Question

How do you find the quadratic function `y=ax^2+bx+c` whose graph passes through the given points (1,-1), (-1, 5), (2,10)?

Answer 1

`y = 14/3x^2-3x-8/3`

Explanation:

There are several ways to do this, but here's a very standard one:

Substitute the given points into the quadratic equation to get three simultaneous equations in `a`, `b` and `c`:

`{ (a+b+c = -1), (a-b+c = 5), (4a+2b+c = 10) :}`

Subtracting the second equation from the first, we get:

`2b = -6`

and hence `b = -3`

Subtracting the first equation from the third equation, we get:

`3a+b = 11`

Substituting `b = -3` and adding `3` to both sides, we get:

`3a = 14`

Hence `a = 14/3`

Then from the first equation:

`c = -1-a-b = -1-14/3+3 = -8/3`

So:

`y = 14/3x^2-3x-8/3`

graph{(y - (14/3x^2-3x-8/3))(9(x-1)^2+(y+1)^2-0.01)(9(x+1)^2+(y-5)^2-0.01)(9(x-2)^2+(y-10)^2-0.01) = 0 [-3, 3, -4, 12]}

Answer 2

`y = 14/3 x^2-6x-8/3`

Explanation:

Here's another method, which is a little more sophisticated and very useful...

Given points: `(1, -1)`, `(-1, 5)`, `(2, 10)`, consider the following polynomials formed from the `x` coordinates:

`f_1(x) = ((x-(color(blue)(-1)))(x-color(blue)(2)))/((color(blue)(1)-(color(blue)(-1)))(color(blue)(1)-color(blue)(2))) = -1/2(x^2-x-2)`

`f_2(x) = ((x-color(blue)(1))(x-color(blue)(2)))/(((color(blue)(-1))-color(blue)(1))((color(blue)(-1))-color(blue)(2))) = 1/6(x^2-3x+2)`

`f_3(x) = ((x-color(blue)(1))(x-(color(blue)(-1))))/((color(blue)(2)-color(blue)(1))(color(blue)(2)-(color(blue)(-1)))) = 1/3(x^2-1)`

These are constructed so that:

`f_1(1) = 1" "f_1(-1) = 0" "f_1(2) = 0`

`f_2(1) = 0" "f_2(-1) = 1" "f_2(2) = 0`

`f_3(1) = 0" "f_3(-1) = 0" "f_3(2) = 1`

We can then add together a linear combination of these polynomials to match the required `y` coordinates, using the `y` coordinates of the three points:

`y = (color(blue)(-1))f_1(x) + color(blue)(5)f_2(x) + color(blue)(10)f_3(x)`

`color(white)(y) = (-1)(-1/2(x^2-x-2)) + 5(1/6(x^2-3x+2)) + 10(1/3(x^2-1))`

`color(white)(y) = 1/2(x^2-x-2) + 5/6(x^2-3x+2) + 10/3(x^2-1)`

`color(white)(y) = 14/3 x^2-6x-8/3`