Question

How do you write a quadratic function in standard form whose graph passes through points (2,-4), (3,-7), (1,-3)?

Answer 1

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

Explanation:

Here's one method...

Putting the points in increasing order of `x` coordinate, we have:

`(1, -3)`, `(2, -4)`, `(3, -7)`

Note that the `x` coordinates are `1, 2, 3`.

Hence we can analyse this as a sequence of `y` values using the method of differences...

Write down the sequence of `y` values:

`color(blue)(-3), -4, -7`

Write down the sequence of differences between pairs of consecutive terms:

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

Write down the sequence of differences of differences:

`color(blue)(-2)`

Having reached a constant sequence (albeit of just one term), we can write a general formula for a term of the initial sequence of `y` values using the intial term of each of the above sequences as coefficients:

`a_n = color(blue)(-3)/(0!)+color(blue)(-1)/(1!)(n-1)+color(blue)(-2)/(2!)(n-1)(n-2)`

`color(white)(a_n) = -3-n+1-n^2+3n-2`

`color(white)(a_n) = -n^2+2n-4" "` for `n = 1,2,3`

Hence our function can be written:

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

Answer 2

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

Explanation:

Here's a general purpose method for matching a polynomial to a set of points...

The formula

Given a set of points:

`{ (x_1, y_1),...,(x_n, y_n) }`

with distinct values `x_1`, `x_2`, ... , `x_n`.

A polynomial that passes through them is:

`((x-x_2)(x-x_3)...(x-x_n))/((x_1-x_2)(x_1-x_3)...(x_1-x_n))y_1 +`

`((x-x_1)(x-x_3)...(x-x_n))/((x_2-x_1)(x_2-x_3)...(x_2-x_n))y_2 + ...`

`+ ((x-x_1)(x-x_2)...(x-x_(n-1)))/((x_n-x_1)(x_n-x_2)...(x_n-x_(n-1)))y_n`

`= sum_(k=1)^n (prod_(j = 1, j != k)^n (x-x_j)/(x_k-x_j)) y_k`

How does it work?

That may look rather complicated, but notice how it works:

Looking at the first expression:

`((x-x_2)(x-x_3)...(x-x_n))/((x_1-x_2)(x_1-x_3)...(x_1-x_n)) = prod_(j = 2)^n (x-x_j)/(x_1-x_j)`

Note that:

• It takes the value `1` when `x=x_1` since the numerator and denominator become identical.

• It takes the value `0` when `x=x_k` for any `k` apart from `1`, since one of the factors in the numerator will be `0`.

• It is a polynomial of degree `n-1`.

So when we multiply this by `y_1` we get a polynomial that takes the value `y_1` when `x=x_1` and `0` at each of `x_2`, `x_3`,...,`x_n`.

Summing this with the similar polynomials for `(x_2, y_2)`,...,`(x_n, y_n)` we get the behaviour we want.

Application

In the given example, with points:

`{ ((x_1, y_1) = (2, -4)), ((x_2, y_2) = (3, -7)), ((x_3, y_3) = (1, -3)) :}`

We can write:

`f(x) = ((x-3)(x-1))/((2-3)(2-1))(-4)+((x-2)(x-1))/((3-2)(3-1))(-7)+((x-2)(x-3))/((1-2)(1-3))(-3)`

`color(white)(f(x)) = 4(x^2-4x+3)-7/2(x^2-3x+2)-3/2(x^2-5x+6)`

`color(white)(f(x)) = -x^2+2x-4`