Answer
The graph is shown below.
Range $[-2,\infty)$.
Work Step by Step
The given function is
$f(x)=(x-1)^2-2$
The standard form of the parabola is
$f(x)=a(x-h)^2+k$
Compare with the original given function.
$a=1,h=1$ and $k=-2$
Step 1:- Parabola opens.
$a>0$, The parabola opens upward.
Step 2:- Vertex.
The value of $h=1$ and $k=-2$. The vertex is $(h,k)=(1,-2)$.
Step 3:- $x-$intercepts.
Replace $f(x)$ with $0$.
$\Rightarrow 0=(x-1)^2-2$
Add $2$ to both sides.
$\Rightarrow 0+2=(x-1)^2-2+2$
Simplify.
$\Rightarrow 2=(x-1)^2$
Apply the square root property.
$\Rightarrow x-1=\sqrt{2}$ or $x-1=-\sqrt{2}$
Add $1$ to both sides in each equation.
$\Rightarrow x-1+1=\sqrt{2}+1$ or $x-1+1=-\sqrt{2}+1$
Simplify.
$\Rightarrow x=\sqrt{2}+1$ or $x=-\sqrt{2}+1$
Hence, the $x-$intercepts are $\sqrt{2}+1$ and $-\sqrt{2}+1$. The parabola passes through $(\sqrt{2}+1,0)$ and $(-\sqrt{2}+1,0)$.
Step 4:- $y-$intercept.
Replace $x$ with $0$.
$\Rightarrow f(0)=(0-1)^2-2$
Simplify.
$\Rightarrow f(0)=1-2$
Simplify.
$\Rightarrow f(0)=-1$
Hence, the $y-$intercept is $-1$. The parabola passes through $(0,-1)$.
Step 5:- Graph.
Use the points vertex, $x-$intercepts and $y-$intercept to draw a parabola.
The axis of symmetry is $x=h=1$.
$A=(1,-2)$
$B=(\sqrt{2}+1,0)$
$C=(-\sqrt{2}+1,0)$
$D=(0,-1)$.
From the graph the range of the function is
$[-2,\infty)$.
