Get Expert Help for Your Rational Functions Questions

Suppose you have a rational function, of the form $H(x)=\frac{B(x)}{A(x)}$. It's differential is of the form $H^{\prime }(x)=\frac{(B^{\prime }(x)\star <!--\; \star \; -->A(x)-<!--\; -\; -->B(x)\star <!--\; \star \; -->A^{\prime }(x))}{(A(x){)}^2}$. It is trivial to prove that, if $k$ is a root of $B(x)$ or $A(x)$ with rank $r$, $r\ge <!--\; \ge \; -->2$, then k is also a root of $H^{\prime }(x)$'s numerator, $B^{\prime }(x)\star <!--\; \star \; -->A(x)-<!--\; -\; -->B(x)\star <!--\; \star \; -->A^{\prime }(x)$, with rank at least $r-<!--\; -\; -->1$. Is there a theorem that states this exact thing?
The reason I need this is, I am going to take an exam soon, and in a specific part of the exam this observation would be very useful. But I will have to prove it by hand during the exam, and time will be very limited, so I was wondering if there is already a name for this to reference it directly and get on with it.
EDIT: Had erroneously named the desired form "polynomial fraction", instead of the right "rational function".

Q1: If you have a rational function, how do you know it will have oblique asymptote?
Q2: if you have a rational function, is long division the best way to find the oblique asymptote?
Any help, please

Suppose we are given a rational function
$f(s)=\frac{25s}{s^4+18s^3+134s^2+472s+680}$
and we need to find the zeros and poles of the function.
Suppose $f(s)=\frac{a(s)}{b(s)}$, then $a(s)=25s$ and $s=0$ is a root of $a(s)=0$ and hence 0 is a zero of f(s). Again b(s)=0 has roots at $s=-<!--\; -\; -->5\pm <!--\; \pm \; -->3i$ and $s=-<!--\; -\; -->4\pm <!--\; \pm \; -->2i$, hence $-<!--\; -\; -->5\pm <!--\; \pm \; -->3i$ and $-<!--\; -\; -->4\pm <!--\; \pm \; -->2i$ are poles of f(s).
Is the solution correct?

We know that given a rational number of the from $p/q$ if we expand it in the decimal base that is an expansion in power of 10 and we know that coefficient is repeating after finite step. In other words, we can say that we need only finitely many coefficients to describe the rest. Let $F(z)$ be a rational function that is $F(z)=\frac{p(z)}{q(z)}$. I take the Taylor series expansion of $F(z)$ at the point away from the pole (maybe in a local coordinate around the point). Say
$F(z){\mid <!--\; \mid \; -->}_z=a\sum <!--\; \sum \; -->t_k(z-<!--\; -\; -->a{)}^k$
My question is there any finite condition on the coefficient of $t_k$ ? That is we need only finitely many $t_k$'s to determine the rest? If so what will be the proof?

Conside Hölder continuous functions $f_1,\dots ,f_m:{\mathbb{R}}^n\to <!--\; \to \; -->\mathbb{R}$ (with Hölder coefficient $\mathrm{\alpha <!--\; \alpha \; -->}$). The claum is now that any rational function of $f_1,\dots ,f_m$ with non vanishing denominator is also Hölder continuous (with coefficient α). The author says this is trivial, but I do not see why this should be trivial!? (I could not even prove the easiest case f(x)/g(x))
Do you have an idea by which it is easily proven?
Best regards.

Express the following rational function in continued-fraction form:
$\frac{4x^2+3x-<!--\; -\; -->7}{2x^3+x^2-<!--\; -\; -->x+5}$
The answer is :
$\begin{array}{}\text{(inline continued fraction)}& \frac{4}{2x-<!--\; -\; -->\frac{1}{2}}+\frac{\frac{23}{8}}{x-<!--\; -\; -->\frac{63}{92}}-<!--\; -\; -->\frac{\frac{406}{529}}{x+\frac{33}{23}}\end{array}$
which means
$\frac{4}{2x-<!--\; -\; -->\frac{1}{2}+\frac{\frac{23}{8}}{x-<!--\; -\; -->\frac{63}{92}-<!--\; -\; -->\frac{\frac{406}{529}}{x+\frac{33}{23}}}}$

I am really confused about the horizontal and slant asymptotes of a function. My textbook says that given a rational function:
$y=f(x)=\frac{a_n{x}^n+a_{n-<!--\; -\; -->1}{x}^{n-<!--\; -\; -->1}+\cdot <!--\; \cdot \; -->\cdot <!--\; \cdot \; -->\cdot <!--\; \cdot \; -->+a_0}{b_m{x}^m+b_{m-<!--\; -\; -->1}{x}^{m-<!--\; -\; -->1}+\cdot <!--\; \cdot \; -->\cdot <!--\; \cdot \; -->\cdot <!--\; \cdot \; -->+b_0}$
The following properties are true. Can you please explain to me why the following properties are true? Do so not too rigorously, I'm still a beginner. And can you not use Calculus in the answers, since I haven't learn it yet? The properties are:
1. If $n<m$, then $y=0$ is a horizontal asymptote. I don't get why it would not be possible for 0 to be in the range of a rational function. Sure, a rational function like $\frac{1}{x}$ cannot produce 0. But how can this be true for all rational functions such that when $n<m$, it can't give an output of 0? Please explain.
2. If $n>m$, there is no horizontal asymptote. Please explain to me why there would be no limit to what the function can produce.
3. If $n=m$, then $y=\frac{a_n}{b_m}$ is a horizontal asymptote. Again, I have no idea why this holds true for all rational functions. Can someone please explain?
4. If $n$ is more than $m$ only by 1 degree, then there is a slant asymptote which can be determined by dividing the denominator into the numerator. Again, can someone please explain?
Thank you so much in advance. Again, please try not to use calculus in your answers.

Assume I am given two polynomials $f(x)$ and $g(x)$ with coefficients from a field ${\mathbb{F}}_p$, where $p$ is a prime. Now I know that the set of these polynomials is a ring and not a field, meaning that not every polynomial has an inverse.
Given evaluation points $f({\mathrm{\alpha <!--\; \alpha \; -->}}_1),g({\mathrm{\alpha <!--\; \alpha \; -->}}_1),f({\mathrm{\alpha <!--\; \alpha \; -->}}_2),g({\mathrm{\alpha <!--\; \alpha \; -->}}_2),\dots <!--\; \dots \; -->$ for some distinct values ${\mathrm{\alpha <!--\; \alpha \; -->}}_i$, one can first compute evaluation points $h({\mathrm{\alpha <!--\; \alpha \; -->}}_i):=f({\mathrm{\alpha <!--\; \alpha \; -->}}_i)/g({\mathrm{\alpha <!--\; \alpha \; -->}}_i)$ and then interpolate the polynomial $h(x)=f(x)/g(x)$ from them such that the interpolated polynomial $h$ contains the correct roots in the numerator and denominator. For example if the polynomial $f$ has roots $a$,$b$,$c$ and $g$ has roots $b$, $c$, and $d$, then the interpolated rational function $h$ will have a numerator with the root $a$ and the denominator will have the root $d$.
My confusion stems from the fact that $g(x)$ may not have an inverse, yet we can interpolate a polynomial $1/g(x)$, which is basically an inverse or not?
P.S. In case it helps, I only care about polynomials of the form $f(x)=(x-<!--\; -\; -->a_1)(x-<!--\; -\; -->a_2)...x(a_k)$, where for all $i\ne <!--\; \ne \; -->j$ it holds that $a_i\ne <!--\; \ne \; -->a_j$.

Compute all the points of discontinuity if $\frac{n^2-<!--\; -\; -->17n+72}{(n^2-<!--\; -\; -->10n+25)(n^2+6n+5)}$.

Calculate all the points of discontinuity of $\frac{v^2-<!--\; -\; -->17v+42}{(v-<!--\; -\; -->2)(v^2-<!--\; -\; -->12v+36)}$.

Write all the points where $\frac{x^2-<!--\; -\; -->18x+56}{x^3-<!--\; -\; -->3x^2-<!--\; -\; -->22x+24}$ is discontinuous.

In the case of rational function, when taking the first derivative we take the numerator and find for which point the function is equal to zero.
Now, for the second derivative, can we take the second derivative just of the numerator? it is not intuitive, as we need to take the second derivative of the whole function, but I did not find a counterexample
For example the function: $x+8+\frac{50}{2x-<!--\; -\; -->4}$

This seems very obvious and I am having a bit of trouble producing a formal proof.
sketch proof that the composition of two polynomials is a polynomial
Let
$$\begin{gathered} p(z_1)=a_n{z}_1^n+a_{n-<!--\; -\; -->1}{z}_1^{n-<!--\; -\; -->1}+...+a_1{z}_1+a_0 \\ q(z_2)=b_n{z}_2^n+b_{n-<!--\; -\; -->1}{z}_2^{n-<!--\; -\; -->1}+...+b_1{z}_2+b_0 \end{gathered}$$
be two complex polynomials of degree $n$ where $a_n,..,a_0\in <!--\; \in \; -->\mathbb{C}$ and $b_n,..,b_o\in <!--\; \in \; -->\mathbb{C}$.
Now,
$\begin{array}{rl}(p\circ <!--\; \circ \; -->q)(z_2)& =p(q(z_2))\text{(by definition)}\\ & =a_n(q(z_2){)}^n+a_{n-<!--\; -\; -->1}(q(z_2){)}^{n-<!--\; -\; -->1}+...+a_1(q(z_2))+a_0\end{array}$
which is clearly a complex polynomial of degree $n^2$.
sketch proof that the composition of two rational functions is a rational function
A rational function is a quotient of polynomials.
Let
$a(z_1)=\frac{p(z_1)}{q(z_1)},b(z_2)=\frac{p(z_2)}{q(z_2)}$
Now,
$\begin{array}{rl}(a\circ <!--\; \circ \; -->b)(z_2)& =a(b(z_2))\text{(by definition)}\\ & =\frac{p\left(\frac{p(z_2)}{q(z_2)}\right)}{q\left(\frac{p(z_2)}{q(z_2)}\right)}\\ & =\frac{a_n{\left(\frac{p(z_2)}{q(z_2)}\right)}^n+a_{n-<!--\; -\; -->1}{\left(\frac{p(z_2)}{q(z_2)}\right)}^{n-<!--\; -\; -->1}+...+a_1\left(\frac{p(z_2)}{q(z_2)}\right)+a_0}{b_n{\left(\frac{p(z_2)}{q(z_2)}\right)}^n+b_{n-<!--\; -\; -->1}{\left(\frac{p(z_2)}{q(z_2)}\right)}^{n-<!--\; -\; -->1}+...+b_1\left(\frac{p(z_2)}{q(z_2)}\right)+b_0}\end{array}$
Notice that ${\left(\frac{p(z_2)}{q(z_2)}\right)}^i(i=n,n-<!--\; -\; -->1,..,0)$ is a polynomial as
$(f\circ <!--\; \circ \; -->g)(z_2)=f(g(z_2))={\left(\frac{p(z_2)}{q(z_2)}\right)}^i$
where
$f(x)=x^i,g(z_2)=\left(\frac{p(z_2)}{q(z_2)}\right)$
are both polynomials. Hence $(a\circ <!--\; \circ \; -->b)(z_2)$ is a rational function as it is the quotient of polynomials.

Write all the points where $\frac{u^2-<!--\; -\; -->27u+182}{u^3+3u^2-<!--\; -\; -->9u+5}$.

Find all the points of discontinuity of the function $\frac{w^2-<!--\; -\; -->22w+117}{(w^2-<!--\; -\; -->1)(w^3-<!--\; -\; -->9w^2+23w-<!--\; -\; -->15)}$.

Write all the points where $\frac{y^2-<!--\; -\; -->10y+16}{(y^2-<!--\; -\; -->3y-<!--\; -\; -->4)(y^3+y^2-<!--\; -\; -->32y-<!--\; -\; -->60)}$ is discontinuous.

Write all the points where $\frac{m^2-<!--\; -\; -->16m+60}{(m-<!--\; -\; -->4)(m^3-<!--\; -\; -->7m^2-<!--\; -\; -->25m+175)}$ is discontinuous.

What are the points of discontinuity of the function $\frac{a^2-<!--\; -\; -->23a+126}{(a+5)(a^3+a^2-<!--\; -\; -->37a+35)}$?

I have a rational function:
$y=\frac{x^2-<!--\; -\; -->3x}{x^2+2x-<!--\; -\; -->48}$
The explanation I was given to find the x-intercepts was: "let y = 0, and solve for x. Basically you just set the numerator of the fraction equal to 0 and factor."
What does "basically" mean? Is this always the case? Do you simply set the numerator expression equal to zero, and always ignore the contents of the denominator? If so, why do you ignore the contents of the denominator?

Write all the points where $\frac{u^2-<!--\; -\; -->24u+135}{(u^2+3u-<!--\; -\; -->18)(u^3-<!--\; -\; -->6u^2-<!--\; -\; -->u+30)}$ is discontinuous.