Question

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

Answer 1

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

Explanation:

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`