Graphing $\frac{(x-3)}{(x^2-3x)}$ and $1/x$ For graphing the first equation, in the solved example given in the textbook, we proceeded as follows:
$\frac{(x-3)}{x^2-3x} \implies \frac{(x-3)}{x(x-3)} \implies \frac{1}{x}$ for $x \neq 3$.
Now the graph of this equation is exactly the same as that of $\frac{1}{x}$ except that there's a hole in $f(3)$ because the equation is not defined for $x=3$. What I don't get is, if we have simplified the equation to $\frac{1}{x}$, why do we still need to consider for the values of $x$ where the value of the original equation was undefined? Both the equations are exactly the same and yet, have different graphs, why so?

$f(x)=\frac{x-3}{x^2-3x}=\frac{(x-3)}{x(x-3)}$, only when $x \neq 3$, as division by $0$ is not possible.


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*If $x \neq 3$, then we can divide both sides by $(x-3)$ to get $f(x)=\frac{1}{x}$. For example, for $x=2$, we have $f(2)=\frac{(2-3)}{2(2-3)}=\frac{1}{2}$(obtained by cancelling $(2-3) \neq 0$ )

*If $x =3$, then $f(3)=\frac{(3-3)}{3(3-3)} \neq \frac{1}{3}$, as we can't cancel $(3-3)=0$ on account of division by zero.

Now, as to why division by $0$ is not possible. Read this
