I am very familiar with constructing functions to have certain properties seen in the image of them. For example, say I wanted a vertical asymptote. Then you just create a division by zero in the function you are creating.
Example: A function with a vertical asymptote at $x=3$ would be $\displaystyle\frac{1}{x-3}$.
Here is what I am having trouble with and is also my question.
Can I construct a rational function $f(x)$ that has at least one vertical asymptote, 1 hole, a horizontal asymptote that is a whole number larger then zero, has a x-intercept but no y intercept. The kicker here is the function can not cross or touch the horizontal asymptote.
Realize that if the function has no $y$-intercept it must have a vertical asymptote at $x=0$:
$$y=\frac 1{(x - 0)}$$
And for it to have a hole we can add a factor to the bottom and top that essentially 'cancel' creating the hole:
$$y=\frac {(x-1)}{(x-0)(x-1)}$$
This would have a hole at $x=1$. Then remember the rules for rational function horizontal asymptote rules.
If the degrees of the denominator and numerator are equal, then the horizontal asymptote is $y = \frac ab$ where $a$ and $b$ are the leading coefficients of the denominator and numerator.
That means, if we add another factor to the top we can have the same degree, we can create a $y=1$ horizontal asymptote:
$$y=\frac {(x-1)(x-2)}{(x-0)(x-1)}$$
This function has a degree of $2$ on both the top and bottom now. That means there will be a horizontal asymptote at $y=\frac ab$, and in this case $a$ and $b$ are $1$ since the leading coefficients are $1$.
And the function does not pass through the horizontal asymptote.