Answer
a.
vertex: $(h,k)=(-1, -4)$
y-intercept: $-1$
x-intercepts: $-1\displaystyle \pm\frac{2\sqrt{3}}{3}$
b. $a>0$,
f has a minimum value at the vertex, $f(-1)=-4$
c.
domain: all reals, $\mathbb{R}$
range: $[-4,\infty)$
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, calcucate 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
Read the graph for range and domain.
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a.
$ f(x)=3x^{2}+6x-1 \quad$... factor out $3$
$ f(x)=3(x^{2}+2x) -1 \quad$... complete the square
$=3(x^{2}+2x+1-1)-1$
$=3(x^{2}+2x+1)-3-1$
$=3(x+1)^{2}-4$
vertex: $(h,k)=(-1, -4)$
y-intercept: f(0) = $-1$
x-intercepts: f(x)=0
$3(x+1)^{2}-4=0$
$3(x+1)^{2}=4$
$(x+1)^{2}=\displaystyle \frac{4}{3}\qquad/\sqrt{...}$
$x+1=\pm\sqrt{\frac{4}{3}}$
$x=-1\displaystyle \pm\frac{2}{\sqrt{3}}=-1\pm\frac{2\sqrt{3}}{3}$
b. $a>0$,
f has a minimum value at the vertex, $f(-1)=-4$
c.
domain: all reals, $\mathbb{R}$
range: $[-4,\infty)$