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@misty1212 how can i find discontinuities for vertical horizontal or oblique asymptotes ?
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you got an example to work with?
i mean i can make one up if you do not
if you had for example \[\frac{x-2}{x+1}\] to find the vertical asymptote, i.e. the discontinuity you would set the denominator equal to zero and solve \[x+1=0\] so \[x=-1\] is the vertical asymptote
you would have a "oblique" asympote, aka a line, if the degree of the numerator was one more than the degree of the deominator like for example \[\frac{x^2+2x-1}{x+1}\]
to find it you have do divide using either long division or synthetic division in this case you would get \[x+1-\frac{2}{x+1}\] the obliques asympote is the line \(y=x+1\)
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