What is the equation of the parabola that has a vertex at # (10, 8) # and passes through point # (5,58) #?

1 Answer
Nghi N.
Nov 24, 2015

Find the equation of a parabola.

Ans: #y = 2x^2 - 40x + 208#

Explanation:

General equation of the parabola: #y = ax^2 + bx + c.#
There are 3 unknowns: a, b, and c. We need 3 equations to find them.
x-coordinate of vertex (10, 8): #x = - (b/(2a)) = 10# --># b = -20a# (1)
y-coordinate of vertex: #y = y(10) = (10)^2a + 10b + c = 8 =#
#= 100a + 10b + c = 8# (2)
Parabola passes through point (5, 58)
y(5) = 25a + 5b + c = 58 (3).
Take (2) - (3) :
75a + 5b = -58. Next, replace b by (-20a) (1)
75a - 100a = -50
-25a = -50 --> #a = 2# --> #b = -20a = -40#
From (3) --> 50 - 200 + c = 58 --> #c = 258 - 50 = 208#
Equation of the parabola: #y = 2x^2 - 40x + 208#.

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