Author:
QUITTNER
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Date Posted: 07:25:39 12/06/04 Mon
>>>> Find an equation for the parabola with horizontal
>axis of symmetry if it passes though points (5,0)
>(14,1) and (5,-2) <<<<
>>> ..... As I remember it, the equation for a parabola with HORIZONTAL axis of symmetry is (y-a) squared = k times (x-b) <<<
..... Normally from this point on you substitute the 3 x and the 3 y of the 3 points given into 3 simultaneous equations, and then solve for the 3 unknown variables a,b,and k.
HOWEVER, in this particular example there is an easier way:
..... Note that two of the given points have the SAME x, with their y being 0 and -2. This means that the axis of symmetry is half-way between these y, namely -1. But the vertex is on this axis, so that the vertex, too, has y=-1
..... In the equation on its left side it says (y-a)^2, and the a represents the y of the vertex, which we know now as being -1 = a. And on the right side of the equation b is the x of the vertex. That means that the new equation, for this example only, is:
(y+1) squared = k times (x-b) which is (y+1)^2=k*(x-b)
Now there are only 2 unknown variables, k and b, and you need now only 2 equations to solve for them:
(0+1)^2 = k*(5-b) for point (5,0)
(1+1)^2 = k*(14-b) for point (14,1)
Always check your results.
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