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Abe_le_Babe
How to find Horizontal Asymptote of: (2x+1)/(x-2)
it's when the function = zero in the numerator... so it's when 2x+1 = 0 for vertical asymptote it's when the function is undefined so when x-2 = 0
you have the correct way to find the vertical asymptote ryan, but not the horizontal
Yes, I know the Vertical, I was inquiring about the Horizontal.
since the degrees are equal, simply divide the leading coefficients to get 2/1 = 2 So the horizontal asymptote is y = 2
If the degree of the denominator is larger, then the horizontal asymptote is simply y = 0
If the degree of the numerator is larger, then you have to use polynomial long division (but at this point, you won't have a horizontal asymptote)
So, if equal degrees, you just divide it out. If denominator degree is greater, the y asymptote is simply 0. and if the numerator degree is greater, you use long division and have a slant asymptote. Is that correct?
you are exactly correct !! Abe!!
that is correct, either a slant asymptote or something a bit more complicated
Thank you very much Jim and Star.