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