Answer
The graph is shown below.
Range $(-\infty,9]$.
Work Step by Step
The given function is a quadratic function:
$f(x)=5-4x-x^2$
Rewrite as
$f(x)=-x^2-4x+5$
The standard form of the quadratic function is
$f(x)=ax^2+bx+c$
Compare both equations $a=-1,b=-4$ and $c=5$.
Step 1:- Parabola opens.
$a<0$, the parabola open downward.
Step 2:- Vertex.
$x-$coordinate of the vertex is $x=-\frac{b}{2a}=-\frac{(-4)}{2(-1)}=-2$.
Substitute the value of $x$ into the function.
$\Rightarrow f(-2)=-(-2)^2-4(-2)+5$
Simplify.
$\Rightarrow f(-2)=-4+8+5$
$\Rightarrow f(-2)=9$
The vertex is $(-2,9)$.
Step 3:- $x-$intercepts.
Replace $f(x)$ with $0$ into the given function.
$\Rightarrow 0=-x^2-4x+5$
Use quadratic formula, we have $a=-1,b=-4$ and $c=5$.
$\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Substitute all values.
$\Rightarrow x=\frac{-(-4)\pm\sqrt{(-4)^2-4(-1)(5)}}{2(-1)}$
Simplify.
$\Rightarrow x=\frac{4\pm\sqrt{16+20}}{-2}$
$\Rightarrow x=\frac{4\pm\sqrt{36}}{-2}$
$\Rightarrow x=\frac{4\pm6}{-2}$
Separate the fractions.
$\Rightarrow x=\frac{4+6}{-2}$ or $x= \frac{4-6}{-2}$
Simplify.
$\Rightarrow x=\frac{10}{-2}$ or $ x=\frac{-2}{-2}$
$\Rightarrow x=-5$ or $ x=1$
The $x-$intercepts are $-5$ and $1$. The parabola passes through $(-5,0)$ and $(1,0)$.
Step 4:- $y-$intercept.
Replace $x$ with $0$ in the given function.
$\Rightarrow f(0)=-(0)^2-4(0)+5$
Simplify.
$\Rightarrow f(0)=5$
The $y-$intercept is $5$. The parabola passes through $(0,5)$.
Step 5:- Graph.
Use the points:- vertex, $x-$intercepts and $y-$intercept to draw a parabola.
The axis of symmetry is $x=-2$.
$A=(-2,9)$
$B=(-5,0)$
$C=(1,0)$
$D=(0,5)$.
From the graph the range of the function is
$(-\infty,9]$.
