Vertical Asymptotes

Discussion in 'Other Pre-University Math' started by nycmathguy, Sep 24, 2021.

  1. nycmathguy

    nycmathguy

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    Section 2.6
    10, 12, 14, 16

    20210924_153258.jpg

    Question 10

    Set denominator to 0 and solve for x.

    (x - 2)^3 = 0

    Cube root both sides.

    cuberoot{(x - 2)^3} = cuberoot{0}

    x - 2 = 0

    x = 2

    Question 12

    Set denominator to 0 and solve for x.

    3 + 2x = 0

    2x = -3

    x = -3/2

    Question 14

    Set denominator to 0 and solve for x.

    x + 2 = 0

    x = -2

    Question 16

    For this question, the quadratic formula is needed.

    See attachment.

    20210924_154526.jpg
     
    nycmathguy, Sep 24, 2021
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  2. nycmathguy

    MathLover1

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    Question 10
    Vertical: x=2-> you did this part
    Horizontal: y=0

    Question 12
    Vertical: x= - 3/2-> you did this part
    Horizontal: y=- 7/2

    Question 14
    Vertical: x=-2
    Horizontal:
    Do polynomial long division 4x^2/(x+2)=4x-8+16/(x+2)
    y=4x-8 (slant asymptote)
    A horizontal asymptote is a special case of a slant asymptote.
    [​IMG]

    Question 16
    upload_2021-9-24_16-5-22.png

    as you can see from the graph, there is no vertical asymptote (function is continuous)


    Horizontal:
    f(x)=(-4x^2+1)/(x^2+x+3) horizontal is quotient of -4x^2/x^2=-4

    =>horizontal asymptote is y=-4
     
    MathLover1, Sep 24, 2021
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  3. nycmathguy

    nycmathguy

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    Why is my work for 16 wrong?
     
    nycmathguy, Sep 25, 2021
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  4. nycmathguy

    MathLover1

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    what you did in 16 is finding roots, and you needed too find asymptotes
     
    MathLover1, Sep 25, 2021
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  5. nycmathguy

    nycmathguy

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    This is what happens when trying to solve math problems with little to no sleep.
     
    nycmathguy, Sep 25, 2021
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