<p>Sorry, I know that this is not a SAT question but it's also urgent and CC is the first place I could think of for smart math people.</p>
<p>For graphs like</p>
<p>(2x-4)/((x^2)-4)</p>
<p>There are vertical asymptotes, but I don't understand in which situations there would be second vertical asymptote or a removable discontinuity. For this particular graph, there is one asymptote at x= -2 and a discontinuity at x= 2. How would I know this is so, and not the other way around, or if there are just two asymptotes? Thank you in advance!</p>
<p>Yeah, not SAT – but kind of interesting. One thing that might help is to do another one where we start backwards:</p>
<p>Say you have a line: y=2x +1…clearly there are no asymptotes and no discontinuites. Now what if you wanted to put a hole in that line at say x = 5…You could do it by building a new function:</p>
<p>y=(2x+1)(x-5)/(x-5)</p>
<p>At all values besides x=5, your new terms would cancel and the new function would be the same line as the old one. But at x=5, the new function is undefined – it has a hole.</p>
<p>Now what if you wanted to hide what you had done? You’d foil out the numerator and write your function this way:</p>
<p>y= (2x^2-9x -5)/(x-5)</p>
<p>It would be less obvious that x=5 was a hole rather than an asymptote. To see what was going on, you’d have to factor.</p>
<p>And that’s your answer: with rational functions, factor the numerator and denominator. Cancel what you can – they were holes (unless they appeared multiply in the denominator). The zeroes remaining in the denominator are asymptotes.</p>
<p>The following theorem is helpful:</p>
<p>For rational functions (one polynomial divided by another), if plugging c into the denominator gives 0, and plugging c into the numerator does NOT give 0, then x=c is a vertical asymptote.</p>
<p>In your example above, notice that plugging -2 into the denominator gives 0, and plugging -2 into the numerator does NOT give 0 (it gives -8). So by the above theorem, x=-2 is a vertical asymptote.</p>
<p>When you plug in 2 you get 0/0, so the theorem does not apply. 0/0 is called an indeterminate form. There may or may not be a vertical asymptote there. If the limit is a finite number, then there is a removable discontinuity (not an asymptote). If the limit is + or - infinity, there is an asymptote.</p>
<p>^^ Both explanations make a lot of sense, thanks so much!</p>