What is the equation of the parabola that has a vertex at # (-11, 6) # and passes through point # (13,36) #?

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Answer 1 · 2017-04-17T16:44:37

#y = 5/96(x+11)^2 + 6#
or
#y = 5/96 x^2 + 55/48x + 1181/96#

The standard form of a parabola is #y = a(x-h)^2 + k#, where #a# is a constant, vertex is #(h, k)# and the axis of symmetry is #x = h#.

Solve for #a# by substituting #h = -11, k = 6 " & " x = 13, y = 36#:

#36 = a(13+11)^2 + 6#

#36 = 576a + 6#

#30 = 576a#

#a = 30/576 = 5/96#

Equation in standard form is #y = 5/96(x+11)^2 + 6#

General form is #y = Ax^2 + Bx +C#

Distribute right side of the equation:
#y = 5/96 (x^2 + 22x + 121) + 6#

#y = 5/96 x^2 + 55/48x + 605/96 + 6#

#y = 5/96 x^2 + 55/48x + 1181/96#