Question

What is the range of a quadratic function?

Answer 1

The range of `f(x) = ax^2+bx+c` is:

`{ ([c-b^2/(4a), oo) " if " a > 0), ((-oo, c-b^2/(4a)] " if " a < 0) :}`

Explanation:

Given a quadratic function:

`f(x) = ax^2+bx+c" "` with `a != 0`

We can complete the square to find:

`f(x) = a(x+b/(2a))^2+(c-b^2/(4a))`

For real values of `x` the squared term `(x+b/(2a))^2` is non-negative, taking its minimum value `0` when `x = -b/(2a)`.

Then:

`f(-b/(2a)) = c - b^2/(4a)`

If `a > 0` then this is the minimum possible value of `f(x)` and the range of `f(x)` is `[c-b^2/(4a), oo)`

If `a < 0` then this is the maximum possible value of `f(x)` and the range of `f(x)` is `(-oo, c-b^2/(4a)]`

Another way of looking at this is to let `y = f(x)` and see if there's a solution for `x` in terms of `y`.

Given:

`y = ax^2+bx+c`

Subtract `y` from both sides to find:

`ax^2+bx+(c-y) = 0`

The discriminant `Delta` of this quadratic equation is:

`Delta = b^2-4a(c-y) = (b^2-4ac)+4ay`

In order to have real solutions, we require `Delta >= 0` and so:

`(b^2-4ac)+4ay >= 0`

Add `4ac-b^2` to both sides to find:

`4ay >= 4ac-b^2`

If `a > 0` then we can simply divide both sides by `4a` to get:

`y >= c-b^2/(4a)`

If `a < 0` then we can divide both sides by `4a` and reverse the inequality to get:

`y <= c-b^2/(4a)`