

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

vertex:

#(2, 5)#