What is the the vertex of $y = 5(x-1)^2+4$ ?

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Answer 1 · 2016-01-03T00:42:47

This equation is already in vertex form ...

$y=a(x-h)^2+k$

where, vertex $=(h,k)$

So, for this problem ...

vertex $=(h,k)=(1,4)$

hope that helped

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What is the the vertex of y = 5(x-1)^2+4y = 5(x-1)^2+4?