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}+4x) +3 \quad$... complete the square
$f(x)=(x^{2}+2(2)x+2^{2}-2^{2})+3$
$f(x)=(x+2)^{2}-4+3$
$f(x)=(x+2)^{2}-1$
b.
vertex: $(h,k)=(-2, -1)$,
a=$+1$, opens up, the vertex is a minimum
y-intercept: f(0) = $3$
x-intercepts: f(x)=0
$(x-2)^{2}-1=0$
$(x-2)^{2}=1\qquad/\sqrt{..}$
$x-2=\pm 1$
$x=-2\pm 1$
x-intercepts: $-3$ and $-1$.
c.
see image
(one pair of additional points, either side of the vertex).
d.
domain: all reals, $\mathbb{R}$
range: $[-1,\infty)$
