Answer
$a.\quad $see image
$ b.\quad$
The domain is $(-\infty,\infty)$
The range is $[-4,\infty)$
$ c.\quad$
Decreasing on $(-\infty,2)$
Increasing on $(2,\infty)$
Work Step by Step
$f(x)=x^{2}-4x$
$a=1,b=-4,c=0$
$ a.\quad$
Leading coefficient is positive - opens up.
Vertex:
$x=\displaystyle \frac{-b}{2a}=\frac{-(-4)}{2(1)}=\frac{4}{2}=2$
$f(2)=-4$
Vertex: $(2,-4)$
Axis of symmetry: the line $x=2$
Zeros $(x-$intercepts):
$x^{2}-4x=0$
$x(x-4)=0$
$x=0$ or $x=4$
$x-$intercepts: $(0,0),(4,0)$
y-intercept: (0,c)$ = (0,0)$
$ b.\quad$
The domain is $(-\infty,\infty)$
The range is $[-4,\infty)$
$ c.\quad$
Decreasing on $(-\infty,2)$
Increasing on $(2,\infty)$
