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What is the equation of a parabola opening upwards that passes through the points...

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lxsptter | Student, Undergraduate | (Level 2) Valedictorian

Posted June 15, 2013 at 5:47 PM via web

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What is the equation of a parabola opening upwards that passes through the points (0,0), (3, 2) and (4, 6)

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted June 15, 2013 at 5:56 PM (Answer #1)

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The equation of a parabola opening upwards is y - h = a*(x - k)^2.

Substituting the co-ordinates of the points through which the parabola passes gives the following equations:

0 - h = a*(0 - k)^2 ...(1)

2 - h = a*(3 - k)^2 ...(2)

6 - h = a*(4 - k)^2 ...(3)

From (1), h = -ak^2

Substituting in (2) and (3)

2 + ak^2 = a(9 + k^2 - 6k) and 6 + ak^2 = a*(16 - 8k + k^2)

=> 2 = 9a - 6ak and 6 = 16a - 8ak

Substitute a = 2/(9 - 6k) in 6 = 16a - 8ak

=> 6 = 16*2/(9 - 6k) - 8*2/(9 - 6k)*k

=> 6*(9 - 6k) = 32 - 16k

=> 54 - 36k = 32 - 16k

=> k = 1.1

a = 5/6 and h = -11/12

The equation of the parabola is y + 11/12 = (5/6)*(x - 1.1)^2

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