How to Find Slant and Vertical Asymptotes
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- Опубліковано 1 жов 2012
- 👉 Learn how to find the slant/oblique asymptotes of a function. A slant (oblique) asymptote usually occurs when the degree of the polynomial in the numerator is higher than the degree of the polynomial in the denominator. To find the slant asymptote of a rational function, we divide the numerator by the denominator using either long division or synthetic division. The quotient obtained when the numerator is divided by the denominator is the slant asymptote of the function.
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I plotted the function in Grapher app and didn't show x=1 as an vertical asymptote. This may be due to the fact that we can factorise the numerator and denominator and simplify the expression.
that is correct, it is actually a hole of the function
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I have to factor the denominator and equal it to zero. Even the numerator. I have to factor them and equal to 0 after that i have to solve them to get my VA. I love your explanation. Professor you are doing great job sir.
I wish you were my teacher...oops you are. Thanks for the videos! Get it you are, bcuz I watch your videos.:)
Thanks!
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Lol the as before this was a “study pod” video that began with “are you really still looking on youtube for math videos? Are those 6 year old videos really still working?? Try study pod.. blah blah” and like yes these videos do help lol
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(x^3 - 1)/(x^2 - x) = (x^2 + x + 1)·(x - 1)/[x·(x - 1)]. Because of the common factor of x - 1 in the numerator and denominator, there is no vertical asymptote there. This is a hole.
As for the oblique asymptote, this is best found by doing division. When you remove the hole at 1, you get (x^2 + x + 1)/x = x + 1 + 1/x. The polynomial part is x + 1, and so the oblique asymptote is given by y = x + 1.
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x=1 is a gap it is not asymptote.
Someone coming over the intercom to announce that students are to be marked tardy if they are not back in their classrooms after a certain number of minutes.... We've heard it once or twice before. XD