Question

What is the equation of the parabola that has a vertex at `(3, -3)` and passes through point `(0, 6)`?

Answer 1

`x^2-9x+18=0`

Explanation:

let's take the equation of the parabola as `ax^2+bx+c=0` `a,b,c in RR`
two points are given as `(3,-3)` and `(0,6)`

just by looking at the two points, we can tell where the parabola intercepts the `y` axis. when the `x` coordinate is `0` the `y` coordinate is `6`.
from this, we can deduce that `c` in the equation we took is `6`

now we only have to find the `a` and `b` of our equation.

since the vertex is `(3,-3)` and the other point is `(0,6)` the graph spreads above the `y=-3` line. hence this parabola has a exact minimum value and goes up to the `oo`. and parabolas which has a minimum value has a `+` value as the `a`.

this is a tip which is useful to remember.
- if the co-efficient of `x^2` is positive then the parabola has a minimum value.
- if the co-efficient of `x^2` is negative then the parabola has a maximum value.

back to our problem,
since the vertex is `(3,-3)` the parabola is symmetrical around `x=3`
so the symmetrical point of (0,6) on the parabola would be (6,6)

so now we have three points altogether. i'm going to substitute these points to the equation we took and then i just have to solve the simultaneous equations i get.

substituting point (3,-3) `9a+3b+6=0`
substituting point (6,6) `36a+6b+6=0`

`3a -1=0`
`a= 1/3`

`b=-3`

so the equation is `1/3x^2-3x+6=0`
make the equation look more nicer,
`x^2-9x+18=0`

graph{x^2-9x+18 [-10, 10, -5, 5]}