How do you graph #f(x)=x/(x^2-1)# using holes, vertical and horizontal asymptotes, x and y intercepts?

Question

Nathan Axel · Accepted Answer

Below

#f(x)=x/(x^2-1)# For horizontal asymptotes, #y=0#. This is because if you replace #x# with different numbers, you will notice that the numerator will always have a smaller value than the denominator. Hence, if you divide a smaller number by a larger number, the answer will be close to #0#. Heads up: If the degree of the numerator is less than the degree of the denominator, then the horizontal asymptote is #y=0# If the degree of the numerator is equal to the degree of the denominator, then the asymptote is the ratio of the leading coefficients. If the degree of the numerator is greater than the degree of the denominator, then you have to use synthetic division to find the oblique asymptote (in most cases, it is an oblique asymptote) For vertical asymptotes, look at the denominator. It cannot equal to #0# as the graph will be undefined. Hence, we let #x^2-1=0# so we can find at what points is the graph undefined. #x^2-1=0# #(x-1)(x+1)=0# #x=+-1# For intercepts, When #y=0#, #x=0# When #x=0#, #y=0# Therefore, after plotting in your intercept and drawing in your vertical and horizontal asymptote, you can hopefully see the outline of your graph. Remember, the asymptotes only influence the end points of your graph and that's it. So the graph can actually cross the asymptotes anywhere else graph{x/(x^2-1) [-10, 10, -5, 5]} Above is the graph

Jameson Allen · Answer

undefined To graph the function f(x) = x/(x^2-1), we can start by identifying the vertical asymptotes, holes, and x-intercepts.

Vertical asymptotes occur when the denominator of the function equals zero. In this case, x^2-1 = 0. Solving this equation, we find x = -1 and x = 1 as the vertical asymptotes.

To find any holes in the graph, we need to simplify the function. Factoring the numerator and denominator, we have f(x) = x/[(x-1)(x+1)]. We can cancel out the common factors of (x-1) in the numerator and denominator, resulting in f(x) = 1/(x+1). Since the canceled factor (x-1) is not equal to zero, there are no holes in the graph.

To determine the x-intercepts, we set f(x) = 0. In this case, 1/(x+1) = 0. However, there are no values of x that make the denominator zero, so there are no x-intercepts.

To find the y-intercept, we substitute x = 0 into the function. f(0) = 0/(0^2-1) = 0/(-1) = 0. Therefore, the y-intercept is at the point (0, 0).

Now, let's summarize the information:

- Vertical asymptotes: x = -1 and x = 1
- Holes: None
- X-intercepts: None
- Y-intercept: (0, 0)

Using this information, we can plot the graph of f(x) = x/(x^2-1).