SOLUTION: sketch a graph of each rational function. Show all asymptotes and/or holes. 1. y=(4)/(x^2-1) 2. y=(3x)/(x^2-3x)

Algebra ->  Rational-functions -> SOLUTION: sketch a graph of each rational function. Show all asymptotes and/or holes. 1. y=(4)/(x^2-1) 2. y=(3x)/(x^2-3x)      Log On


   



Question 143151: sketch a graph of each rational function. Show all asymptotes and/or holes.
1. y=(4)/(x^2-1)
2. y=(3x)/(x^2-3x)

Answer by nabla(475) About Me  (Show Source):
You can put this solution on YOUR website!
First, simplify each equation a little:
f%28x%29=4%2F%28x%5E2-1%29=4%2F%28%28x%2B1%29%28x-1%29%29
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.
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:
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.
graph%28+300%2C+300%2C+-10%2C+10%2C+-10%2C+10%2C+4%2F%28x%5E2-1%29%29
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.

#2. f%28x%29=3x%2F%28x%5E2-3x%29=3x%2F%28x%28x-3%29%29=3%2F%28x-3%29
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.
graph%28+300%2C+300%2C+-10%2C+10%2C+-10%2C+10%2C+3x%2F%28x%5E2-3x%29%29