Monthly Archives: November 2014

Week 4: Rational Functions – Day 4

Horizontal Asymptotes (end behavior, symmetry)

If the degree of the numerator of a rational function is less than the degree of the denominator, the function has a horizontal asymptote. Use division by the highest power of x to determine the equation of the horizontal asymptote.

If the degree of the numerator is one more than the denominator, there is a slant asymptote. Use long division to find the equation of the slant asymptote.

If the degree of the numerator is two or more than the denominator, the end behavior of the function does not follow a linear asymptote. Instead, it is similar to that of a polynomial of the same degree as the difference between numerator and denominator.

What is the end behavior of each of these functions? If there is a horizontal or slant asymptote, what is its equation?

1. f(x) = (2x + 1) / (x – 4)

2. f(x) = (x2 + 3) / (x – 1)

3. f(x) = (3x4 – x3 + 5) / (x2 + 7)

4. f(x) = 17 / (x2 + 4x – 5)

5. f(x) = (6x2 + 1) / (3x2 – 1)


Week 4: Rational Functions – Day 5

Sign (intervals of definition, increasing, decreasing)

The graph of a rational function can be broken into disconnected segments by the presence of vertical asymptotes. Remember, at a vertical asymptote the function approaches either positive or negative infinity (sometimes both). To graph a rational function correctly it helps to determine where it is positive or negative.

Here is an organized way to do this.

A. Find the zeros and undefined x-values of the function (places where numerator or denominator are zero).
B. Arrange these values in increasing order.
C. Divide the x-axis into intervals around these values. For example, if the values are -1 and 0, the intervals are:
-1 < x, -1 < x < 0, and 0 < x. If there are N values, there will be N+1 intervals.
D. Test the sign of the function on each interval. You don’t have to compute exact values of the function, just check to see if it is positive or negative.

Here is a set of problems that guide through the process:

1. f(x) = (x2 + x) / (x2 + 6x + 5)
Factor the numerator and denominator of f(x).

2. Where does f(x) have zero value(s)?

3. Where does f(x) have vertical asymptote(s)?

4. Where does f(x) have a hole?

5. Using the information from problems 1-4, test the sign of f(x) over the appropriate intervals.


Week 4: Rational Functions – Day 3

Vertical Asymptotes

Asymptotes are lines that the graph of a function approaches. A vertical line in the x-y plane is a line that goes to plus and minus infinity. So vertical asymptotes indicate that the values of a function approach infinity (or minus infinity) as x approaches a finite value.

Example notes: The function f(x) = 1/(x – 1) has a vertical asymptote at x = 1. But the function g(x) = (x2 – 1)/(x – 1) does not have a vertical asymptote at x = 1. As long as x is not equal to 1, the function g(x) is equivalent to x + 1, which has no vertical asymptote. (Factor and simplify g(x) to make sure that this is true.)

Which functions below have vertical asymptotes? If there are any vertical asymptotes, what are their equations?

1. f(x) = 3 / (x – 2)

2. f(x) = (x3 + x)/x

3. f(x) = x / (x2 + 5)

4. f(x) = (2x3 – 32) / (x2 – 6x + 5)

5. f(x) = 1 – 1/x + 2/(x – 1)


Week 4: Rational Functions – Day 1

Definition, Domain

The definition of a rational function is: it can be written as a ratio of two polynomials.

1. Suppose f(x) = 1/x and g(x) = 1/(x + 1). Are f(x) and g(x) both rational functions?

2. Is h(x) = f(x) + g(x) from problem 1 above a rational function? Explain.

3. What is the domain of f(x) = 1/x?

4. What is the domain of the rational function f(x) = (x – 1) / (x2 + 1) ?

5. What is the domain of h(x) = 1/x + 1/(x + 1)?