Question

Given three points (-2,2), (-1,-1),(2,6) how do you write a quadratic function in standard form with the points?

Answer 1

`f(x) =4/3x^2+x-4/3`

Explanation:

Here's an alternative method that I think is quite fun.

Suppose the quadratic function we want to find is `f(x)`

We are told:

`f(-2) = 2`

`f(-1) = -1`

`f(2) = 6`

Also let:

`f(0) = p`

`f(1) = q`

Then form a sequence with the values of `f(-2), f(-1), f(0), f(1), f(2)`:

`2, -1, p, q, 6`

Since these are values of a quadratic function for an equally spaced sequence of values of `x`, then if we form the sequence of differences between pairs of terms, then the sequence of differences of those differences, then the sequence of differences of those differences, both of the terms of the final sequence will be `0`.

So starting with:

`2, -1, p, q, 6`

write down the sequence of differences:

`-3, p+1, q-p, 6-q`

then the sequence of differences of those differences:

`p+4, q-2p-1, p-2q+6`

then the sequence of differences of those differences:

`q-3p-5, 3p-3q+7`

Both of these last two terms must be `0`, so we have:

`{ (q-3p-5 = 0), (3p-3q+7 = 0) :}`

If we add these two equations together we find:

`-2q+2 = 0`

Hence:

`q=1`

Then from the second equation we find:

`3p-3+7 = 0`

Hence:

`p = -4/3`

So our original sequence `f(-2), f(-1), f(0), f(1), f(2)` looks like this:

`2, -1, -4/3, 1, 6`

Next write out just the sequence `f(0), f(1), f(2)` with the values we have found:

`color(blue)(-4/3), 1, 6`

then the sequence of differences:

`color(blue)(7/3), 5`

then the sequence of differences of differences:

`color(blue)(8/3)`

We can then use the initial term of each of these sequences to write out a direct formula for the function:

`f(x) = color(blue)(-4/3)(1/(0!)) + color(blue)(7/3)(x/(1!)) + color(blue)(8/3)((x(x-1))/(2!))`

`=-4/3+7/3x+4/3x^2-4/3x`

`=4/3x^2+x-4/3`