document.write( "Question 143151: sketch a graph of each rational function. Show all asymptotes and/or holes.
\n" ); document.write( "1. y=(4)/(x^2-1)
\n" ); document.write( "2. y=(3x)/(x^2-3x) \n" ); document.write( "
Algebra.Com's Answer #104190 by nabla(475)\"\" \"About 
You can put this solution on YOUR website!
First, simplify each equation a little:\r
\n" ); document.write( "\n" ); document.write( "\"f%28x%29=4%2F%28x%5E2-1%29=4%2F%28%28x%2B1%29%28x-1%29%29\"
\n" ); document.write( "This tells us right away that x cannot equal +/- 1!!! (No dividing by zero.) Moreover, the lines x=1 and x=-1 will be vertical asymptotes. Now, for horizontal asymptotes, we must consider the degree of the x term in the numerator compared to the denominator. That is, 0 compared with 2. Since 0<2 we will have a horizontal asymptote at y=0.\r
\n" ); document.write( "\n" ); document.write( "Now, having already said that the denominator cannot be zero, we can say that there will be NO x-intercepts (because the denominator is the only part with x-terms). Y-intercept is easily found, however:
\n" ); document.write( "f(0)=4/(-1)=-4. So (0,-4) is the y-intercept. I now have enough information to plot, however you may want to write out some points to test.
\n" ); document.write( "\"graph%28+300%2C+300%2C+-10%2C+10%2C+-10%2C+10%2C+4%2F%28x%5E2-1%29%29\"
\n" ); document.write( "Note that the method this site uses to graph inaccurately draws vertical lines where the vertical asymptotes are. Do not include this on your graph.\r
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\n" ); document.write( "\n" ); document.write( "#2. \"f%28x%29=3x%2F%28x%5E2-3x%29=3x%2F%28x%28x-3%29%29=3%2F%28x-3%29\"
\n" ); document.write( "Much of the same applies here process-wise as in#1. x=3 is the vertical asymptote, y=0 is the horizontal asymptote, (0,-1) is the y-intercept.
\n" ); document.write( "\"graph%28+300%2C+300%2C+-10%2C+10%2C+-10%2C+10%2C+3x%2F%28x%5E2-3x%29%29\" \n" ); document.write( "
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