opinaj
2022-09-29
Answered

How do you graph $f\left(x\right)=\frac{2}{x-3}+1$ using holes, vertical and horizontal asymptotes, x and y intercepts?

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Emilia Boyle

Answered 2022-09-30
Author has **10** answers

For vertical asymptote, look at the denominator. It cannot equal to 0 as the graph will be undefined at that point. Hence, you let denominator equal to 0 to find at what point the graph cannot equal to 0.

x−3=0

x=−3

For horizontal asymptote, imagine what happens to the graph when $x\to \infty$. As $x\to \infty$, $\frac{2}{x-3}\to 0$ so f(x)=1+0=1 ie y=1

For intercepts,

When y=0, x=1

When x=0, $y=\frac{1}{3}$

Below is what the graph looks like

graph{1+2/(x-3) [-10, 10, -5, 5]}

x−3=0

x=−3

For horizontal asymptote, imagine what happens to the graph when $x\to \infty$. As $x\to \infty$, $\frac{2}{x-3}\to 0$ so f(x)=1+0=1 ie y=1

For intercepts,

When y=0, x=1

When x=0, $y=\frac{1}{3}$

Below is what the graph looks like

graph{1+2/(x-3) [-10, 10, -5, 5]}

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Let

$f(z)={\displaystyle \frac{z-a}{z-b}},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}z\ne b\ne a$

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How can I show that, if $|a|,|b|<1,$, then there is a complex number ${z}_{0}$ satisfying $|{z}_{0}|=1$ and $f({z}_{0})\in \mathbb{R}$ ?

I have tried in many ways, but on success. Basically I tried to show that there is a unimodular complex number such that

$\frac{a-b}{z-b}}={\displaystyle \frac{\overline{a}-\overline{b}}{\overline{z}-\overline{b}}}.$

I could make a quadratic equation by using the fact that $\overline{z}={\displaystyle \frac{1}{z}},\phantom{\rule{thinmathspace}{0ex}}\mathrm{\forall}z\in \mathrm{\partial}\mathbb{D}.$ Unfortunately I could not solve this question using that. So, I would like to see different (and somewhat general) approach.

Also, I would like to know that what happen if (both or atleast one) $a,b\notin \mathbb{D}.$Any comment or hint will be welcome. Thank you.

$f(z)={\displaystyle \frac{z-a}{z-b}},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}z\ne b\ne a$

be a complex valued rational function.

How can I show that, if $|a|,|b|<1,$, then there is a complex number ${z}_{0}$ satisfying $|{z}_{0}|=1$ and $f({z}_{0})\in \mathbb{R}$ ?

I have tried in many ways, but on success. Basically I tried to show that there is a unimodular complex number such that

$\frac{a-b}{z-b}}={\displaystyle \frac{\overline{a}-\overline{b}}{\overline{z}-\overline{b}}}.$

I could make a quadratic equation by using the fact that $\overline{z}={\displaystyle \frac{1}{z}},\phantom{\rule{thinmathspace}{0ex}}\mathrm{\forall}z\in \mathrm{\partial}\mathbb{D}.$ Unfortunately I could not solve this question using that. So, I would like to see different (and somewhat general) approach.

Also, I would like to know that what happen if (both or atleast one) $a,b\notin \mathbb{D}.$Any comment or hint will be welcome. Thank you.

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More importantly, if this is false, then is there a nice description of the set of pairs $(({a}_{1},...,{a}_{n}),({b}_{1},...,{b}_{n}))\in {\mathbb{C}}^{n}\times {\mathbb{C}}^{n}$ for which there is a rational function $f\in \mathbb{C}(x)$ such that $f({a}_{i})={b}_{i}$?

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