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. factor out 3,
$ f(x)=3(x^{2}+2x) \quad$... complete the square
$f(x)=3(x^{2}+2x+1-1)$
$f(x)=3(x+1)^{2}-3$
b.
vertex: $(h,k)=(-1, -3)$,
a=$+3$, opens up, the vertex is a minimum
y-intercept: f(0) = $0$
x-intercepts: f(x)=0
$3x^{2}+6x=0$
$3x(x+2)=0$
x-intercepts: 0 and $-2$.
c.
see image
(one pair of additional points, either side of the vertex).
d.
domain: all reals, $\mathbb{R}$
range: $[-3,\infty)$