Question

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

Answer 1

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`.