Recall that division by zero is not defined. Therefore, the rational function is undefined where x+1=0 and x-1=0.
c|c
x+1=0 & x-1=0
⇕ & ⇕
x=- 1 & x=1
This means that x=- 1 and x=1 are not included in the domain.
Domain
All real numbers except x=- 1 and x=1
Once again, let's consider the given function.
f(x)=x^4-2/x^2-1
Note that we cannot cancel out common factors. Therefore, there are no holes. Also, if the real number a is not included in the domain, there is a vertical asymptote at x=a. In this case we have vertical asymptotes at x=- 1 and x=1.
Horizontal and Oblique Asymptotes
To find the horizontal and oblique asymptotes we can use a set of rules. To properly use these rules, in the following m and n must be the highest degree of the numerator and the denominator.
y=ax^m/bx^n
Asymptote
Asymptote Type
m
y=0
Horizontal
m>n
None
None
m=n
y=a/b
Horizontal
m-n=1
y= the quotient of the polynomials with no remainder
Oblique
Let's look at the degrees of the numerator and denominator for our function.
f(x)=x^4-2/x^2-1
We can see that the degree of the numerator is higher than the degree of the denominator. Therefore, there are no horizontal or oblique asymptotes.
Intercepts
The intercepts of the function are the points at which the graph intersects the axes.
x-intercepts
The x-intercepts are the points where the graph intersects the x-axis. At these points the value of the y-coordinate is zero. Let's substitute 0 for f(x) in the given function and solve for x.
There are x-intercepts at (1.19,0) and (- 1.19,0).
y-intercept
The y-intercept is the point where the graph intersects the y-axis. At this point, the value of the x-coordinate is zero. Let's substitute 0 for x in the given function and solve for f(x).
