Answer
The graph is shown below.
Range $(-\infty,1]$.
Work Step by Step
Rearrange the given function.
$f(x)=-(x-3)^2+1$
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 downward.
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 $0$.
$\Rightarrow 0=-(x-3)^2+1$
Add $(x-3)^2$ to both sides.
$\Rightarrow 0+(x-3)^2=-(x-3)^2+1+(x-3)^2$
Simplify.
$\Rightarrow (x-3)^2=1$
Apply the square root property.
$\Rightarrow x-3=\sqrt{1}$ or $x-3=-\sqrt{1}$
Simplify.
$\Rightarrow x-3=1$ or $x-3=-1$
Add $3$ to both sides in each equation.
$\Rightarrow x-3+3=1+3$ or $x-3+3=-1+3$
Simplify.
$\Rightarrow x=4$ or $x=2$
Hence, the $x-$intercepts are $4$ and $2$. The parabola passes through $(4,0)$ and $(2,0)$.
Step 4:- $y-$intercept.
Replace $x$ with $0$.
$\Rightarrow f(0)=-(0-3)^2+1$
Simplify.
$\Rightarrow f(0)=-9+1$
Simplify.
$\Rightarrow f(0)=-8$
Hence, the $y-$intercept is $-8$. The parabola passes through $(0,-8)$.
Step 5:- Graph.
Use the points vertex, $x-$intercepts and $y-$intercept to draw a parabola.
The axis of symmetry is $x=h=3$.
$A=(3,1)$
$B=(4,0)$
$C=(2,0)$
$D=(0,-8)$.
From the graph the range of the function is
$(-\infty,1]$.
