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(24785)   (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
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