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}-2x) +2 \quad$... complete the square
$f(x)=(x^{2}-2(1)x+1^{2}-1^{2})+2$
$f(x)=(x-1)^{2}-1+3$
$f(x)=(x-1)^{2}+2$
b.
vertex: $(h,k)=(1, 1)$,
a=$+1$, opens up, the vertex is a minimum
y-intercept: f(0) = $2$
x-intercepts: f(x)=0
$(x-1)^{2}+2=0$
$(x-1)^{2}=-2$
(square can not be negative, no solutions)
x-intercepts: none
c.
see image
(two pairs of additional points, either side of the vertex).
d.
domain: all reals, $\mathbb{R}$
range: $[1,\infty)$