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Dodo1

How do you find horizontal asymptote ?

  • one year ago
  • one year ago

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  1. ash2326
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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

    • one year ago
  2. pottersheep
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    This website helped me a lot when I was learning about horizontal aymptotes. http://www.purplemath.com/modules/asymtote2.htm

    • one year ago
  3. pottersheep
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    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. "

    • one year ago
  4. ash2326
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    @Dodo1 do you understand?

    • one year ago
  5. Dodo1
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    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. \[\lim_{x \rightarrow \infty}10+\frac{ 3x }{ x^2-12x+3 }\]

    • one year ago
  6. Dodo1
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    Yes, the basic concept! thank you i will take note.

    • one year ago
  7. pottersheep
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    Let me get my grade 12 notes

    • one year ago
  8. pottersheep
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    If y = [ax^n + ...............]/[Ax^N +.................] the horizontal asymptote = 0

    • one year ago
  9. pottersheep
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    Is the exponent on the bottom is greater than the one on the top, then it approaches zero

    • one year ago
  10. ash2326
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    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?

    • one year ago
  11. pottersheep
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    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!

    • one year ago
  12. pottersheep
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    That's how I understood it :)

    • one year ago
  13. Dodo1
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    ok, thsnk you potterssheep. :) how do i find limit?

    • one year ago
  14. ash2326
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    \[\lim_{x \rightarrow \infty}10+\frac{ 3x }{ x^2-12x+3 }\] 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

    • one year ago
  15. Dodo1
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    I see but why x-12= infinity?

    • one year ago
  16. ash2326
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    \[x-12\] \[\infty-12\] Infinity is a very big no. subtracting 12 won't change it \[\infty-12 \longrightarrow \infty\]

    • one year ago
  17. Dodo1
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    Oh i see thank you!! how about \[\lim_{x \rightarrow -\infty} \frac{ 5-8x }{ 7+x }+\frac{ (6x^2+8) }{ (14x-12)^2 }\]

    • one year ago
  18. Dodo1
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    Do I mutliply the ()2 first then add?

    • one year ago
  19. ash2326
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    \[\lim_{x \rightarrow -\infty} \frac{ 5-8x }{ 7+x }+\frac{ (6x^2+8) }{ (14x-12)^2 }\] 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?

    • one year ago
  20. Dodo1
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    OK, i got it so far.

    • one year ago
  21. Dodo1
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    :) its fun!

    • one year ago
  22. ash2326
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    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?

    • one year ago
  23. Dodo1
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    mmm, 6/(196-336)?

    • one year ago
  24. ash2326
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    x/x^2=1/x \[x\to -\infty \]\[\frac 1 x \to 0, \frac 1 {x^2} \to 0\]

    • one year ago
  25. ash2326
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    Try again now :)

    • one year ago
  26. Dodo1
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    6/196?

    • one year ago
  27. ash2326
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    yes, limit is the combined limit

    • one year ago
  28. Dodo1
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    whats combined limit?

    • one year ago
  29. ash2326
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    \[-8+\frac 6 {196}\]

    • one year ago
  30. Dodo1
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    Oh i see! - infinity and infinity are really matter? beacuse it seems that it does nt matter.?

    • one year ago
  31. ash2326
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    It does matter, in this question it doesn't

    • one year ago
  32. Dodo1
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    oh, not horizentally matter?

    • one year ago
  33. ash2326
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    nope

    • one year ago
  34. Dodo1
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    got it! thank you, i have other 4 questions but i will try and if i stuck can i ask you?

    • one year ago
  35. ash2326
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    Okay, try them. If I'm here, I'll help you But ask them in a new question. Close this

    • one year ago
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