Question

Given three points (1,-2), (2,-4) and (3,-4), how do you write a quadratic function in standard form with the points?

Answer 1

`f(x) = x^2 -5x + 2`

Explanation:

Since the `x` coordinates of the points are `1`, `2` and `3`, this problem is equivalent to finding a quadratic formula for the terms of the sequence:

`color(blue)(-2), -4, -4`

Write down the sequence of differences of this sequence:

`color(blue)(-2), 0`

Write down the sequence of differences of this sequence:

`color(blue)(2)`

Then we can construct a formula for `a_n` (`n = 1, 2, 3`) using the initial term of each of these sequences as coefficients:

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

`=-2-2n+2+n^2-3n+2`

`=n^2-5n+2`

This corresponds to the standard quadratic function:

`f(x) = x^2-5x+2`

Answer 2

`f(x) = x^2-5x+2`

Explanation:

Alternatively, consider the general problem of finding the equation of a parabola with vertical axis passing through points `(x_1, y_1)`, `(x_2, y_2)` and `(x_3, y_3)`.

Consider a definition of the form:

`f(x) = a + b (x-x_1) + c (x-x_1)(x-x_2)`

where we want to determine `a`, `b` and `c`.

Since `f(x_1) = y_1` we must have `a = y_1`

Then `y_2 = f(x_2) = y_1 + b(x_2 - x_1)`

So `b = (y_2 - y_1)/(x_2 - x_1)`

Then `y_3 = f(x_3) = y_1 + (y_2 - y_1)/(x_2 - x_1)(x_3-x_1) + c(x_3-x_1)(x_3-x_2)`

Hence `c = (y_3 - y_1- (y_2 - y_1)/(x_2 - x_1)(x_3-x_1))/((x_3-x_1)(x_3-x_2))`

`=(y_3-y_1)/((x_3-x_1)(x_3-x_2)) - (y_2-y_1)/((x_2-x_1)(x_3-x_2))`

`color(white)()`
Turning to our example: `(x_1, y_1) = (1, -2)`, `(x_2, y_2) = (2, -4)` and `(x_3, y_3) = (3, -4)`

Plugging these values into the formulae for `a, b, c` we find:

`{ (a = -2), (b = -2), (c = 1) :}`

So:

`f(x) = -2-2(x-1)+(x-1)(x-2)`

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

`=x^2-5x+2`

Answer 3

`f(x) = x^2-5x+2`

Explanation:

We want to find a quadratic function `f(x)` passing through the points `(1, -2)`, `(2, -4)` and `(3, -4)`.

Since `f(2) = f(3) = -4` the vertical axis must lie half way between these `x` coordinates, at `x = 5/2`.

So `f(x) = a(x - 5/2)^2 + b` for some `a` and `b`.

Then from `f(1) = -2` we get:

`-2 = a(1-5/2)^2 + b = 9/4 a + b`

and from `f(2) = -4` we get:

`-4 = a(2-5/2)^2 + b = 1/4 a + b`

Subtracting this second equation from the first we find:

`2a = 2`

So `a=1`

and `b = -2 - 9/4a = -2 - 9/4 = -17/4`

So:

`f(x) = (x-5/2)^2-17/4 =x^2-5x+25/4-17/4 = x^2-5x+2`