Question

How do you find the equation of a parabola given the coordinates of the vertex: (2,3) point : (-1,6)?

Answer 1

Assuming the axis of symmetry is vertical
then the axis of symmetry is `x=2`
and since `(-1,6) =(2-3,6)` is a given point on the parabola
then so is `(2+3,6)=(5,6)`

The general form of the equation for a parabola (with a vertical axis of symmetry) is
`ax^2+bx+c = y`

We can substitute our three points that we know into this general form:

`a(-1)^2+b(-1)+c = 6`

`a(2)^2+b(2)+c = 3`

`a(5)^2+b(5)+c=6`

and solve for `a, b, " and " c`

obtaining
`a=1/3, b=-4/3, " and " c=13/3`

So the equation is
`y = 1/3x^2-4/3x+13/3`

Note if the axis of symmetry is horizontal, a similar method can be used; but if the axis of symmetry is other than vertical or horizontal there is insufficient information to derive the parabolic equation.

Answer 2

Assuming that it is a vertical parabola, the equation looks like:

`y=a(x-h)^2+k` where the vertex is `(h, k)`.

So this parabola has equation:

y=a(x-2)^2+3#

Knowing that `(-1,6)` lies on the parabola also tells us that when I put in `-1` for `x` and `6` for `y`, the equation will be true. That lets us find `a`

`6=a(-1-2)^2+3`

`6=9a+3`

`9a = 3`, so

`a=1/3`

The desired equation is: `y=1/3 (x-2)^2+3`

If the parabola is horizontal, the answer is different.