What is the vertex of $y=2(x-2)^2-11$ ?

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Answer 1 · 2016-02-06T09:08:08

Vertex is at $(2, -11)$

This is a parabola which opens upward

of the form $(x-h)^2=4p(y-k)$ where vertex is $(h, k)$

from the given $y=2(x-2)^2-11$

transform first to the form

$y=2(x-2)^2-11$

$y+11=2(x-2)^2$

$(y+11)/2=(2(x-2)^2)/2$

$(y+11)/2=(cancel2(x-2)^2)/cancel2$

$1/2*(y+11)=(x-2)^2$

$(x-2)^2=1/2*(y+11)$

$(x-2)^2=1/2*(y--11)$

so that $h=2$ and $k=-11$

vertex is at $(2, -11)$

Kindly see the graph
graph{y=2(x-2)^2-11[-5,40,-15,10]}

Have a nice day ! from the Philippines...

What is the vertex of y=2(x-2)^2-11 y=2(x-2)^2-11 ?