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)=-3x^{2}+6x-2 \qquad $factor out $-3$,
$ f(x)=-3(x^{2}-2x) -2 \quad$... complete the square
$f(x)=-3(x^{2}-2(1)x+1^{2}-1^{2}) -2 $
$f(x)=-3(x-1)^{2}+3-2$
$f(x)=-3(x-1)^{2}+1$
b.
vertex: $(h,k)=(1, 1)$,
a=$-3$, opens down, the vertex is a maximum
y-intercept: f(0) = $-2$
x-intercepts: f(x)=0
$-3(x-1)^{2}+1=0$
$1=3(x-1)^{2}\qquad/\div 3$
$(x-1)^{2}=\displaystyle \frac{1}{3}\qquad/\sqrt{..}$
$x-1=\pm\sqrt{\frac{1}{3}}$
$x=1\pm\sqrt{\frac{1}{3}}$
c.
see image
(one pair of additional points, either side of the vertex).
d.
domain: all reals, $\mathbb{R}$
range: $(-\infty,1]$
