Answer
The axis of symmetry of the given function is $x=1$ , domain is $\left( -\infty ,\infty \right)$ , and the range is $\left[ -4,\infty \right)$.
Work Step by Step
The graph of the given function can be drawn by following the steps as given below.
On comparing the function $f\left( x \right)={{x}^{2}}-2x-3$ with the standard form of a parabola $f\left( x \right)=a{{x}^{2}}+bx+c$ , it can be observed that,
$a=1,b=-2,\text{ and }c=-3$
The vertex for the quadratic form of the parabola, is given by
$\left( h,k \right)=\left( \frac{-b}{2a},f\left( \frac{-b}{2a} \right) \right)$
Thus, the x-coordinate of the vertex is $x=\frac{-b}{2a}=-\frac{\left( -2 \right)}{2\left( 1 \right)}=1$.
And, the y-coordinate is calculated as:
$\begin{align}
& f\left( 1 \right)={{\left( 1 \right)}^{2}}-2\left( 1 \right)-3 \\
& =1-2-3 \\
& =-4
\end{align}$
So, the vertex of the given parabola is at $\left( 1,-4 \right)$.
Now, to find the $y$ -intercept, find $f\left( 0 \right)$ as follows:
$\begin{align}
& f\left( 0 \right)={{\left( 0 \right)}^{2}}-2\left( 0 \right)-3 \\
& =-3
\end{align}$
The $y$ -intercept is $-3$.
And, to calculate the x-intercept, put $f\left( x \right)=0$
$\begin{align}
& f\left( x \right)={{x}^{2}}-2x-3 \\
& 0={{x}^{2}}-2x-3
\end{align}$
Then, simplifying the above equation, we get:
$\begin{align}
& {{x}^{2}}-2x-3=0 \\
& {{x}^{2}}-3x+x-3=0 \\
& x\left( x-3 \right)+1\left( x-3 \right)=0 \\
& \left( x-3 \right)\left( x+1 \right)=0
\end{align}$
Thus, the x-intercepts of the given parabola are 3 and $-1$.
Since the coefficient of ${{x}^{2}}$ is $1$, which is positive, therefore, the parabola opens upwards.
As can be observed from above graph, the axis of symmetry of the parabola will be $x=1$.
Also, since the parabola opens upwards, its minimum value is $-4$ and its maximum value is $\infty $. Thus, the range of the function is $\left[ -4,\infty \right)$.
And, since the function is defined for all real values of x, its domain will be given by $\left( -\infty ,\infty \right)$.
