# 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

Algebra ->  Algebra  -> Quadratic-relations-and-conic-sections -> 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      Log On

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 Click here to see ALL problems on Quadratic-relations-and-conic-sections 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(3464)   (Show Source): You can put this solution on YOUR website!y = (1/3)*x^2 - 2*x + 7 x = -b/2a ------------------ a = (1/3) b = -2 c = 7 ------------------ -b/2a = -(-2)/(2/3) = 2/(2/3) = 2*3/2 = 6/2 = 3 ------------------- 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 --------------------- your vertex should be at the point pair of (x,y) = 3,4) ----------------------- 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)