## iNeedhelpwithSchool 2 years ago What kinds of discontinuities are possible for a rational function?

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1. AKANicole

i think um,

2. AKANicole

um i forgot i know this somewhere,

3. jim_thompson5910

Case 1) The function is completely continuous over the entire domain, so there are no discontinuities Ex: y = 1/(x^2 + 1) The denominator is ALWAYS positive, so it can NEVER be zero ----> no vertical asymptotes ----> no discontinuities ------------------------------------------------------- Case 2) There is at least one asymptotic discontinuity, so it's discontinuous at this point or points. Ex: y = 1/(x+1) ... has a vertical asymptote at x = -1 ------------------------------------------------------- Case 3) There is a removable discontinuity (or a point discontinuity) Ex: y = ((x+1)(x+2))/(x+2) simplifies to y = x+1, but it has a removable discontinuity at x = -2

4. oldrin.bataku

Adding on to @jim_thompson5910's last note -- when you "simplify" $$\frac{(x+1)(x+2)}{x+2}=x+1$$, you're dividing both our numerator and denominator by $$x+2$$. Remember that we're never allowed to divide by $$0$$, so by simplifying we're essentially assuming that $$x+2\ne0$$ -- and this assumption is good for almost all $$x$$ -- except for $$x=-2$$. We'll just be left with a little hole there, with the rest of the function looking identical to $$x+1$$, but there's still that *removable* discontinuity, removable because it's just a single little hole that we could "fill in" to be $$y+1$$.