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-1 $
$ f(x)=(x^{2}+4x) -1 \quad$... complete the square
$f(x)=(x^{2}+4x+2^{2}-2^{2})-1$
$f(x)=(x^{2}+4x+2^{2})-4-5$
$f(x)=(x+2)^{2}-5$
b.
vertex: $(h,k)=(-2, -5)$,
a=1, opens up, the vertex is a minimum
y-intercept: f(0) = $-1$
x-intercepts: f(x)=0
$(x+2)^{2}-5=0$
$(x+2)^{2}=5\qquad.../\sqrt{..}$
$x+2=\pm\sqrt{5}$
$x=-2\pm\sqrt{5}$
c.
see image
(one pair of additional points, either side of the vertex).
d.
domain: all reals, $\mathbb{R}$
range: $[-5,\infty)$