Question

Given three points (1,7) (4, -2) (5, -1) how do you write a quadratic function in standard form with the points?

Answer 1

`f(x) = x^2-8x+14`

Explanation:

The following method is slightly non-standard, but I find it fun:

Suppose the quadratic function is `f(x)`

Let `p = f(2)` and `q = f(3)`.

Consider the quadratic sequence: `f(1), f(2), f(3), f(4), f(5)`:

`7, p, q, -2, -1`

The sequence of differences of this sequence is an arithmetic progression:

`p - 7, q - p, -2-q, 1`

The sequence of differences of this sequence is a constant sequence:

`q - 2p + 7, p-2q-2, 3+q`

The sequence of differences of this sequence is all `0`:

`3p-3q-9, 3q-p+5`

So:

`{ (3p-3q-9 = 0), (3q-p+5 = 0) :}`

Hence `p = 2` and `q = -1`

So our original quadratic sequence becomes:

`color(blue)(7), 2, -1, -2, -1`

with sequence of differences:

`color(blue)(-5), -3, -1, 1`

with sequence of differences:

`color(blue)(2), 2, 2`

Having reached a constant sequence we can use the first term of each of these sequences as coefficients to find a formula for `f(x)`:

`f(x) = color(blue)(7)/(0!) + color(blue)(-5)/(1!)(x-1) + color(blue)(2)/(2!)(x-1)(x-2)`

`=7-5x+5+x^2-3x+2`

`=x^2-8x+14`

Answer 2

`f(x) = x^2-8x+14`

Explanation:

Here's a more standard way to find the quadratic:

`f(x) = ax^2+bx+c`

Each of the three points satisfies this equation, so we find:

`{ (7 = a+b+c), (-2 = 16a+4b+c), (-1 = 25a+5b+c) :}`

This can be rearranged into a matrix:

`((1, 1, 1, 7), (16, 4, 1, -2), (25, 5, 1, -1))`

Perform some row operations to change the left hand side into a `3xx3` identity matrix:

Subtract `16xx"row"1` from `"row"2` and `25xx"row"1` from `"row"3` to get:

`((1, 1, 1, 7), (0, -12, -15, -114), (0, -20, -24, -176))`

Divide `"row"2` by `-12` and `"row3"` by `-20` to get:

`((1, 1, 1, 7), (0, 1, 5/4, 19/2), (0, 1, 6/5, 44/5))`

Subtract `"row"2` from `"row"3` to get:

`((1, 1, 1, 7), (0, 1, 5/4, 19/2), (0, 0, -1/20, -7/10))`

Multiply `"row"3` by `-20` to get:

`((1, 1, 1, 7), (0, 1, 5/4, 19/2), (0, 0, 1, 14))`

Subtract `"row"2` from `"row"1` to get:

`((1, 0, -1/4, -5/2), (0, 1, 5/4, 19/2), (0, 0, 1, 14))`

Add `1/4"row"3` to `"row"1` and subtract `5/4"row"3` from `"row"2` to get:

`((1, 0, 0, 1), (0, 1, 0, -8), (0, 0, 1, 14))`

Having got to an identity matrix on the left hand side, the right hand column gives us the values of `a, b, c`, namely `1, -8, 14`

So `f(x) = x^2-8x+14`

Answer 3

`p(x) = 14 - 8 x + x^2`

Explanation:

There is a technique known as Lagrange's interpolating polynomials (published in 1795) that consists in build a polynomial, using a set of basis functions `l_k(x)` such that this polynomial interpolates a given a table of distinct points. Or mathematically

`p(x) = sum_{k=1}^n y_k l_k(x)`

The interpolating basis functions have the following properties.

`l_k(x_k) = 1` and `l_k(x_j)= 0 forall j ne k`

having this in mind we assure that

`p(x_k) = y_k, k=1,2,cdots,n`

The basis construction is elegant and has the structure

`l_k(x) = ((x-x_1)(x-x_2)cdots(x-x_{k-1})(x-x_{k+1})cdots(x-x_n))/((x_k-x_1)(x_k-x_2)cdots(x_k-x_{x-1})(x_k-x_{k+1})cdots(x-x_n))`

or in short notation

`l_k(x) = (Pi_{j ne k}(x-x_j))/(Pi_{j ne k}(x_k-x_j))`

In our case we have the table of points

`x_1 = 1, y_1 = 7`
`x_2 = 4, y_2 = -2`
`x_3 = 5, y_3 = -1`

The basis functions are

`l_1(x) = ((x-x_2)(x-x_3))/((x_1-x_2)(x_1-x_3)) = ((x-4)(x-5))/((1-4)(1-5))`
`l_2(x) = ((x-x_1)(x-x_3))/((x_2-x_1)(x_2-x_3)) = ((x-1)(x-5))/((4-1)(4-5))`
`l_3(x) =((x-x_1)(x-x_2))/((x_3-x_1)(x_3-x_2))= ((x-1)(x-4))/((5-1)(5-4))`

and finally

`p(x) = 7 l_1(x)-2l_2(x)-1l_3(x)`

or simplifying

`p(x) = 14 - 8 x + x^2`