f(x) = 1/(x^2-x-2)
does f(x) have two or three asymptotes
does f(x) have one or two discontinuities
We have f(x) = 1/(x^2-x-2) or
f(x)=1/[(x-2)(x+1)]
f has vert. asympt. at the roots of the denominator.
As x becomes unbounded in either direction you should be able to see that it approaches the x-axis.
So 2 vertical and 1 horizntal asymptote = 3 asymptotes, and 2 roots = 2 discontinuities.
To determine the number of asymptotes for the function f(x) = 1/(x^2-x-2), we need to first find the roots of the denominator. The denominator can be factored as (x-2)(x+1).
Therefore, the roots of the denominator are x = 2 and x = -1. These are the values of x where the function f(x) could potentially have vertical asymptotes.
Since the function approaches the x-axis as x becomes unbounded in either direction, we can conclude that there are two vertical asymptotes for f(x) at x = 2 and x = -1.
Additionally, a rational function can also have a horizontal asymptote if the degree of the numerator is less than or equal to the degree of the denominator. In this case, the degree of the numerator is 0 and the degree of the denominator is 2.
Since the degree of the numerator is less than the degree of the denominator, we can conclude that there is also a horizontal asymptote for f(x).
Therefore, f(x) has two vertical asymptotes at x = 2 and x = -1, and one horizontal asymptote.
To determine the number of discontinuities, we need to look at the roots of the denominator again. In this case, there are two roots: x = 2 and x = -1.
Since these roots are also the values where the function f(x) could potentially have vertical asymptotes, we can conclude that there are two discontinuities in f(x) at x = 2 and x = -1.
In summary, f(x) has three asymptotes (two vertical and one horizontal) and two discontinuities.