Question

How do you find domain and range of a quadratic function?

Answer 1

The domain is all real numbers and the range is all the reals at or above the vertex y coordinate (if the coefficient on the squared term is positive) or all the reals at or below the vertex (if said coefficient is negative).

Answer 2

See explanation...

Explanation:

Suppose:

`f(x) = ax^2+bx+c \` where `a != 0`

First note that `f(x)` is well defined for any value of `x`. So the domain of `f(x)` is `RR`.

We can complete the square and find:

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

This is vertex form for a parabola with vertex at `(-b/(2a), c-b^2/(4a))` and multiplier `a`.

Note that for real values of `x`, we have:

`(x+b/(2a))^2 >= 0`

with equality when `x = -b/(2a)`.

Hence if `a > 0` then `f(x) >= c - b^2/(4a)`.

In fact for any `y >= c - b/(2a)` we have:

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

So the range of `f(x)` is `[c-b^2/(4a), oo)`

Similarly if `a < 0` then `f(x) <= c - b^2/(4a)` and for any `y < c - b/(2a)` we have:

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

So the range of `f(x)` is `(-oo, c-b^2/(4a)]`