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}-6x) +4 \quad$... complete the square
$f(x)=-(x^{2}-2(3)x+3^{2}-3^{2})+4$
$f(x)=-(x-3)^{2}+9+4$
$f(x)=-(x-3)^{2}+13$
b.
vertex: $(h,k)=(3, 13)$,
a=$-1$, opens down, the vertex is a maximum
y-intercept: f(0) = $4$
x-intercepts: f(x)=0
$-(x-3)^{2}+13=0$
$(x-3)^{2}=13\qquad/\sqrt{..}$
$x-3=\pm\sqrt{13}$
$x=3\pm\sqrt{13}$
c.
see image
(one pair of additional points, either side of the vertex).
d.
domain: all reals, $\mathbb{R}$
range: $(-\infty,13]$
