Question

How do you write the quadratic function in standard form given points `(-2,-1)`, `(1,11)`, `(2, 27)`?

Answer 1

`f(x) = 3x^2+7x+1`

Explanation:

Given points `(x_1, y_1)`, `(x_2, y_2)` and `(x_3, y_3)`, the quadratic function passing through them can be written as:

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

This works because each rational expression like:

`((x-x_2)(x-x_3))/((x_1-x_2)(x_1-x_3))`

is constructed so that it takes the value `1` at the selected `x` coordinate (`x_1` in this case) and `0` at the two other `x` coordinates (`x_2` and `x_3` in this case).

See https://socratic.org/s/aAEj5Fvc for another example of this method.

In our example:

`(x_1, y_1) = (-2, -1)`

`(x_2, y_2) = (1, 11)`

`(x_3, y_3) = (2, 27)`

So:

`f(x) = (-1)((x-1)(x-2))/(((-2)-1)((-2)-2))+(11)((x-(-2))(x-2))/((1-(-2))(1-2))+(27)((x-(-2))(x-1))/((2-(-2))(2-1))`

`color(white)(f(x)) = (-1)(x^2-3x+2)/12+(11)(x^2-4)/(-3)+(27)(x^2+x-2)/4`

`color(white)(f(x)) = 1/12(-x^2+3x-2-44x^2+176+81x^2+81x-162)`

`color(white)(f(x)) = 1/12(36x^2+84x+12)`

`color(white)(f(x)) = 3x^2+7x+1`

Answer 2

`f(x) = 3x^2+7x+1`

Explanation:

Given points:

`(-2, -1)`, `(1, 11)`, `(2, 27)`

We want to find a function of the form:

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

passing through all three points.

Substitute the coordinates of the three points into the formula for `f(x)` to get `3` simultaneous linear equations:

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

`11 = a(1)^2+b(1)+c = a+b+c`

`27 = a(2)^2+b(2)+c = 4a+2b+c`

Subtracting the first of these three equations from the third, we get:

`28 = 4b`

So:

`b = 7`

Adding the first and third equations, we get:

`26 = 8a+2c`

Then subtracting twice the second equation from this one, we get:

`4 = 6a-2b = 6a-14`

Add `14` to both ends to get:

`18 = 6a`

Hence:

`a = 3`

Then from the second equation:

`c = 11 - a - b = 11-7-3 = 1`

So:

`f(x) = 3x^2+7x+1`