What are the asymptotes of $y=\frac{2}{x}+3$ and how do you graph the function?

Natalya Mayer
2022-09-13
Answered

What are the asymptotes of $y=\frac{2}{x}+3$ and how do you graph the function?

You can still ask an expert for help

Ashlynn Cox

Answered 2022-09-14
Author has **12** answers

I tend to think of this function as a transformation of the function $f\left(x\right)=\frac{1}{x}$, which has a horizontal asymptote at y=0 and a vertical asymptote at x=0.

The general form of this equation is $f\left(x\right)=\frac{a}{x-h}+k$

In this transformation, h=0 and k=3, so the vertical asymptote is not shifted left or right, and the horizontal asymptote is shifted up three units to y=3.

graph{2/x+3 [-9.88, 10.12, -2.8, 7.2]}

The general form of this equation is $f\left(x\right)=\frac{a}{x-h}+k$

In this transformation, h=0 and k=3, so the vertical asymptote is not shifted left or right, and the horizontal asymptote is shifted up three units to y=3.

graph{2/x+3 [-9.88, 10.12, -2.8, 7.2]}

asked 2021-02-25

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

asked 2022-06-27

I have the function: $f(x)=2x-{2}^{x}+2$

I know that this function has an oblique asymptote, but all the tutorials I can find on google, are with rational functions with the form:

$f(x)=\frac{P(x)}{Q(x)}$

Where they simply just divide the denominator with the numerator.

But I can't do that, because my equation doesn't contain any fractions. So my question is: how do I find the function to the oblique asymptote for my $f(x)$?

I know that this function has an oblique asymptote, but all the tutorials I can find on google, are with rational functions with the form:

$f(x)=\frac{P(x)}{Q(x)}$

Where they simply just divide the denominator with the numerator.

But I can't do that, because my equation doesn't contain any fractions. So my question is: how do I find the function to the oblique asymptote for my $f(x)$?

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Which is the direct linear variation equation for the relationship given y varies directly with x and y = 12 when x = 3?

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I sometimes have some trouble finding the range of a rational function. What is an easy or good way to do this?

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How to show that the two rational functions $a(t),b(t)\in \mathbb{K}(t)$ when $\mathrm{Char}(\mathbb{K})\ne 2$ are algebraically dependent.

(there is polynomial $P(x,y)\in \mathbb{K}[x,y]$ such that $P(a(t),b(t))=0$)?

(there is polynomial $P(x,y)\in \mathbb{K}[x,y]$ such that $P(a(t),b(t))=0$)?

asked 2022-06-06

Can anyone help me prove the convexity of this rational function? The man who proved the convexity of function used these facts. But I don't know this fact is correct or not. Here are the facts and function:

1. As N increases, f(N) goes to infinity, That implies that there must be a minima (either at N=0 or somewhere else with a finite N)

2. There cannot be more than one (positive) minima since we're dealing with second order equation.

$f(N)=+-c{N}^{4}+-d{N}^{3}+-e{N}^{2}+-fN+-g/+-a{N}^{2}+-bN$

a,b,c,d,e,f,g is constants. and $N\ge 1$. I guess the second order equation means that between the leading coefficient of the numerator and the denominator is 2.

Are these facts correct? I think fact 1 is no problem, but fact 2 is correct or not.

I am waiting for any answers.

1. As N increases, f(N) goes to infinity, That implies that there must be a minima (either at N=0 or somewhere else with a finite N)

2. There cannot be more than one (positive) minima since we're dealing with second order equation.

$f(N)=+-c{N}^{4}+-d{N}^{3}+-e{N}^{2}+-fN+-g/+-a{N}^{2}+-bN$

a,b,c,d,e,f,g is constants. and $N\ge 1$. I guess the second order equation means that between the leading coefficient of the numerator and the denominator is 2.

Are these facts correct? I think fact 1 is no problem, but fact 2 is correct or not.

I am waiting for any answers.

asked 2022-05-21

Could someone please let me know if my steps are correct? I am trying to rewrite the first line as a rational function, $\frac{p(x)}{q(x)}$

$\begin{array}{rl}& 3{x}^{1}+3{x}^{2}+9{x}^{3}+9{x}^{4}+27{x}^{5}+27{x}^{6}+\cdots \\ =& 3x(1+x)+9{x}^{3}(1+x)+27{x}^{5}(1+x)+\cdots \\ =& 3x(1+x)(1+3{x}^{2}+9{x}^{4}+\cdots )\\ =& (3x+3{x}^{2})\frac{1}{1-3{x}^{2}}\\ =& \frac{3x+3{x}^{2}}{1-3{x}^{2}}\end{array}$

$\begin{array}{rl}& 3{x}^{1}+3{x}^{2}+9{x}^{3}+9{x}^{4}+27{x}^{5}+27{x}^{6}+\cdots \\ =& 3x(1+x)+9{x}^{3}(1+x)+27{x}^{5}(1+x)+\cdots \\ =& 3x(1+x)(1+3{x}^{2}+9{x}^{4}+\cdots )\\ =& (3x+3{x}^{2})\frac{1}{1-3{x}^{2}}\\ =& \frac{3x+3{x}^{2}}{1-3{x}^{2}}\end{array}$