How do you graph $f\left(x\right)=\frac{3}{{x}^{2}(x+5)}$ using holes, vertical and horizontal asymptotes, x and y intercepts?

Hrefnui9
2022-09-12
Answered

How do you graph $f\left(x\right)=\frac{3}{{x}^{2}(x+5)}$ using holes, vertical and horizontal asymptotes, x and y intercepts?

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Leon Webster

Answered 2022-09-13
Author has **17** answers

For vertical asymptotes, look at the denominator. ${x}^{2}(x+5)\ne 0$ because the graph will be undefined

${x}^{2}(x+5)\ne 0$ means that the vertical asymptotes are x=0 and x=−5 when solving for x

For the horizontal asymptote, look at the degree of the numerator and denominator

If the degree of the numerator is less than the degree of the denominator, then the horizontal asymptote is y=0.

Or you can think of it as if you put some numbers into your f(x), you will notice that ${x}^{2}(x+5)$ will be a lot bigger than 3. Hence, when you divide a small number by a big number, $\frac{3}{{x}^{2}(x+5)}\to 0$

For your x and y intercepts,

sub y=0 for x intercepts,

$0=\frac{3}{{x}^{2}(x+5)}$

0=3 which isn't true so no x-intercepts

sub x=0 for y intercepts which cannot occur since x=0 is an asymptote

Therefore, there are no intercepts

Below is the graph. You can see that the endpoints of the graphs approaches the asymptotes y=0, x=0 and x=−5

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

${x}^{2}(x+5)\ne 0$ means that the vertical asymptotes are x=0 and x=−5 when solving for x

For the horizontal asymptote, look at the degree of the numerator and denominator

If the degree of the numerator is less than the degree of the denominator, then the horizontal asymptote is y=0.

Or you can think of it as if you put some numbers into your f(x), you will notice that ${x}^{2}(x+5)$ will be a lot bigger than 3. Hence, when you divide a small number by a big number, $\frac{3}{{x}^{2}(x+5)}\to 0$

For your x and y intercepts,

sub y=0 for x intercepts,

$0=\frac{3}{{x}^{2}(x+5)}$

0=3 which isn't true so no x-intercepts

sub x=0 for y intercepts which cannot occur since x=0 is an asymptote

Therefore, there are no intercepts

Below is the graph. You can see that the endpoints of the graphs approaches the asymptotes y=0, x=0 and x=−5

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

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

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Let $f:\mathbb{Q}\u27f6\mathbb{Q}$ be a continuous function. Is there necessarily an interval $(a,b)$, with $a,b\in \mathbb{Q}$ such that $a<b$, such that the restriction of $f$ to $(a,b)\cap \mathbb{Q}$ is a rational function?

My guess is that the answer is negative. I tried to prove it as follows: I took a countable and dense set $\{{r}_{n}\mid n\in \mathbb{N}\}$ of irrational numbers and defined, for each $x\in \mathbb{Q}$, $f(x)=\sum _{{r}_{n}<x}{2}^{-n}$. This will work, in the sense that $f$ is continuous and that the restriction of $f$ to any interval $(a,b)$ is never a rational function. The problem is that, of course, in general, $f(x)\notin \mathbb{Q}$.

My guess is that the answer is negative. I tried to prove it as follows: I took a countable and dense set $\{{r}_{n}\mid n\in \mathbb{N}\}$ of irrational numbers and defined, for each $x\in \mathbb{Q}$, $f(x)=\sum _{{r}_{n}<x}{2}^{-n}$. This will work, in the sense that $f$ is continuous and that the restriction of $f$ to any interval $(a,b)$ is never a rational function. The problem is that, of course, in general, $f(x)\notin \mathbb{Q}$.

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For a field $k$, let $V$ be an affine variety over $k$. Denote by $k(V)$ the function field of $V$, containing all rational functions $r:V\u21e2{\mathbb{A}}_{k}^{1}$. My question is, if a rational function $f\in k(V)$ has a pole at $p\in V$, is there an expression $f=\frac{g}{h}$ where $g,h\in k[V]$ are regular functions, and $g(p)\ne 0$, $h(p)=0$?

When $V\subseteq {\mathbb{A}}_{k}^{1}$, this is clear, since if we have $f=\frac{g}{h}$ where $g(p)=h(p)=0$, we can simply reduce the expression of $g$ and $h$ and eliminate the factor $(x-p)$ until we get $f=\frac{{g}^{\prime}}{{h}^{\prime}}$ such that ${g}^{\prime}(p)\ne 0$, ${h}^{\prime}(p)=0$. But when $g,h$ are multivariate functions, I wonder how to get such a reduced expression?

When $V\subseteq {\mathbb{A}}_{k}^{1}$, this is clear, since if we have $f=\frac{g}{h}$ where $g(p)=h(p)=0$, we can simply reduce the expression of $g$ and $h$ and eliminate the factor $(x-p)$ until we get $f=\frac{{g}^{\prime}}{{h}^{\prime}}$ such that ${g}^{\prime}(p)\ne 0$, ${h}^{\prime}(p)=0$. But when $g,h$ are multivariate functions, I wonder how to get such a reduced expression?

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I am trying to decompose the following rational function: $\frac{1}{({x}^{2}-1{)}^{2}}$ in partial fractions (in order to untegrate it later).

I have notices that $({x}^{2}-1{)}^{2}=(x+1{)}^{2}(x-1{)}^{2}$

Therefore $\mathrm{\exists}A,B,C,D$ s.t:

$\frac{1}{({x}^{2}-1{)}^{2}}}={\displaystyle \frac{1}{(x-1{)}^{2}(x+1{)}^{2}}}={\displaystyle \frac{Ax+B}{(x+1{)}^{2}}}+{\displaystyle \frac{Cx+D}{(x-1{)}^{2}}$

We then have

$Ax+B={{\displaystyle \frac{1}{(x-1{)}^{2}}}|}_{x=-1}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}B-A=1/4$

Same thing for $Cx+D$:

$Cx+D={{\displaystyle \frac{1}{(x+1{)}^{2}}}|}_{x=1}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}C+D=1/4$

How do I find A, B, C, D from here?

I have notices that $({x}^{2}-1{)}^{2}=(x+1{)}^{2}(x-1{)}^{2}$

Therefore $\mathrm{\exists}A,B,C,D$ s.t:

$\frac{1}{({x}^{2}-1{)}^{2}}}={\displaystyle \frac{1}{(x-1{)}^{2}(x+1{)}^{2}}}={\displaystyle \frac{Ax+B}{(x+1{)}^{2}}}+{\displaystyle \frac{Cx+D}{(x-1{)}^{2}}$

We then have

$Ax+B={{\displaystyle \frac{1}{(x-1{)}^{2}}}|}_{x=-1}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}B-A=1/4$

Same thing for $Cx+D$:

$Cx+D={{\displaystyle \frac{1}{(x+1{)}^{2}}}|}_{x=1}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}C+D=1/4$

How do I find A, B, C, D from here?

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where $\{a,b,c\}\in \mathbb{R}$. I have looked in a table of integrals for rational functions, but with no luck. Is there a smart trick I can utilize?

${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}\frac{a+x}{{b}^{2}+(a+x{)}^{2}}\frac{1}{1+c(a-x{)}^{2}}dx$

where $\{a,b,c\}\in \mathbb{R}$. I have looked in a table of integrals for rational functions, but with no luck. Is there a smart trick I can utilize?

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