# SOLUTION: With a y-intercept 10, x-intercept 2, and equation of axis of symmetry x-3 = 0, find the graph equation in standard form. So I've got two points (0, 10) (2, 0) and half of the ver

Algebra ->  Algebra  -> Coordinate Systems and Linear Equations -> SOLUTION: With a y-intercept 10, x-intercept 2, and equation of axis of symmetry x-3 = 0, find the graph equation in standard form. So I've got two points (0, 10) (2, 0) and half of the ver      Log On

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 Click here to see ALL problems on Linear-systems Question 622726: With a y-intercept 10, x-intercept 2, and equation of axis of symmetry x-3 = 0, find the graph equation in standard form. So I've got two points (0, 10) (2, 0) and half of the vertex (3, k). I don't know how to get this. The equation is y = a(x-h)^2 + k Either of the two points could be labelled x and y. I'll use the x-intercept (2, 0). The vertex is used for h, k. So I've got 0 = a(2 - 3)^2 + k Two unsolved for variables. Not sure what to do.Answer by ewatrrr(10682)   (Show Source): You can put this solution on YOUR website! ``` Hi, parabola: with axis of symmetry x = 3, passing thru (0,10) and (2,0) y = a(x-h)^2 + k |Note: Parabola with this format.. always opens along its axis of symmetry x = h y = a(x-3)^2 + k Pt(2,0) 0 = a(-1)^2 + k -k = a 10 = a(-3)^2 + k Pt(0,10) (10-k)/9 = a (10-k)/9 = -k 10-k = -9k 8k = -10 k = -10/8 = -5/4 and a = 5/4 ```