Rational Function With Slant Asymptote...3

Discussion in 'Other Pre-University Math' started by nycmathguy, Sep 29, 2021.

  1. nycmathguy

    nycmathguy

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    Section 2.6
    Question 52

    20210925_001418.jpg

    Part (a)

    x - 1 = 0

    x = 1

    Domain = All real numbers except for x = 1.

    Part (b)

    h(x) = x^2/(x - 1)

    Let h(x) = 0

    0 = x^2/(x - 1)

    0(x - 1) = x^2

    0 = x^2

    0 = x

    h(0) = 0^2/(0 - 1)

    h(0) = 0/-1

    h(0) = 0

    The x-intercept = 0 = y-intercept.

    Part (c)

    The vertical asymptote is x = 1.
    The slant asymptote is the line h(x) = x + 1.

    Part (d)

    x............h(x)
    0.............1
    -1............0

    Before graphing the given function, is my work here right?
     
    nycmathguy, Sep 29, 2021
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  2. nycmathguy

    MathLover1

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    a, b, and c are correct

    Part (d)
    should be
    x............h(x)
    0.............0 ->h(0) = 0^2/(0 - 1)=0/-1=0
    -1............-1/2 ->h(-1) = (-1)^2/(-1 - 1) =1/-2=-1/2
     
    MathLover1, Sep 29, 2021
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    nycmathguy likes this.
  3. nycmathguy

    nycmathguy

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    Thank you. I will post the graph tomorrow. On the 4 train now to work. Maybe some word problems later. We are done with Section 2.6.
     
    nycmathguy, Sep 30, 2021
    #3
  4. nycmathguy

    nycmathguy

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    Here are some more additional points to make our graph:

    x.........h(x)
    -2.........-4/3
    -1.........-1/2
    0..............0
    2..............4

    20210930_110324.jpg
     
    nycmathguy, Sep 30, 2021
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  5. nycmathguy

    MathLover1

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    correct
     
    MathLover1, Oct 1, 2021
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  6. nycmathguy

    nycmathguy

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    Wish I had more time to graph by hand. Lots of fun for sure.
     
    nycmathguy, Oct 1, 2021
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