Question

What is the equation of the parabola that has a vertex at `(7, 9)` and passes through point `(0, 2)`?

Answer 1

`y = -1/7 (x - 7)^2 + 9`

Explanation:

This problem requires that we understand how a function can be shifted around and stretched to meet particular parameters. In this case, our basic function is `y =x^2`. This describes a parabola which has its vertex at `(0,0)`. However we can expand it as:

`y = a(x + b)^2 + c`

In the most basic situation:
`a = 1`
`b = c = 0`

But by altering these constants, we can control the shape and position of our parabola. We will start with the vertex. Since we know it needs to be at `(7,9)` we need to shift the default parabola to the right by `7` and up by `9`. That means manipulating the `b` and `c` parameters:

Obviously `c = 9` because that will mean all `y` values will increase by `9`. But less obviously, `b = -7`. This is because when we add a factor to the `x` term, the shift will be opposite that factor. We can see that here:
`x + b = 0`
`x = -b`

When we add `b` to `x`, we move the vertex to `-b` in the `x` direction.

So our parabola so far is:
`y = a(x - 7)^2 + 9`

But we need to stretch it to pass through point `(0,2)`. This is as simple as plugging in those values:
`2 = a(-7)^2 +9`
`2 = 49a + 9`
`-7 = 49a`
`a = -1/7`

That means our parabola will have this equation:
`y = -1/7 (x - 7)^2 + 9`