SOLUTION: y=1/3x^2-2x+7 I need to find the vertex. I know I first use the axis of symmetry which is x=-b/2a. I did that and I got 2/2/3. I don't know what to do next. Please help. Thank you

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Question 388126: y=1/3x^2-2x+7
I need to find the vertex. I know I first use the axis of symmetry which is x=-b/2a. I did that and I got 2/2/3. I don't know what to do next. Please help. Thank you.
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
y = (1/3)*x^2 - 2*x + 7
x = -b/2a
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a = (1/3)
b = -2
c = 7
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-b/2a = -(-2)/(2/3) = 2/(2/3) = 2*3/2 = 6/2 = 3
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you found x
now you need to find f(x) to get the y value
when x = 3, (1/3)*x^2 - 2*x + 7 = (1/3)*3^2 - 2*3 + 7
this equals 3 - 6 + 7 = 4
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your vertex should be at the point pair of (x,y) = 3,4)
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the graph of this equation looks like this:

you can see from the graph that the vertex is at the point (x,y) = 3,4)