Answer
The graph is shown below.
Range $(-\infty,-1]$.
Work Step by Step
The given function is a quadratic function:
$f(x)=2x-x^2-2$
Rewrite as.
$f(x)=-x^2+2x-2$
The standard form of the quadratic function is
$f(x)=ax^2+bx+c$
Compare both equations $a=-1,b=2$ and $c=-2$.
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.
$\Rightarrow f(1)=-(1)^2+2(1)-2$
Simplify.
$\Rightarrow f(1)=-1+2-2$
Simplify.
$\Rightarrow f(1)=-1$
The vertex is $(1,-1)$.
Step 3:- $x-$intercepts.
Replace $f(x)$ with $0$ into the given function.
$\Rightarrow 0=-x^2+2x-2$
Use quadratic formula, we have $a=-1,b=2$ and $c=-2$.
$\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Substitute all values.
$\Rightarrow x=\frac{-(2)\pm\sqrt{(2)^2-4(-1)(-2)}}{2(-1)}$
Simplify.
$\Rightarrow x=\frac{-2\pm\sqrt{4-8}}{-2}$
$\Rightarrow x=\frac{-2\pm\sqrt{-4}}{-2}$
The equation has imaginary solutions, there are no $x-$intercepts.
Step 4:- $y-$intercept.
Replace $x$ with $0$ in the given function.
$\Rightarrow f(0)=-(0)^2+2(0)-2$
Simplify.
$\Rightarrow f(0)=-2$
The $y-$intercept is $-2$. The parabola passes through $(0,-2)$.
Step 5:- Graph.
Use the points:- vertex, $x-$intercepts and $y-$intercept to draw a parabola.
The axis of symmetry is $x=1$.
$A=(1,-1)$
$B=(0,-2)$.
From the graph the range of the function is
$(-\infty,-1]$.
