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Puzzles A Problematic Parabola A parabola opens downward and has its vertex on the line y = 2. It passes through the points (0,1) on the y-axis and (2,0) on the x-axis. If its vertex is the point (a,2), find the value of a.
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Comments | cosmo145 |  12:08:55, 18 Dec 07 |  Newbie
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| I get a=2. starting from the standard vertex form of y-k=(-1/4p)*(x-h)^2
| | | | terveloc | 12:57:29, 17 Nov 07 | Newbie
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| It's ( -2 + 2*sqrt(2) ). Start from standard vertex form of the parabola ( y = s*(x - a)^2 +2 ) and work from there.
| | | | Vuong | 02:33:52, 09 Nov 07 |  Dr Maths
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| I no, the answer is wrong, at least i tried right? lol
| | | | Vuong | 02:33:18, 09 Nov 07 | Dr Maths
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| I have an answer, not sure if its right:
We have f(2) = 0 and f(0) = 1, so we need the constant term to be 1, and since it is a upside down parabola, we need a -x^2 term. Also differentiating twice would give us a minus number; hence we would get a maxima. f(2)=0 so we need an x term such that 2^2-1+x = 0.
First I tried f(x) = -(x^2) + (3x)/2 + 1, which gives f(0) = 1, f(2) = 0, but this did not work since differentiating would give x=3, i.e. a=3, which does not make sense.
So I tried f(x) = -(x^2)/2 + x/2 + 1, which gives f(0) = 1, f(2) = 0, differentiating gives f'(x) = -x + ½, putting f'(x) = 0, we get x = ½, and differentiating again gives f''(½) = -1, so x = ½ is a maxima, which is right. We also have to check the roots, f(x) = 0 gives x = 2 and x = -1 which seems correct.
Hence a = ½ right?
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