Question

How do you write a quadratic function in standard form whose graph passes through points (1,-1/2), (0,-3), (2,3)?

Answer 1

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

Explanation:

Notice that the distinct `x` coordinates are `{ 0, 1, 2 }`, which are in arithmetic progression with common difference `1`.

Let us write down the corresponding `y` coordinates in ascending order of `x` coordinate:

`color(blue)(-3)`, -1/2, 3#

Next, write down the sequence of differences between consecutive terms:

`color(blue)(5/2), 7/2`

Next, write down the sequence of differences between consecutive terms of that sequence:

`color(blue)(1)`

Having arrived at a constant sequence (albeit of just one term), we can use the first term of each of these sequences to write a formula for the desired function:

`f(x) = color(blue)(-3)/(0!) + color(blue)(5/2)/(1!)x + color(blue)(1)/(2!)x(x-1)`

`color(white)(f(x)) = -3+5/2x+1/2x^2-1/2x`

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

graph{(y - (1/2x^2+2x-3))(2x^2+(y+3)^2-0.02)(2(x-1)^2+(y+1/2)^2-0.02)(2(x-2)^2+(y-3)^2-0.02) = 0 [-9.59, 6, -6, 4.84]}