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 $-1$,
$ f(x)=-(x^{2}+4x) +4 \quad$... complete the square
$f(x)=-(x^{2}+3(3)x+2^{2}-2^{2}) +4$
$f(x)=-(x+2)^{2}+4+4$
$f(x)=-(x+2)^{2}+8$
b.
vertex: $(h,k)=(-2, 8)$,
a=$-1$, opens down, the vertex is a maximum
y-intercept: f(0) = $4$
x-intercepts: f(x)=0
$-(x+2)^{2}+8=0$
$(x+2)^{2}=8\qquad/\sqrt{..}$
$x+2=\pm 2\sqrt{2}$
$x=-2\pm 2\sqrt{2}$
c.
see image
(one pair of additional points, either side of the vertex).
d.
domain: all reals, $\mathbb{R}$
range: $(-\infty,8]$