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

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

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

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

What is the vertex form of 4(y-1)=(x-2)^24(y-1)=(x-2)^2?