Question

How do you find horizontal asymptote ?

Answer 1

Suppose you have a function
\[y=f(x)\]
Express x in terms of y,
now find value of y for which x goes to infinity or - infinity
That value of y is your horizontal asymptote

Answer 2

This website helped me a lot when I was learning about horizontal aymptotes. http://www.purplemath.com/modules/asymtote2.htm

Answer 3

Btw, your line CAN touch a horizontal aymptote.

"Whereas vertical asymptotes are sacred ground, horizontal asymptotes are just useful suggestions. Whereas you can never touch a vertical asymptote, you can (and often do) touch and even cross horizontal asymptotes. Whereas vertical asymptotes indicate very specific behavior (on the graph), usually close to the origin, horizontal asymptotes indicate general behavior far off to the sides of the graph. "

Answer 4

@Dodo1 do you understand?

Answer 5

This is a question
A function is said to have a horizontal asymptote if either the limit at infinity exists or the limit at negative infinity exists.
Show that each of the following functions has a horizontal asymptote by calculating the given limit.
\[\lim_{x \rightarrow \infty}10+\frac{ 3x }{ x^2-12x+3 }\]

Answer 6

Yes, the basic concept! thank you i will take note.

Answer 7

Let me get my grade 12 notes

Answer 8

If y = [ax^n + ...............]/[Ax^N +.................]

the horizontal asymptote = 0

Answer 9

Is the exponent on the bottom is greater than the one on the top, then it approaches zero

Answer 10

Yes this is a way too, let's find the limit here
\[\lim_{x \rightarrow \infty}10+\frac{ 3x }{ x^2-12x+3 }\]
Can you find the limit?

Answer 11

Because eventually, the number on the bottom will become HUGEEEEE compared to the top. And a number / a hugeeeeeeeee number is close to zero, a veryy small decimal!

Answer 12

That's how I understood it :)

Answer 13

ok, thsnk you potterssheep.
:)
how do i find limit?

Answer 14

\[\lim_{x \rightarrow \infty}10+\frac{ 3x }{ x^2-12x+3 }\]
First divide the numerator and denominator by x
\[\lim_{x \rightarrow \infty}10+\frac{ 3\frac x x }{ \frac{x^2}{x}-12\frac x x +\frac 3 x}\]
We'll get
\[\lim_{x\to \infty} 10+\frac 3 {x-12+\frac 3 x }\]
3/x =0 so we get
\[\lim_{x\to \infty} 10+\frac 3 {x-12}\]
as \(x\to \infty\)
we get
\\[10+\frac 3 \infty\]
\[=10+0=>10\]
so that's the limit

Answer 15

I see but why x-12= infinity?

Answer 16

\[x-12\]
\[\infty-12\]
Infinity is a very big no.
subtracting 12 won't change it
\[\infty-12 \longrightarrow \infty\]

Answer 17

Oh i see thank you!!
how about
\[\lim_{x \rightarrow -\infty} \frac{ 5-8x }{ 7+x }+\frac{ (6x^2+8) }{ (14x-12)^2 }\]

Answer 18

Do I mutliply the ()2 first then add?

Answer 19

\[\lim_{x \rightarrow -\infty} \frac{ 5-8x }{ 7+x }+\frac{ (6x^2+8) }{ (14x-12)^2 }\]
Let's split the limits
\[\lim_{x \rightarrow -\infty} \frac{ 5-8x }{ 7+x }\]
First step divide by x , numerator and denominator
\[\lim_{x \rightarrow -\infty} \frac{\frac 5 x -8}{\frac 7 x +1}\]
1/x terms will become 0, when x goes to + or - infinity. We'll get
\[\frac{-8}{1}\]
Do you get this?

Answer 20

OK, i got it so far.

Answer 21

:) its fun!

Answer 22

Great, now the second part
\[\lim_{x \rightarrow -\infty}\frac{ (6x^2+8) }{ (14x-12)^2 }\]
Let's expand the denominator
\[\lim_{x \rightarrow -\infty}\frac{ (6x^2+8) }{ (196x^2-336x+144) }\]
Divide numerator and denominator by x^2,
\[\lim_{x \rightarrow -\infty}\frac{ (6+\frac8{x^2}) }{ (196-336\frac{x}{x^2}+\frac{144}{x^2}) }\]Can you find the limit from here?

Answer 23

mmm, 6/(196-336)?

Answer 24

x/x^2=1/x
\[x\to -\infty \]\[\frac 1 x \to 0, \frac 1 {x^2} \to 0\]

Answer 25

Try again now :)

Answer 26

6/196?

Answer 27

yes, limit is the combined limit

Answer 28

whats combined limit?

Answer 29

\[-8+\frac 6 {196}\]

Answer 30

Oh i see!
- infinity and infinity are really matter? beacuse it seems that it does nt matter.?

Answer 31

It does matter, in this question it doesn't

Answer 32

oh, not horizentally matter?

Answer 33

nope

Answer 34

got it! thank you, i have other 4 questions but i will try and if i stuck
can i ask you?

Answer 35

Okay, try them. If I'm here, I'll help you
But ask them in a new question. Close this