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)=(2x^{2}+4x) +3\quad $ factor out $2$,
$ f(x)=2(x^{2}+2x) +3 \quad$... complete the square
$f(x)=2(x^{2}+2(1)x+1^{2}-1^{2})+3$
$f(x)=2(x+1)^{2}-2+3$
$f(x)=2(x+1)^{2}+1$
b.
vertex: $(h,k)=(-1, 1)$,
a=$+2$, opens up, the vertex is a minimum
y-intercept: f(0) = $3$
x-intercepts: f(x)=0
$(x-1)^{2}+1=0$
$(x-1)^{2}=-1$
(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: $[1,\infty)$