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Rational Functions

does the following function have a hole or a vertical asymptote or both? how do you know? find the y-value at that point.


I factored this out to

y = (x+3)(x+4)/(x+5)(x+3)
The x+3's cancelled out. I don't get the whole verticle and horizontal asymptote stuff, how do u know w/o using technology whether there is an asymptote and if there is a hole?

  • Pre-Calculus -

    If a pair of factors cancel, then you would get a hole
    in your case, there is a hole when x = -3
    If you have a factor in the denominator which does not cancel, it will cause a vertical asymptote
    in your case there will be a vertical asymptote at x = -5

    BTW, you should use brackets when typing a function like yours
    = (x+3)(x+4)/( (x+5)(x+3) )

  • Pre-Calculus -

    there is a vertical asymptote if the denominator is zero and the numerator is not zero

    there is a horizontal asymptote if the degree of the numerator is less than or equal to the degree of the denominator

    There is a slant asymptote if the degree of the numerator is 1 more than the degree of the denominator.

    I can't believe these facts were not covered in your book.


    y = (x+4)/(x+5) as long as x ≠ -3

    If x = -3, y = 0/0 which is not defined. There is a "hole" at (-3,1/2)

    So, you should be able to figure the asymptotes present.

    Visit and type in

    plot (x^2+7x+12)/(x^2+8x+15)

    you can play around with lots of stuff there

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