Question

What is the equation of the parabola that has a vertex at `(-2, -5)` and passes through point `(2,6)`?

Answer 1

The equation of parabola is `y = 11/16(x+2)^2 -5`

Explanation:

Vertex `(h= -2,k= -5)` The equation of parabola is `y=a(x-h)^2+k or y = a(x+2)^2 -5`
The point `(2,6)` lies on parabola. `:. 6 = a*(2+2)^2 -5 or 16a =11 or a =11/16`
Hence the equation of parabola is `y = 11/16(x+2)^2 -5` graph{11/16(x+2)^2-5 [-10, 10, -5, 5]}[Ans]