What is the vertex form of #4(y-1)=(x-2)^2#?

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Answer 1 · 2017-05-02T22:52:48

#y=1/4(x-2)^2+1#

Vertex form has the form #y=a(x-b)^2+c#

Divide both sides by #4#

#4*4(y-1)=1/4(x-2)^2#

#cancel4*cancel4(y-1)=1/4(x-2)^2#

#y-1=1/4(x-2)^2#

Add #1# to both sides

#y-1+1=1/4(x-2)^2+1#

#y-cancel1+cancel1=1/4(x-2)^2+1#

#y=1/4(x-2)^2+1#