Answer
The graph is shown below.
Range $[2,\infty)$.
Work Step by Step
The given function is
$f(x)=(x-3)^2+2$
The standard form of the parabola is
$f(x)=a(x-h)^2+k$
Compare with the original given function.
$a=1,h=3$ and $k=2$
Step 1:- Parabola opens.
$a>0$, The parabola opens upward.
Step 2:- Vertex.
The value of $h=3$ and $k=2$. The vertex is $(h,k)=(3,2)$.
Step 3:- $x-$intercepts.
Replace $f(x)$ with $0$.
$\Rightarrow 0=(x-3)^2+2$
Subtract $2$ from both sides.
$\Rightarrow 0-2=(x-3)^2+2-2$
Simplify.
$\Rightarrow -2=(x-3)^2$
Apply the square root property.
$\Rightarrow x-3=\sqrt{-2}$ or $x-3=-\sqrt{-2}$
Hence, the equation has imaginary solutions, there are no $x-$intercepts.
Step 4:- $y-$intercept.
Replace $x$ with $0$.
$\Rightarrow f(0)=(0-3)^2+2$
Simplify.
$\Rightarrow f(0)=9+2$
Simplify.
$\Rightarrow f(0)=11$
Hence, the $y-$intercept is $11$. The parabola passes through $(0,11)$.
Step 5:- Graph.
Use the points vertex and $y-$intercept to draw a parabola.
The axis of symmetry is $x=h=3$.
$A=(3,2)$
$B=(0,11)$.
From the graph the range of the function is
$[2,\infty)$.
