What is the vertex of $y = \frac{1}{5} {(x + 1)}^{2} + 4$ $y= 1/5(x+1)^2+4$ ?

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Answer 1 · 2015-12-16T16:56:01

This equation for a parabola is already in vertex form ...

$y = a {(x - h)}^{2} + k$ $y=a(x-h)^2+k$ where vertex $= (h, k)$ $=(h,k)$

Using the vertex form above,

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

hope that helped

What is the vertex of y=15(x+1)2+4 y= 1/5(x+1)^2+4 ?