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