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

Answer:

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)]#