anonymous
  • anonymous
How do you find horizontal asymptote ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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ash2326
  • ash2326
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
pottersheep
  • pottersheep
This website helped me a lot when I was learning about horizontal aymptotes. http://www.purplemath.com/modules/asymtote2.htm
pottersheep
  • pottersheep
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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ash2326
  • ash2326
@Dodo1 do you understand?
anonymous
  • anonymous
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. \[\lim_{x \rightarrow \infty}10+\frac{ 3x }{ x^2-12x+3 }\]
anonymous
  • anonymous
Yes, the basic concept! thank you i will take note.
pottersheep
  • pottersheep
Let me get my grade 12 notes
pottersheep
  • pottersheep
If y = [ax^n + ...............]/[Ax^N +.................] the horizontal asymptote = 0
pottersheep
  • pottersheep
Is the exponent on the bottom is greater than the one on the top, then it approaches zero
ash2326
  • ash2326
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?
pottersheep
  • pottersheep
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!
pottersheep
  • pottersheep
That's how I understood it :)
anonymous
  • anonymous
ok, thsnk you potterssheep. :) how do i find limit?
ash2326
  • ash2326
\[\lim_{x \rightarrow \infty}10+\frac{ 3x }{ x^2-12x+3 }\] 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
anonymous
  • anonymous
I see but why x-12= infinity?
ash2326
  • ash2326
\[x-12\] \[\infty-12\] Infinity is a very big no. subtracting 12 won't change it \[\infty-12 \longrightarrow \infty\]
anonymous
  • anonymous
Oh i see thank you!! how about \[\lim_{x \rightarrow -\infty} \frac{ 5-8x }{ 7+x }+\frac{ (6x^2+8) }{ (14x-12)^2 }\]
anonymous
  • anonymous
Do I mutliply the ()2 first then add?
ash2326
  • ash2326
\[\lim_{x \rightarrow -\infty} \frac{ 5-8x }{ 7+x }+\frac{ (6x^2+8) }{ (14x-12)^2 }\] 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?
anonymous
  • anonymous
OK, i got it so far.
anonymous
  • anonymous
:) its fun!
ash2326
  • ash2326
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?
anonymous
  • anonymous
mmm, 6/(196-336)?
ash2326
  • ash2326
x/x^2=1/x \[x\to -\infty \]\[\frac 1 x \to 0, \frac 1 {x^2} \to 0\]
ash2326
  • ash2326
Try again now :)
anonymous
  • anonymous
6/196?
ash2326
  • ash2326
yes, limit is the combined limit
anonymous
  • anonymous
whats combined limit?
ash2326
  • ash2326
\[-8+\frac 6 {196}\]
anonymous
  • anonymous
Oh i see! - infinity and infinity are really matter? beacuse it seems that it does nt matter.?
ash2326
  • ash2326
It does matter, in this question it doesn't
anonymous
  • anonymous
oh, not horizentally matter?
ash2326
  • ash2326
nope
anonymous
  • anonymous
got it! thank you, i have other 4 questions but i will try and if i stuck can i ask you?
ash2326
  • ash2326
Okay, try them. If I'm here, I'll help you But ask them in a new question. Close this

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