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

2 Answers
Jun 28, 2018

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).

Jun 28, 2018

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)]