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

1 Answer
Nghi N.
Nov 24, 2015

Find the equation of a parabola.

Ans: y = 2x^2 - 40x + 208y=2x2−40x+208

Explanation:

General equation of the parabola: y = ax^2 + bx + c.y=ax2+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)) = 10x=−(b2a)=10 --> b = -20ab=−20a (1)
y-coordinate of vertex: y = y(10) = (10)^2a + 10b + c = 8 =y=y(10)=(10)2a+10b+c=8=
= 100a + 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 = 2a=2 --> b = -20a = -40b=−20a=−40
From (3) --> 50 - 200 + c = 58 --> c = 258 - 50 = 208c=258−50=208
Equation of the parabola: y = 2x^2 - 40x + 208y=2x2−40x+208.

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