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:
$y = 5/96 (x^2 + 22x + 121) + 6$

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

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

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