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logynklode

  • 3 years ago

Pre Calc. Can anyone help me?

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  1. 3psilon
    • 3 years ago
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    What's the question?

  2. findme
    • 3 years ago
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    "Can anyone help me?"

  3. findme
    • 3 years ago
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    YES! :D

  4. logynklode
    • 3 years ago
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  5. findme
    • 3 years ago
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    \[f(x) = \frac{ 2 }{ x^2-2x-3 }=\frac{ 2 }{ (x+1)(x-3) }\]

  6. logynklode
    • 3 years ago
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    okay

  7. findme
    • 3 years ago
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    That means the x cannot be -1, 3 so the domain (-infi, -1)U(-1, 3)U(3, infi)

  8. findme
    • 3 years ago
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    or x <-1, -1<x<3, x>3

  9. logynklode
    • 3 years ago
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    okay

  10. findme
    • 3 years ago
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    for x-intercept: set f(x)=0, then: \[0=\frac{ 2 }{ (x+1)(x-3) }\] since no x exist that satisfy the above equation, there is no x-intercept

  11. findme
    • 3 years ago
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    for y-intercept: set x = 0, then: \[f(0) = \frac{ 2 }{ (0+1)(0-3) }=\frac{ -2 }{ 3 }\] so the y-intercept is at the point (0, -2/3)

  12. findme
    • 3 years ago
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    To get the horizontal asymptote: divide every individual block in the equation by the highest power of x, which is x^2

  13. logynklode
    • 3 years ago
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    no x inct? no range?

  14. findme
    • 3 years ago
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    o lol, forgot about range and there's no x-int as I've explained

  15. logynklode
    • 3 years ago
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    okay

  16. findme
    • 3 years ago
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    sorry, I need to a lecture now, but here is the graph for reference. good luck :D

  17. logynklode
    • 3 years ago
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    okay.

  18. logynklode
    • 3 years ago
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    anyone?

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