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Dodo1

  • 2 years ago

How do you find horizontal asymptote ?

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  1. ash2326
    • 2 years ago
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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

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

  3. pottersheep
    • 2 years ago
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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. "

  4. ash2326
    • 2 years ago
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    @Dodo1 do you understand?

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

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

  7. pottersheep
    • 2 years ago
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    Let me get my grade 12 notes

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

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

  10. ash2326
    • 2 years ago
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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?

  11. pottersheep
    • 2 years ago
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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!

  12. pottersheep
    • 2 years ago
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    That's how I understood it :)

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

  14. ash2326
    • 2 years ago
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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

  15. Dodo1
    • 2 years ago
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    I see but why x-12= infinity?

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

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

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

  19. ash2326
    • 2 years ago
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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?

  20. Dodo1
    • 2 years ago
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    OK, i got it so far.

  21. Dodo1
    • 2 years ago
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    :) its fun!

  22. ash2326
    • 2 years ago
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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?

  23. Dodo1
    • 2 years ago
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    mmm, 6/(196-336)?

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

  25. ash2326
    • 2 years ago
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    Try again now :)

  26. Dodo1
    • 2 years ago
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    6/196?

  27. ash2326
    • 2 years ago
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    yes, limit is the combined limit

  28. Dodo1
    • 2 years ago
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    whats combined limit?

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

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

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

  32. Dodo1
    • 2 years ago
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    oh, not horizentally matter?

  33. ash2326
    • 2 years ago
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    nope

  34. Dodo1
    • 2 years ago
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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?

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

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