Question

What is the equation of the parabola that has a vertex at `(-3, 6)` and passes through point `(1,9)`?

Answer 1

`f(x) = 3/16x^2+ 9/8x + 123/16`

Explanation:

The parabola `f` is written as `ax^2 + bx + c` such that `a != 0`.

1st of all, we know this parabol has a vertex at `x=-3` so `f'(-3) = 0`. It already gives us `b` in function of `a`.

`f'(x) = 2ax + b` so `f'(-3) = 0 iff -6a + b = 0 iff b = 6a`

We now have to deal with two unknown parameters, `a` and `c`. In order to find them, we need to solve the following linear system :

`6 = 9a - 18a + c; 9 = a + 6a + c iff 6 = -9a + c;9 = 7a + c`

We now substract the 1st line to the 2nd one in the 2nd line :

`6 = -9a + c;3 = 16a` so we now know that `a = 3/16`.

We replace `a` by its value in the 1st equation :

`6 = -9a + c iff c = 6 + 9*(3/16) iff c = 123/16` and `b = 6a iff b = 9/8`.