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

1 Answer
Omar T.
Oct 31, 2015

Vertex #=(5,2)#

Explanation:

In a parabolic function #y=a(x-p)^2+q#, the vertex is #(-p,q)#.
So in your function of #y=(x-5)^2+2#,

#p=-(-5)#
#q=(+2)#

So the point of the vertex would be #(5,2)#

Here is the graph:
graph{y=(x-5)^2+2 [-10, 10, -5, 5]}

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