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anonymous

  • 5 years ago

Determine the equations of the asymptotes of the graph for the function f(x)=x/5x+2

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  1. amistre64
    • 5 years ago
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    va = -2/5 ha = 1/5

  2. amistre64
    • 5 years ago
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    assuming that 5x+2 is the denominator....

  3. anonymous
    • 5 years ago
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    yes it is , thank you ! could you show me how you achieved that. Im clueless.

  4. amistre64
    • 5 years ago
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    va..vertical asymptote...is whatever turns the bottom to zero; 5x+2 =0 5x = -2 x = -2/5 is the va

  5. amistre64
    • 5 years ago
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    the ha is found by dividing top and bottom by the highest power of x in the bottom; in this case that is just 'x' x/x 1 -------- = ------- 5x/x + 2/x 5 + (2/x)

  6. amistre64
    • 5 years ago
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    with it like this; we see that when x get very large; 1 ------------ = .0000...00001 - is very tiny 10000...0000

  7. amistre64
    • 5 years ago
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    so 2/x gets close to zero and fades away... in fact, anything with an x left on bottom goes to zero and we are left with: 1/5

  8. anonymous
    • 5 years ago
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    ohh right , so to find the the ha you always have to take the variable x into consideration

  9. amistre64
    • 5 years ago
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    yes

  10. anonymous
    • 5 years ago
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    The Rule for horizontals is that it depends on the degree of the top as compared to the degree of the bottom. If the degree of the top is smaller, then you have a horizontal asymptote on the x-axis. If the degrees are the same, then you have a horizontal asymptote on the line y=a/b where a is the leading coefficient on the top and b is the leading coefficient on the bottom. If the degree of the bottom is smaller, then you don't have a horizontal asymptote - you have an oblique one instead. I teach my students to find those using synthetic division and ignoring the remainder but you can just use regular polynomial long-division

  11. amistre64
    • 5 years ago
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    and if you want to read a novel..there it is :)

  12. anonymous
    • 5 years ago
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    thank you both for your help !

  13. amistre64
    • 5 years ago
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    :) youre welcome.

  14. anonymous
    • 5 years ago
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    sorry for butting in - someone told me about this site and I was looking and saw this question. I am a teacher and gave a final exam that had this question on it yesterday so thought I'd try to help. Cool site - I'm going to tell my students about it!

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