Answer
a.
vertex: $(h,k)=(1, -3)$
y: intercept: $-1$
x-intercepts: $1\pm\sqrt{\frac{3}{2}}$
b.
f has a minimum value at the vertex, $f(-2)=-3$
c.
domain: all reals, $\mathbb{R}$
range: $[-3,\infty)$
Work Step by Step
Rewrite f(x) in standard form, $f(x)=a(x-h)^{2}+k$,
read the vertex, (h,x)
For the x-intercept, calcucate f(0)
For the y- 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)=2x^{2}-4x-1 \quad$... factor out $2$
$ f(x)=2(x^{2}-2x) -1 \quad$... complete the square ...
$=2(x^{2}-2x+1-1)-1$
$=2(x^{2}-2x+1)-2-1$
$=2(x-1)^{2}-3$
vertex: $(h,k)=(1, -3)$
x-intercept: f(0) = $-1$
y-intercepts: f(x)=0
$2(x-1)^{2}-3=0$
$2(x-1)^{2}=3$
$(x-1)^{2}=\displaystyle \frac{3}{2}\qquad/\sqrt{...}$
$x-1=\pm\sqrt{\frac{3}{2}}$
$x=1\pm\sqrt{\frac{3}{2}}$
b.
f has a minimum value at the vertex, $f(1)=-3$
c.
domain: all reals, $\mathbb{R}$
range: $[-3,\infty)$