Graphing a rational function with a slant asymptote

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  • Опубліковано 6 січ 2025

КОМЕНТАРІ • 15

  • @googleuser1937
    @googleuser1937 4 роки тому +1

    One of the best teachers out there on the internet.

  • @TheMissAlira
    @TheMissAlira 6 років тому +1

    Thank you so much for this example I just tried to understand similar, it helped!

  • @mussienewaymitiku3792
    @mussienewaymitiku3792 6 років тому

    its simple and easy ways of explanation and i am satisfy thank you so much!!!

  • @7elas241
    @7elas241 6 років тому +4

    I don't speak English very well but I understand your explain thank you

  • @onethkahandawa1909
    @onethkahandawa1909 4 роки тому +1

    Perfect👍🏻

  • @jaia846
    @jaia846 6 років тому +2

    What if you have 1/3x -1/3 as your slant ..do you start at -1/3

    • @brianmclogan
      @brianmclogan  6 років тому +2

      that would be the y-intercept, yes

    • @jaia846
      @jaia846 6 років тому

      Thanks sooo much 👍🏾

    • @brianmclogan
      @brianmclogan  6 років тому

      you are very welcome!

  • @maricarcruz7556
    @maricarcruz7556 5 років тому

    I just notice that getting the x intercept is like getting the Vertical asymptote but instead of getting the zero of denominator we will use the numerator🤔🤔 to get x-int?

    • @angelmendez-rivera351
      @angelmendez-rivera351 3 роки тому

      Yes, that is fact what it is. The vertical asymptote of the numerator is the zero of the denominator, and vice versa.

  • @segayanmx4442
    @segayanmx4442 2 роки тому

    Sir, why you don't calculate the derivative of the function to study the overall curve of the function? I've seen many videos of yours ,but in these you avoid to calculate !

  • @tohidalamkhan4081
    @tohidalamkhan4081 6 років тому

    Tq sir

  • @irwandasaputra9315
    @irwandasaputra9315 2 роки тому

    2x+x^-1