Answer
Vertex: $(1,-4)$ $;$ Axis of symmetry: $x=1$
$x$-intercepts: $x=3$ and $x=-1$ $;$ $y$-intercept: $y=-3$
Domain: $(-\infty,\infty)$ $;$ Range: $[-4,\infty)$
Work Step by Step
$f(x)=x^{2}-2x-3$
Group the terms with $x$ together:
$f(x)=(x^{2}-2x)-3$
Complete the square for the expression inside parentheses by adding $\Big(\dfrac{b}{2}\Big)^{2}$ inside. Also, subtract $\Big(\dfrac{b}{2}\Big)^{2}$ outside the parentheses.
In this case, $b=-2$
$f(x)=\Big[x^{2}-2x+\Big(-\dfrac{2}{2}\Big)^{2}\Big]-3-\Big(-\dfrac{2}{2}\Big)^{2}$
$f(x)=(x^{2}-2x+1)-3-1$
$f(x)=(x^{2}-2x+1)-4$
Factor the expression inside parentheses, which is a perfect square trinomial:
$f(x)=(x-1)^{2}-4$
The equation is obtained is now in standard form and its vertex is:
Vertex: $(1,-4)$
The axis of symmetry is the vertical line that passes through the vertex. This axis is the line $x=1$
Find the $x$-intercepts by setting $f(x)$ equal to $0$ and solving for $x$:
$(x-1)^{2}-4=0$
$(x-1)^{2}=4$
$\sqrt{(x-1)^{2}}=\sqrt{4}$
$x-1=\pm2$
$x=1\pm2$
$x=3$ and $x=-1$
Find the $y$-intercept by setting $x$ equal to $0$ and solving for $f(x)$:
$f(0)=(0-1)^{2}-4$
$f(0)=1-4$
$f(0)=-3$
From the function's graph, the domain can be identified as $(-\infty,\infty)$ and the range as $[-4,\infty)$
