Question

What is the equation of the parabola that has a vertex at `(3, -5)` and passes through point `(1,-2)`?

Answer 1

`8y = x^2 - 6x - 11`

Explanation:

Set up simultaneous equations using the coordinates of the two points, and then solve.
`y = ax^2 +bx + c` is the general formula of a parabola
The vertex is(`-b/(2a)`, `(4ac - b^2)/(2a)`)
Therefore `-b/(2a)` = 3 and `(4ac - b^2)/(2a) = -5`
and from the other point `-2 = a.1^2 +b.1 + c`
Hence`a + b + c = -2`
`c = -2 - a - b`
`b = -6a`
`c = -2 - a +6a` = -2 +5a#

`-5 = (4a(-2+5a)- (-6a)^2)/(2a)`
`-5 = 2(-2+5a) -18a`
`-5 =-4 -8a`
`8a = 1`
`a = 1/8`
`b = -6/8`
`c = -2 +5/8 = -11/8`
`8y = x^2 -6x -11`