McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
4. Graphing Rational Functions
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Exercise 1 Page 557

Division by zero is not defined. This means that the denominator cannot be zero.

Practice makes perfect

To graph the given rational function we will find its domain, asymptotes, and intercepts. Then we will find points using a table of values. Finally, we will plot and connect those points.

Domain

Consider the given function. f(x)=x^4-2/x^2-1 This first thing we need to do is factor the denominator of this function.
f(x)=x^4-2/x^2-1
f(x)=x^4-2/(x+1)(x-1)
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

Asymptotes

Asymptotes can be vertical, horizontal, or oblique lines.

Vertical Asymptotes

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.
f(x)=x^4-2/x^2-1
0=x^4-2/x^2-1
Solve for x
0=x^4-2
2=x^4
x^4=2
x=± sqrt(2)
x=± 1.19
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).
f(x)=x^4-2/x^2-1
f(x)=0^4-2/0^2-1
Solve for f(x)
f(x)=0-2/0-1
f(x)=- 2/- 1
f(x)=2
There is a y-intercept at (0,2).

Graph

Let's make a table of values to graph the given function. Make sure to only use values included in the domain of the function.

x x^4-2/x^2-1 f(x)=x^4-2/x^2-1
- 2 ( - 2)^4-2/( - 2)^2-1 ≈ 4.67
- 0.9 ( - 0.9)^4-2/( - 0.9)^2-1 ≈ 7
- 0.5 ( - 0.5)^4-2/( - 0.5)^2-1 ≈ 2.58
0.5 0.5^4-2/0.5^2-1 ≈ 2.58
0.9 0.9^4-2/0.9^2-1 ≈ 7
2 2^4-2/2^2-1 ≈ 4.67

Finally, let's plot and connect the points. Do not forget to draw the asymptotes and to plot the intercepts and holes, if any.