SOLUTION: find the equation of the parabola With the vertex on the y axis, axis of symmetry parallel to the x axis, and passes through (2, 2), (8,-1)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: find the equation of the parabola With the vertex on the y axis, axis of symmetry parallel to the x axis, and passes through (2, 2), (8,-1)      Log On


   



Question 401658: find the equation of the parabola With the vertex on the y axis, axis of symmetry parallel to the x axis, and passes through (2, 2), (8,-1)
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
find the equation of the parabola With the vertex on the y axis, axis of symmetry parallel to the x axis, and passes through (2, 2), (8,-1)
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If vertex on y-axis it passes through (0,y)
If axis of symmetry parallel to x-axis it passes thru
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Form of equation: y = a(x-h)^2+k
(0,y) implies y = a(x)^2+k
(2,2) implies 2 = a(2)^2+k = 4a+k
(8,-1) implies -1 = a(8)^2+k = 64a+k
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Solve the system:
4a+k = 2
64a+k = -1
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Subtract top from bottom to get:
60a = -3
a = -1/20
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Solve for "k":
4a+k = 2
4(-1/20) + k = 2
(-1/5) + k = 2
k = 11/5
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Equation:
y = (-1/20)x^2 + (11/5)
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Cheers,
Stan H.