What are the asymptotes of $y=\frac{1}{x-2}$ and how do you graph the function?

Jonah Jacobson
2022-09-12
Answered

What are the asymptotes of $y=\frac{1}{x-2}$ and how do you graph the function?

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Lorenzo Aguilar

Answered 2022-09-13
Author has **18** answers

$y=\frac{1}{x-2}$

y is defined for $x\in (-\infty ,2)\cup (2,+\infty )$

Consider $\underset{x\to {2}^{+}}{lim}y=+\infty$

And $\underset{x\to {2}^{-}}{lim}y=-\infty$

Hence, y has a vertical asymptote x=2

Now, consider $\underset{x\to \infty}{lim}y=0$

Hence, y has a horizontal asymptote y=0

y is a rectangular hyperbola with the graph below.

graph{1/(x-2) [-10, 10, -5, 5]}

y is defined for $x\in (-\infty ,2)\cup (2,+\infty )$

Consider $\underset{x\to {2}^{+}}{lim}y=+\infty$

And $\underset{x\to {2}^{-}}{lim}y=-\infty$

Hence, y has a vertical asymptote x=2

Now, consider $\underset{x\to \infty}{lim}y=0$

Hence, y has a horizontal asymptote y=0

y is a rectangular hyperbola with the graph below.

graph{1/(x-2) [-10, 10, -5, 5]}

asked 2021-02-25

True or False. The graph of a rational function may intersect a horizontal asymptote.

asked 2022-05-13

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

$p({z}_{1})={a}_{n}{z}_{1}^{n}+{a}_{n-1}{z}_{1}^{n-1}+...+{a}_{1}{z}_{1}+{a}_{0}\phantom{\rule{0ex}{0ex}}q({z}_{2})={b}_{n}{z}_{2}^{n}+{b}_{n-1}{z}_{2}^{n-1}+...+{b}_{1}{z}_{2}+{b}_{0}$

be two complex polynomials of degree $n$ where ${a}_{n},..,{a}_{0}\in \mathbb{C}$ and ${b}_{n},..,{b}_{o}\in \mathbb{C}$.

Now,

$\begin{array}{rl}(p\circ q)({z}_{2})& =p(q({z}_{2}))\text{}\text{}\text{}\text{}\text{}\text{(by definition)}\\ & ={a}_{n}(q({z}_{2}){)}^{n}+{a}_{n-1}(q({z}_{2}){)}^{n-1}+...+{a}_{1}(q({z}_{2}))+{a}_{0}\end{array}$

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

$a({z}_{1})=\frac{p({z}_{1})}{q({z}_{1})},\text{}b({z}_{2})=\frac{p({z}_{2})}{q({z}_{2})}$

Now,

$\begin{array}{rl}(a\circ b)({z}_{2})& =a(b({z}_{2}))\text{}\text{}\text{}\text{}\text{}\text{(by definition)}\\ & =\frac{p\left(\frac{p({z}_{2})}{q({z}_{2})}\right)}{q\left(\frac{p({z}_{2})}{q({z}_{2})}\right)}\\ & =\frac{{a}_{n}{\left(\frac{p({z}_{2})}{q({z}_{2})}\right)}^{n}+{a}_{n-1}{\left(\frac{p({z}_{2})}{q({z}_{2})}\right)}^{n-1}+...+{a}_{1}\left(\frac{p({z}_{2})}{q({z}_{2})}\right)+{a}_{0}}{{b}_{n}{\left(\frac{p({z}_{2})}{q({z}_{2})}\right)}^{n}+{b}_{n-1}{\left(\frac{p({z}_{2})}{q({z}_{2})}\right)}^{n-1}+...+{b}_{1}\left(\frac{p({z}_{2})}{q({z}_{2})}\right)+{b}_{0}}\end{array}$

Notice that ${\left(\frac{p({z}_{2})}{q({z}_{2})}\right)}^{i}\text{}\text{}\text{}\text{}(i=n,n-1,..,0)$ is a polynomial as

$(f\circ g)({z}_{2})=f(g({z}_{2}))={\left(\frac{p({z}_{2})}{q({z}_{2})}\right)}^{i}$

where

$f(x)={x}^{i},\text{}\text{}g({z}_{2})=\left(\frac{p({z}_{2})}{q({z}_{2})}\right)$

are both polynomials. Hence $(a\circ b)({z}_{2})$ is a rational function as it is the quotient of polynomials.

sketch proof that the composition of two polynomials is a polynomial

Let

$p({z}_{1})={a}_{n}{z}_{1}^{n}+{a}_{n-1}{z}_{1}^{n-1}+...+{a}_{1}{z}_{1}+{a}_{0}\phantom{\rule{0ex}{0ex}}q({z}_{2})={b}_{n}{z}_{2}^{n}+{b}_{n-1}{z}_{2}^{n-1}+...+{b}_{1}{z}_{2}+{b}_{0}$

be two complex polynomials of degree $n$ where ${a}_{n},..,{a}_{0}\in \mathbb{C}$ and ${b}_{n},..,{b}_{o}\in \mathbb{C}$.

Now,

$\begin{array}{rl}(p\circ q)({z}_{2})& =p(q({z}_{2}))\text{}\text{}\text{}\text{}\text{}\text{(by definition)}\\ & ={a}_{n}(q({z}_{2}){)}^{n}+{a}_{n-1}(q({z}_{2}){)}^{n-1}+...+{a}_{1}(q({z}_{2}))+{a}_{0}\end{array}$

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

$a({z}_{1})=\frac{p({z}_{1})}{q({z}_{1})},\text{}b({z}_{2})=\frac{p({z}_{2})}{q({z}_{2})}$

Now,

$\begin{array}{rl}(a\circ b)({z}_{2})& =a(b({z}_{2}))\text{}\text{}\text{}\text{}\text{}\text{(by definition)}\\ & =\frac{p\left(\frac{p({z}_{2})}{q({z}_{2})}\right)}{q\left(\frac{p({z}_{2})}{q({z}_{2})}\right)}\\ & =\frac{{a}_{n}{\left(\frac{p({z}_{2})}{q({z}_{2})}\right)}^{n}+{a}_{n-1}{\left(\frac{p({z}_{2})}{q({z}_{2})}\right)}^{n-1}+...+{a}_{1}\left(\frac{p({z}_{2})}{q({z}_{2})}\right)+{a}_{0}}{{b}_{n}{\left(\frac{p({z}_{2})}{q({z}_{2})}\right)}^{n}+{b}_{n-1}{\left(\frac{p({z}_{2})}{q({z}_{2})}\right)}^{n-1}+...+{b}_{1}\left(\frac{p({z}_{2})}{q({z}_{2})}\right)+{b}_{0}}\end{array}$

Notice that ${\left(\frac{p({z}_{2})}{q({z}_{2})}\right)}^{i}\text{}\text{}\text{}\text{}(i=n,n-1,..,0)$ is a polynomial as

$(f\circ g)({z}_{2})=f(g({z}_{2}))={\left(\frac{p({z}_{2})}{q({z}_{2})}\right)}^{i}$

where

$f(x)={x}^{i},\text{}\text{}g({z}_{2})=\left(\frac{p({z}_{2})}{q({z}_{2})}\right)$

are both polynomials. Hence $(a\circ b)({z}_{2})$ is a rational function as it is the quotient of polynomials.

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Is it true that $\mathbb{C}(x,y)$ is not a rational function field? In other words, there is no $z\in \mathbb{C}(x,y)$ such that $\mathbb{C}(x,y)=\mathbb{C}(z)$

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Confirm that if x is rational and $x\ne 0$ , then $\frac{1}{x}$ is rational.

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