Answer
The graph is shown below.
Range $[-4,\infty)$.
Work Step by Step
The given function is a quadratic function:
$f(x)=x^2-2x-3$
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 upward.
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 radicals.
$\Rightarrow x=\frac{2+4}{2}$ or $ \frac{2-4}{2}$
Simplify.
$\Rightarrow x=\frac{6}{2}$ or $ x=\frac{-2}{2}$
$\Rightarrow x=3$ or $ x=-1$
The $x-$intercepts are $3$ and $-1$. The parabola passes through $(3,0)$ and $(-1,0)$.
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 $x=1$.
$A=(1,-4)$
$B=(3,0)$
$C=(-1,0)$
$D=(0,-3)$.
From the graph the range of the function is
$[-4,\infty)$.
