Answer
The graph is shown below.
Range $(-\infty,4]$.
Work Step by Step
The given function is a quadratic function:
$f(x)=2x-x^2+3$
Rewrite as
$f(x)=-x^2+2x+3$
The standard form of the quadratic function is
$f(x)=ax^2+bx+c$
Compare both equations $a=-1,b=2$ and $c=3$.
Step 1:- Parabola opens.
$a<0$, the parabola open downward.
Step 2:- Vertex.
$x-$coordinate of the vertex is $x=-\frac{b}{2a}=-\frac{2}{2(-1)}=1$.
Substitute the value of $x$ into the function.
$\Rightarrow f(1)=-(1)^2+2(1)+3$
Simplify.
$\Rightarrow f(1)=-1+2+3$
$\Rightarrow f(1)=4$
The vertex is $(1,4)$.
Step 3:- $x-$intercepts.
Replace $f(x)$ with $0$ into the given function.
$\Rightarrow 0=-x^2+2x+3$
Use quadratic formula, we have $a=-1,b=2$ and $c=3$.
$\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Substitute all values.
$\Rightarrow x=\frac{-(2)\pm\sqrt{(2)^2-4(-1)(3)}}{2(-1)}$
Simplify.
$\Rightarrow x=\frac{-2\pm\sqrt{4+12}}{-2}$
$\Rightarrow x=\frac{-2\pm\sqrt{16}}{-2}$
$\Rightarrow x=\frac{-2\pm4}{-2}$
Separate the fractions.
$\Rightarrow x=\frac{-2+4}{-2}$ or $x= \frac{-2-4}{-2}$
Simplify.
$\Rightarrow x=\frac{2}{-2}$ or $ x=\frac{-6}{-2}$
$\Rightarrow x=-1$ or $ x=3$
The $x-$intercepts are $-1$ and $3$. The parabola passes through $(-1,0)$ and $(3,0)$.
Step 4:- $y-$intercept.
Replace $x$ with $0$ in the given function.
$\Rightarrow f(0)=-(0)^2+2(0)+3$
Simplify.
$\Rightarrow f(0)=3$
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 $x=1$.
$A=(1,4)$
$B=(-1,0)$
$C=(3,0)$
$D=(0,3)$.
From the graph the range of the function is
$(-\infty,4]$.
