SOLUTION: Find the vertex, line of symmentry, and the max/min value of f(x)=1/5(x+2)^2+2, is the valuef(=2)=2 min or max and graph. I am having problems trying to solve this.

SOLUTION: Find the vertex, line of symmentry, and the max/min value of f(x)=1/5(x+2)^2+2, is the valuef(=2)=2 min or max and graph. I am having problems trying to solve this.

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Question 408202: Find the vertex, line of symmentry, and the max/min value of f(x)=1/5(x+2)^2+2, is the valuef(=2)=2 min or max and graph. I am having problems trying to solve this.
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!

I can say , then

In an equation of the form
, the z-coordinate of
the vertex is at , but
here, so (0,f(0)) is the vertex, and to find

The vertex is at (0,2)
Now I translate this to (x,f(x))

(x,f(x)) = (-2,2)
The line of symmetry is
The graph is a min at
Here is the plot:

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