Simplify Asymptotes Learning with Plainmath's Expert Tips and Real-World Examples

Describe the vertical asymptotes) and holes) for the graph of $y=\frac{(x+2)(x+4)}{(x+4)(x+1)}$
A. asymtotes: x=-1 and hole: x=4
B. asymtotes: x=2 and hole: x=-1
C. asymtotes: x= -1 and hole: x=-4
D. asymtotes: x=1 and hole: x=4

Which of the following statements is true for the function
f(x)=x 2 +x−6 2x 2 −2 ?
a)
y=1 2 is a horizontal asymptote and x=2 is a vertical asymptote.
b)
y=1 2 is a horizontal asymptote and x=1 is a vertical asymptote.
c)
y=3 is a horizontal asymptote and x=-1 is a vertical asymptote.
d)
y=3 is a horizontal asymptote and x=2 is a vertical asymptote

How do you find vertical and horizontal asymptotes
Whenever i am doing curve sketching in calculus class i find it incredibly difficult to find the vertical and horizontal asymptotes and as such i always get up to that point correct and anything from asymptotes and further wrong. Sometimes even my curve is a bit off

Asymptotes of a function
I have the function $f(x)=x\cdot <!--\; \cdot \; -->\arctan <!--\; \; -->(x)$ And i need to find all asymptotes, I am stuck at $\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}x\cdot <!--\; \cdot \; -->\arctan <!--\; \; -->(x)$, please help me.

Finding the horizontal asymptotes of f
The question I must answer is:
"Find the horizontal asymptotes of ${\displaystyle f(x)=\frac{\sqrt{10x^2+11}}{12x+10}}$"
However, I am unsure how to find the horizontal asymptote or what this means. Can anyone explain this?

Find the asymptotes of the Parametric equation?
Consider
$x(t)=2e^{-<!--\; -\; -->t}+3e^{2t}$
$y(t)=5e^{-<!--\; -\; -->t}+2e^{2t}$
which represents a non rectilinear paths
Horizontal and Verical Asymptotes :
If $t\to <!--\; \to \; -->+\mathrm{\infty <!--\; \infty \; -->}or-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}$ and $x(t)andy(t)\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$, So there are no asymptotes parallel to coordinate axis
oblique Asymptotes:
Please tell me how to find the Oblique asymptotes

Infinite amount of vertical asymptotes
Is it possible that the graph of function has infinitely many vertical asymptotes?
I suppose, that it is not possible, because such function would not exist. But I need to prove it in a math-fashioned-way, and I'm clueless how to do it.

What are the vertical asymptotes for this function?
$F(x)=\frac{5x^2}{4x^2}+9$
Okay so when i graphed this function there are no vertical asymptotes but why is that ? Becuase if u set the denominator equal to zero then u can solve and get the vertical asymptotes

Calculating slant asymptotes of radical function
I'm trying to calculate the slant asymptotes of the function $\sqrt{x^2+2x+2}$. I've found out that the gradients of the asymptotes are 1 for $x\to <!--\; \to \; -->+\mathrm{\infty <!--\; \infty \; -->}$ and -1 for $x\to <!--\; \to \; -->-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}$. I've also found out that the constant of the positive asymptote is 1. Intuitively, I know the constant of the negative asymptote is -1, but I'm struggling to show it through calculation. I need to evaluate this to find it:
$\underset{x\to <!--\; \to \; -->-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\sqrt{x^2+2x+2}+x$
without using l'Hôpital's rule (for the purposes of the assignment I'm not supposed to know how to use it.) I have tried rationalizing the numerator using the conjugate but I just end up with an undefined value.

Asymptotes and uniform continuity
Let $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ be a continuous function with oblique asymptotes for $x\to <!--\; \to \; -->\pm <!--\; \pm \; -->\mathrm{\infty <!--\; \infty \; -->}$. Prove that the function is uniformly continuous.
Proof: Let . Since the function has obliquous asymptotes then
$\mathrm{\exists <!--\; \exists \; -->}M>0:|f(x)-<!--\; -\; -->(ax+b)|<\mathrm{ϵ<!--\; ϵ\; -->}\mathrm{\forall <!--\; \forall \; -->}x>M$
and
$\mathrm{\exists <!--\; \exists \; -->}m>0:|f(x)-<!--\; -\; -->(cx+d)|<\mathrm{ϵ<!--\; ϵ\; -->}\mathrm{\forall <!--\; \forall \; -->}x<-<!--\; -\; -->m$
Now, in the interval the function is uniformly continuous because of the Heine-Cantor theorem. We deal with the case $+\mathrm{\infty <!--\; \infty \; -->}$ since the other one is formally identical. We know that lines are uniformly continuous. So, using the that certainly exists for the line we have:
$\begin{array}{rl}|f(x)-<!--\; -\; -->f(x_0)|& =|f(x)-<!--\; -\; -->(a{x}_0+b)+(a{x}_0+b)-<!--\; -\; -->f(x_0)|\\ & \le <!--\; \le \; -->|f(x)-<!--\; -\; -->(a{x}_0+b)|+|(a{x}_0+b)-<!--\; -\; -->f(x_0)|\\ & <|f(x)-<!--\; -\; -->(a{x}_0+b)|+\mathrm{ϵ<!--\; ϵ\; -->}\end{array}$
But
$|f(x)-<!--\; -\; -->(a{x}_0+b)|\le <!--\; \le \; -->|f(x)-<!--\; -\; -->(ax+b)|+|(ax+b)-<!--\; -\; -->(a{x}_0+b)|<2\mathrm{ϵ<!--\; ϵ\; -->}$
So we are finished because if a continuous function is uniformly continuous in and , then it is uniformly continuous in .

Vertical asymptotes if any? How do you find it?
$$f(x)=\frac{4x}{4x^2-<!--\; -\; -->2}$$

Oblique asymptotes?
A rational function, $\frac{p(x)}{q(x)}$ has an oblique asymptote only when the degree of p(x)= degree of q(x)−1.
What "causes" the "slant" of the asymptote? Most asymptotes are caused by a function approaching an undefined value - I assume this is the same, but why (unlike others) would these asymptotes be slanted?
Why does this only work with a difference of 1 between degrees?

Identifying Asymptotes of a Hyperbola
basic hyperbolic functionHow would I find the vertical and horizontal asymptotes of a $y=\frac{1}{x}$ function algebraically? For example, $y=-<!--\; -\; -->\frac{2}{x+3}-<!--\; -\; -->1$ (as you would type into a calculator). Simply, how do I find the x and y values by looking at this equation? In other words, the middle point where the asymptotes in the picture has moved and the whole graph has been vertically stretched, where are the asymptotes now?

Finding the horizontal asymptotes of a function
Here is the function I am trying to find the horizontal asymptotes for:
$y=\frac{1-<!--\; -\; -->2^x}{1+2^x}$
Could you please explain how the horizontal asymptotes can be found for this function?

Finding the asymptotes of a hyperbola
Given the hyperbola $y^2/9$ what would the equation of the asymptotes be?

Obliques asymptotes of a function
$f:[0,\mathrm{\infty <!--\; \infty \; -->})\to <!--\; \to \; -->R,f(x)=\sqrt{x^2+x\ln <!--\; \; -->(e^x+1)}$
I have this function and i need to find out the asymptotes to $+\mathrm{\infty <!--\; \infty \; -->}$ (+infinity)
i calculate the horizontal ones, and they are $+\mathrm{\infty <!--\; \infty \; -->}$
i can't calculate the limits at the obliques asymptotes

Do all functions with vertical asymptotes also have oblique asymptotes?
I just started learning about asymptotes in my Advanced Functions class, and as I was taking a look at all this stuff, a question came up. Do all rational functions that have vertical asymptotes also have an oblique asymptote? Or is an oblique asymptote only formed when the degree of the numerator is 1 higher than the degree of the denominator, and so only functions with a vertical asymptote with a degree of 1 can also have an oblique asymptote?

Finding total number of asymptotes
I was solving some problems Related to asymptotes, I have a question Can we know number of asymptotes of a 1.rational function 2.trignometric function 3. Implicit function
Without calculating them how can we know how many total number of asymptotes function will have? is we guess this from degree of denominator? but in implicit functions we cannot separate y to put in fraction? plz help me

Graphic intersecting asymptotes
Sometimes graphics intersect the asymptotes(horizontal) of the function we plot and then they tend to the asymptote to infinity.What gives us the information whether the graph only tends to the asymptote and does not intersect it or intersects the asymptote a then tends to her for a given point?

Asymptotes of $\arctan <!--\; \; -->2x$
My book tells me the horizontal asymptotes of $\arctan <!--\; \; -->2x$ is either at positive or negative $\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}$, yet the vertical asymptotes of $\tan <!--\; \; -->2x$ occurs at positive or negative $x=\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}$, so obviously the horizontal asymptotes of $\arctan <!--\; \; -->2x$ should be at either positive or negative $y=\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}$.
Can someone tell me what I'm doing wrong?