Question

Find the equation of parabola whose axis of symmetry is `x=-3` and passes through points `(-1,3)` and `(2,-5)`?

Answer 1

Please see below.

Explanation:

Given that the axis is `x = -3`, we know the vertex is at `x = -3`, so the equation can be written in the form

`y = a(x+3)^2 +k`

Now use the two points given to get:

From `(-1,3)`, we have `3 = a(2)^2+k` so `4a+k = 3` and `k = 3-4a`

From `(2,-5)`, we get `-5 = a(5)^2+k` so `25a+k=-5`

Replace `k` in the last equation with `3-4a` to get

`25a+(3-4a) = -5`, so

`21a = -8` and `a = -8/21`

Using `k = 3-4a` again we get `k = 3-(-32/21) = 63/21+32/21 = 95/21`.

The equation is `y = -8/21(x+3)^2 + 95/21`

If course, there are other ways to write the answer.

graph{(y+8/21(x+3)^2-95/21)(x+3)((x+1)^2+(y-3)^2-0.02)((x-2)^2+(y+5)^2-0.02)=0 [-12, 12, -6, 6]}