How do I convert the equation $f\left( x \right) = {x^2} - 2x - 3$ to vertex form?

Hint: We have to convert the equation $f\left( x \right) = {x^2} - 2x - 3$ to vertex form. For this, complete the square for ${x^2} - 2x - 3$. Use the form $a{x^2} + bx + c$, to find the values of $a$, $b$, and $c$. Consider the vertex form of a parabola to be (I). Next, substitute the values of $a$ and $b$ into the formula (II) and simplify the right side. Next, find the value of $e$ using the formula (III). Next, substitute the values of $a$, $d$, and $e$ into the vertex form (I). Finally, set $y$ equal to the new right side and get the required result.
Formula used:
I). Vertex form of a parabola: $a{\left( {x + d} \right)^2} + e$
II). $d = \dfrac{b}{{2a}}$
III). $e = c - \dfrac{{{b^2}}}{{4a}}$
IV). Vertex form: $y = a{\left( {x - h} \right)^2} + k$
V). Vertex: $\left( {h,k} \right)$
VI). $p = \dfrac{1}{{4a}}$
VII). Focus: $\left( {h,k + p} \right)$
VIII). Directrix: $y = k - p$

Complete step-by-step solution:
We have to convert the equation $f\left( x \right) = {x^2} - 2x - 3$ to vertex form.
For this, complete the square for ${x^2} - 2x - 3$.
Use the form $a{x^2} + bx + c$, to find the values of $a$, $b$, and $c$.
$a = 1,b = - 2,c = - 3$
Consider the vertex form of a parabola.
$a{\left( {x + d} \right)^2} + e$
Now, substitute the values of $a$ and $b$ into the formula $d = \dfrac{b}{{2a}}$.
$d = \dfrac{{ - 2}}{{2 \times 1}}$
Simplify the right side.
$ \Rightarrow d = - 1$
Find the value of $e$ using the formula $e = c - \dfrac{{{b^2}}}{{4a}}$.
$e = - 3 - \dfrac{{{{\left( { - 2} \right)}^2}}}{{4 \times 1}}$
$ \Rightarrow e = - 4$
Now, substitute the values of $a$, $d$, and $e$ into the vertex form $a{\left( {x + d} \right)^2} + e$.
${\left( {x - 1} \right)^2} - 4$
Set $y$ equal to the new right side.
$y = {\left( {x - 1} \right)^2} - 4$
Hence, the vertex form of the given function is $y = {\left( {x - 1} \right)^2} - 4$.

Note: We can also convert the equation $f\left( x \right) = {x^2} - 2x - 3$ to vertex form by directly completing the square.
Vertex form can be represented as $y = {\left( {x - h} \right)^2} + k$
where the point $\left( {h,k} \right)$ is the vertex.
To do that, we should complete the square of
$y = {x^2} - 2x - 3$
First, we should try to change the last number in a way so we can factor the entire thing
⇒ we should aim for $y = {x^2} - 2x + 1$
to make it look like $y = {\left( {x - 1} \right)^2}$
If we notice, the only difference between the original $y = {x^2} - 2x - 3$ and the factor-able $y = {x^2} - 2x + 1$ is simply changing the $ - 3$ to $1$.
Since we can't randomly change the $ - 3$ to a 1, we can add 1 and subtract a 1 to the equation at the same time to keep it balanced.
So, we get
$y = {x^2} - 2x + 1 - 3 - 1$
Organizing
$y = \left( {{x^2} - 2x + 1} \right) - 3 - 1$
Add like terms
$ - 3 - 1 = - 4$
Factor
$y = {\left( {x - 1} \right)^2} - 4$
Hence, the vertex form of the given function is $y = {\left( {x - 1} \right)^2} - 4$.