McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
6. Solving Rational Equations and Inequalities
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Exercise 48 Page 578

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^2-5x-14/x-7 Recall that division by zero is not defined. Therefore, the rational function is undefined where x-7=0. x-7=0 ⇔ x=7 This means that x=7 is not included in the domain. Domain All real numbers except x=7

Asymptotes

Asymptotes can be vertical, horizontal, or oblique lines.

Vertical Asymptotes

Let's consider the given function. f(x)=x^2-5x-14/x-7 Notice that the numerator can be factored. Let's factor this expression and cancel out any common factors between the numerator and the denominator.

x^2-5x-14
x^2-7x+2x-14
x(x-7)+2x-14
x(x-7)+2(x-7)
(x-7)(x+2)

We found that the denominator can be written as (x-7)(x+2). Let's use this to cancel common factors.

f(x)=(x-7)(x+2)/x-7
f(x)=(x-7)(x+2)/x-7
f(x)=x+2

We canceled out the factor x-7. Therefore, there is a hole in the graph at x=7. Since there are no other factors in the denominator, there are no vertical asymptotes.

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 table, m and n are the degree of the numerator and denominator, respectively.

f(x)=ax^m/bx^n Asymptote Asymptote Type
m y=0 Horizontal
m>n none -
m=n y=a/b Horizontal
m-n=1 y= the quotient of the polynomials with no remainder Oblique

Now, consider the function one more time. Let's look at the degrees of the numerator and denominator for our function. f(x)=x^2-5x-14/x^1-7 We can see that the degree of the numerator is higher than the degree of the denominator. Therefore, there are no 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-7)(x+2)/x-7
0=(x-7)(x+2)/x-7
â–¼
Solve for x
0=(x-7)(x+2)
(x-7)(x+2)=0

By the Zero Product Property, the equation equals 0 when either of the factors is equal to 0. c|c x-7=0 & x+2=0 ⇕ & ⇕ x=7 & x=-2 We found that if f(x)=0, the value of x is 7 or -2. However, x=7 is not included in the domain of the function. Therefore, the only x-intercept occurs at the point (-2,0).

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 simplify.

f(x)=(x-7)(x+2)/x-7
f(x)=( 0-7)( 0+2)/0-7
â–¼
Evaluate right-hand side
f(x)=-7(2)/-7
f(x)=-7(2)/-7
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^2-5x-14/x-7 f(x)=x^2-5x-14/x-7
1 1^2-5( 1)-14/1-7 3
5 5^2-5( 5)-14/5-7 7

Finally, let's plot and connect the points. Do not forget to plot the intercepts and hole.