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How do you find horizontal asymptote ?

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

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Other answers:

@Dodo1 do you understand?
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. \[\lim_{x \rightarrow \infty}10+\frac{ 3x }{ x^2-12x+3 }\]
Yes, the basic concept! thank you i will take note.
Let me get my grade 12 notes
If y = [ax^n + ...............]/[Ax^N +.................] the horizontal asymptote = 0
Is the exponent on the bottom is greater than the one on the top, then it approaches zero
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?
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!
That's how I understood it :)
ok, thsnk you potterssheep. :) how do i find limit?
\[\lim_{x \rightarrow \infty}10+\frac{ 3x }{ x^2-12x+3 }\] 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
I see but why x-12= infinity?
\[x-12\] \[\infty-12\] Infinity is a very big no. subtracting 12 won't change it \[\infty-12 \longrightarrow \infty\]
Oh i see thank you!! how about \[\lim_{x \rightarrow -\infty} \frac{ 5-8x }{ 7+x }+\frac{ (6x^2+8) }{ (14x-12)^2 }\]
Do I mutliply the ()2 first then add?
\[\lim_{x \rightarrow -\infty} \frac{ 5-8x }{ 7+x }+\frac{ (6x^2+8) }{ (14x-12)^2 }\] 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?
OK, i got it so far.
:) its fun!
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?
mmm, 6/(196-336)?
x/x^2=1/x \[x\to -\infty \]\[\frac 1 x \to 0, \frac 1 {x^2} \to 0\]
Try again now :)
6/196?
yes, limit is the combined limit
whats combined limit?
\[-8+\frac 6 {196}\]
Oh i see! - infinity and infinity are really matter? beacuse it seems that it does nt matter.?
It does matter, in this question it doesn't
oh, not horizentally matter?
nope
got it! thank you, i have other 4 questions but i will try and if i stuck can i ask you?
Okay, try them. If I'm here, I'll help you But ask them in a new question. Close this

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