Answer
please see "step by step"
Work Step by Step
Rewrite f(x) in standard form, $f(x)=a(x-h)^{2}+k$,
read the vertex, (h,x)
For the y-intercept, calculate f(0)
For the x- intercept, solve f(x) = 0 for x.
If $a>0$, parabola opens up, the vertex is a minimum point,
If $a<0$, parabola opens down, the vertex is a maximum.
With this information (and possible additional points) sketch a graph
Read the graph for range and domain.
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a.
$ f(x)=x^{2}-6x \quad$... complete the square
$f(x)=x^{2}-2(3)x+3^{2}-3^{2}$
$f(x)=(x^{2}-6x+3^{2})-9$
$f(x)=(x-3)^{2}-9$
b.
vertex: $(h,k)=(3, -9)$,
a=1, opens up, the vertex is a minimum
y-intercept: f(0) = $0$
x-intercepts: f(x)=0
$x^{2}-6x =0$
$x(x-6)=0$
x-intercepts: 0 and 6.
c.
see image
(one pair of additional points, either side of the vertex).
d.
domain: all reals, $\mathbb{R}$
range: $[-9,\infty)$
