Answer
The graph is shown below.
Range $[1,\infty)$.
Work Step by Step
Add $1$ to both sides of the given equation.
$y-1+1=(x-3)^2+1$
Simplify.
$y=(x-3)^2+1$
Let $y=f(x)$.
$f(x)=(x-3)^2+1$
The standard form of the parabola is
$f(x)=a(x-h)^2+k$
Compare with the original given function.
$a=1,h=3$ and $k=1$
Step 1:- Parabola opens.
$a>0$, The parabola opens upward.
Step 2:- Vertex.
The value of $h=3$ and $k=1$. The vertex is $(h,k)=(3,1)$.
Step 3:- $x-$intercepts.
Replace $f(x)$ with $0$.
$\Rightarrow 0=(x-3)^2+1$
Subtract $1$ from both sides.
$\Rightarrow 0-1=(x-3)^2+1-1$
Simplify.
$\Rightarrow -1=(x-3)^2$
Apply the square root property.
$\Rightarrow x-3=\sqrt{-1}$ or $x-3=-\sqrt{-1}$
Hence, the equation has imaginary solutions, there are no $x-$intercepts.
Step 4:- $y-$intercept.
Replace $x$ with $0$.
$\Rightarrow f(0)=(0-3)^2+1$
Simplify.
$\Rightarrow f(0)=9+1$
Simplify.
$\Rightarrow f(0)=10$
Hence, the $y-$intercept is $10$. The parabola passes through $(0,10)$.
Step 5:- Graph.
Use the points vertex and $y-$intercept to draw a parabola.
The axis of symmetry is $x=h=3$.
$A=(3,1)$
$B=(0,10)$.
From the graph the range of the function is
$[1,\infty)$.
