Question

find horizontal asymptote of (x-2)/(x^+1) +2

Answer 1

wouldn't it be just zero?

Answer 2

sorry I made a mistake! the bottom is \[x ^{2}-1\]

Answer 3

does the +2 at the end change anything?

Answer 4

\(\bf \cfrac{x-2}{(x^2-1)+2} \ \ ?\)

Answer 5

no the +2 is outside

Answer 6

so its just 2?

Answer 7

\(\bf \cfrac{x-2}{(x^2-1)}+2\implies \cfrac{x-2+2x^2-2}{x^2-1} \implies
\cfrac{3x^2-4}{x^2-1}\)

Answer 8

so the horizontal asymptote for a rational whose num/den are same degree is the leading coefficients of n/d

Answer 9

the +2 will make the distinction that you'd need to consolidate first, into one rational

Answer 10

wait how did that top become that?

Answer 11

Whoops... jdoe shouldn't have added the x to the \(\large 2x^2\)
Top should be as follows:
\[\Large \frac{\color{green}{2x^2 + x-4}}{x^2 -1}\]

Answer 12

why are you adding 2x^2

Answer 13

or how did that top change?

Answer 14

from x-2???

Answer 15

Is this the problem?
$$\tt
\begin{align*} \\
\frac{x-2}{x^2 -1}\\
\end{align*}$$

Answer 16

yeah and +2

Answer 17

$$\tt
\begin{align*} \\
\text{Horizontal Asymptote:}~
lim_{x\to\infty}\frac{x-2}{x^2-1}=0\\
\end{align*}$$

Answer 18

uh

Answer 19

its \[[(x-2)/(x ^{2}-1)] +2\]

Answer 20

\[\frac{ x-2 }{x^2+1}+2\]

Answer 21

sorry that's the correct one

Answer 22

np
$$\tt
\dfrac{ x-2 }{x^2+1}+2\\\\
\text{Horizontal Asymptote:}~~
lim_{x\to\infty}\dfrac{ x-2 }{x^2+1}+2=2\\
$$

Answer 23

horizontal asymptote is 2?

Answer 24

why?

Answer 25

Yes. The asymptote is where the function will "settle" as x gets larger. In the long-run, the first part of your equation converges (or "settles") to 0, while as x gets larger, the second part of your equation, the "2", which is of course a constant, keeps the entire function at 2.

Answer 26

The first part of your equation has x with degree 1 on the numerator and this is the dominant term as x gets larger. In the denominator, the x has a degree 2. As x gets largerer, x^2 in the denominator will become grow faster than "x" in the numerator. So, when x gets big enough, this whole fractional part looks like x/x^2 or 1/x. So 1/x tends to zero as x gets larger, leaving the constant "2" to dominate and thus creating the horizontal asymptote. The whole thing tends to 2 as x gets larger, never quite getting there.

Answer 27

ok thanks!