Answer
(a) Critical number: $x=2$
(b) Increasing: $(2,\infty)$
Decreasing: $(-\infty,2)$
(c) Relative minimum: $(2,-4)$
Work Step by Step
To find critical numbers, find f'(x) and set it equal to zero:
$f(x)=x^2-4x$
$f'(x)=2x-4$
$0=2x-4$
$x=2$
Test numbers above and below 2
$f'(3)=2(3)-4$
$f'(3)=2>0$
Therefore increasing on $(2,\infty)$
$f'(1)=2(1)-4$
$f'(1)=-2<0$
Therefore decreasing on $(-\infty,2)$
Since the graph is increasing on $(2,\infty)$ and decreasing on $(-\infty,2)$ there is a relative minimum at 2, now find f(2):
$f(2)=(2)^2-4(2)$
$f(2)=4-8$
$f(2)=-4$
Therefore the relative minimum is $(2,-4)$
