Simplify Asymptotes Learning with Plainmath's Expert Tips and Real-World Examples

Asymptotes of function
How to find a if the function below doesn't have a vertical asymptotes.
$f(x)=\frac{x^2-<!--\; -\; -->ax+2}{x-<!--\; -\; -->2}$

Find eccentricity when asymptotes are given
What is the Eccentricity of hyperbola whose asymptotes are $x+2y=3$ and $2x-<!--\; -\; -->y=1$.
I know for any hyperbola ${\displaystyle \frac{(x-<!--\; -\; -->h{)}^2}{a^2}}-<!--\; -\; -->{\displaystyle \frac{(y-<!--\; -\; -->k{)}^2}{b^2}}=1$
The asymptotes are $y-<!--\; -\; -->k=\pm <!--\; \pm \; -->{\displaystyle \frac{b}{a}}(x-<!--\; -\; -->h)$
But how to solve this problem

Parallel Asymptotes
By definition, any lines that are not parallel will intersect eventually. But to be parallel lines, both lines must have the same slope. However, consider this situation:
There is a exponential graph that is an asymptote on the y-axis and on the same graph there is a reflection across the y-axis of the first exponential curve.
Now, we know that since these lines are asymptotes they will never actually reach the y-axis (where they would intersect). But also, they do not have the same slope. So - my question is - are these lines parallel?

Asymptotes In Pre-Calculus
I am currently learning about the asymptotes of rational functions in precalculus. However, 4 things about it confuse me, specifically about the asymptotes of it, and how to calculate them.
Why is it when given a rational function with 2 polynomials of the same degree on their numerators and denominators, the asymptote can be found by dividing the coefficients of the highest degree terms?
Why is it if the polynomial in the numerator is a lower degree than the denominator, the x-axis (y = 0) is the horizontal asymptote?
Why is it if the polynomial in the numerator is a higher degree than the denominator, there is no horizontal asymptote?
Why is it when the polynomial in the numerator is a higher degree than the polynomial in the denominator, there will be a slant asymptote which you can find through polynomial long division?
Can you explain all of this using simple algebra, without complicated techniques such as Euler's division of polynomials? I don't understand any complicated techniques and theorems beyond the quadratic formula. Can you also show and explain your working, so it is easier for me to follow through? I am still a beginner, so that would help very much

Can asymptotes be curved?
When I was first introduced to the idea of an asymptote, I was taught about horizontal asymptotes (of form y=a) and vertical ones ( of form x=b).
I was then shown oblique asymptotes-- slanted asymptotes which are not constant (of the form y=ax+b).
What happens, though, if we've got a function such as
$f(x)=e^x+\frac{1}{x}?$
Is $y=e^x$ considered an asymptote in this example?
Another example, just to show you where I'm coming from, is
$g(x)=x^2+\sin <!--\; \; -->(x)$
-- is $y=x^2$ an asymptote in this case?
The reason that I ask is that I don't really see the point in defining oblique asymptotes and not curved ones; surely, if we want to know the behaviour of y as $x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$, we should include all types of functions as asymptotes.
If asymptotes cannot be curves, then why arbitrarily restrict asymptotes to lines?

Find the equation of the horizontal and vertical asymptotes
Find the equations of the asymptotes for the following function:
$\frac{x^2+8}{x^2-<!--\; -\; -->9}$
My solution is the asymptotes are first to find the vertical asymptotes. To do this, I have to find the value that make expression undefined. As a result, the vertical asymptotes are x=−3 and x=3. To find the horizontal asymptotes, I have the compare the degrees of the numerator and the denominator, which result in y=1
My question is, how do I find the equations for the vertical and horizontal asymptotes? Thank you.

Find all asymptotes of a function
Find all asymptotes of a function:
$f(x)=\log <!--\; \; -->(x^2-<!--\; -\; -->4)$
Domain: $x\in <!--\; \in \; -->(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->2)\cup <!--\; \cup \; -->(2,\mathrm{\infty <!--\; \infty \; -->})$
Vertical asymptotes are x=−2 (left) and x=2(right):
$\underset{x\to <!--\; \to \; -->-<!--\; -\; -->2^{-<!--\; -\; -->}}{lim}\log <!--\; \; -->(x^2-<!--\; -\; -->4)=-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}$
$\underset{x\to <!--\; \to \; -->2^{+}}{lim}\log <!--\; \; -->(x^2-<!--\; -\; -->4)=-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}$
I calculate the limits in +/- infinity:
$\underset{x\to <!--\; \to \; -->+\mathrm{\infty <!--\; \infty \; -->}}{lim}\log <!--\; \; -->(x^2-<!--\; -\; -->4)=+\mathrm{\infty <!--\; \infty \; -->}$
$\underset{x\to <!--\; \to \; -->-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\log <!--\; \; -->(x^2-<!--\; -\; -->4)=+\mathrm{\infty <!--\; \infty \; -->}$
So I'm looking for the oblique asymptotes of a form y=Ax+B:
$A_{+}=\underset{x\to <!--\; \to \; -->+\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{\log <!--\; \; -->(x^2-<!--\; -\; -->4)}{x}=\underset{x\to <!--\; \to \; -->+\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{\frac{2x}{x^2-<!--\; -\; -->4}}{1}=0$
$B_{+}=\underset{x\to <!--\; \to \; -->+\mathrm{\infty <!--\; \infty \; -->}}{lim}\log <!--\; \; -->(x^2-<!--\; -\; -->4)-<!--\; -\; -->A_{+}x=+\mathrm{\infty <!--\; \infty \; -->}$
The same for $-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}$. How should I interpret this? There are no oblique asymptotes?

Vertical/horizontal asymptotes
We know that for rational functions
$f(x)/g(x)$
vertical asymptotes are defined as the lines $x=x_0$ where $g(x_0)=0$, and for horizontal asymptotes it is the limit as x approaches $\mathrm{\infty <!--\; \infty \; -->}$ or x approaches $-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}$. My question is the following: what is the relationship between vertical and horizontal asymptotes? For instance, if we have
$y=f(x)$
and we express x as function of y, or
$x=f^{-<!--\; -\; -->1}(y),$
is the horizontal asymptote of inverse function the same as the vertical one of the original?
Thanks in advance.

Does cancellation impact vertical asymptotes?
Question: Let $r(x)=\frac{(x^2+x)}{(x+1)(2x-<!--\; -\; -->4)}$. Does the graph has x=1 as one of its asymptotes?
Answer: No.
My reasoning: $\frac{(x^2+x)}{(x+1)(2x-<!--\; -\; -->4)}=\frac{x(x+1)}{(x+1)(2x-<!--\; -\; -->4)}=\frac{x}{(2x-<!--\; -\; -->4)}$ and so, it cannot have x=−1 as one of its asymptotes.
However, what if I don't calcel and then say that −1 is a vertical asymptote? Will I be wrong?

Finding the asymptotes
Find the asymptotes for $F(x)=ln(x^2=-<!--\; -\; -->3)$. I realise there is no vertical asymptote since $X^2=-<!--\; -\; -->3$. I want to know if the horizontal asymptote exist