2-08 Graphs of Rational Functions

Find the Intercepts of a Rational Function

Find the intercepts of \(f(x) = \frac{(x + 1)(x - 3)}{3x-2}\).

Solution

To find the y-intercept, substitute 0 for x and simplify.

$$ f(0) = \frac{(0 + 1)(0 - 3)}{3(0)-2} $$

$$ f(0) = \frac{3}{2} $$

The y-intercept is \(\left(0, \frac{3}{2}\right)\).

To find the x-intercepts, substitute 0 for y, or f(x), and solve for x. Because a fraction can only equal zero if the numerator is zero, set the numerator equal to zero and solve for x.

(x + 1)(x − 3) = 0

x + 1 = 0 or x − 3 = 0

x = −1 or x = 3

The x-intercepts are (−1, 0) and (3, 0).

Graph a Rational Function

Sketch a graph of \(k(x) = \frac{x^2 - 9}{x^3 - 3x + 2}\).

Solution

1. Start by finding the y-intercept. Substitute x = 0 and simplify.

   $$ k(0) = \frac{0^2 - 9}{0^3 - 3(0) + 2} = -\frac{9}{2} $$

   The y-intercept is \(\left(0, -\frac{9}{2}\right)\).

2. Next, factor the numerator and denominator. The numerator, x² − 9, is a difference of squares and becomes (x − 3)(x + 3). The denominator, x³ − 3x + 2, is cubic and cannot be factored easily. The Rational Zero Theorem will need to be used.

   The p's are factors of the constant term, 2.

   p = ±1, ±2

   The q's are factors of the leading coefficient, 1. So, the \(\frac{p}{q}\) are the same as p.

   $$ \frac{p}{q} = ±1, ±2 $$

   Pick one such as −2 and use synthetic division to check to see if it is a factor.

   $$ \begin{array}{rrrrr} \underline{-2}| & 1 & 0 & -3 & 2 \\ & & -2 & 4 & -2 \\ \hline & 1 & -2 & 1 & |\underline{\phantom{0}0} \end{array} $$

   Since the remainder is zero, (x + 2) is a factor. The depressed polynomial is x² − 2x + 1 which factors into (x − 1)². Thus, the factored form of the function is

   $$ k(x) = \frac{(x - 3)(x + 3)}{(x - 1)^2(x + 2)} $$

3. There are no factors common to the numerator and denominator to cancel, so there are also no removable discontinuities.

4. The x-intercepts are found by setting the numerator equal to zero and solving.

   (x − 3)(x + 3) = 0

   x = 3 or x = −3

   The x-intercepts are (3, 0) and (−3, 0).

5. Then, find the vertical asymptotes by setting the denominator equal to zero and solving.

   (x − 1)²(x + 2) = 0

   x = 1 (multiplicity 2) or x = −2

   The vertical asymptotes are x = 1 with a multiplicity 2 and x = −2.

6. There no removable discontinuities because no factors canceled in step 3.

7. Find the horizontal or slant asymptote by comparing the degrees of the numerator and denominator. The degree of the numerator, N, is 2. The degree of the denominator, D, is 3. N < D, so the horizontal asymptote is y = 0.

8. Make a table of values so that accurate points can be plotted. This will make the sketch more accurate. Notice a few extra points were found between the two vertical asymptotes so that there are several points to make an accurate graph.

   x	−5	−4	−3	−2.5	−2	−1.5	−1	−0.5	0	0.5	1	2	3	4	5
   y	−0.15	−0.14	0	0.45	UND	−2.16	−2	−2.59	−4.5	−14	UND	−1.25	0	0.13	0.14

9. Start the graph by plotting the intercepts and points from the table. Then add the asymptotes drawn as dashed lines.

   

   Then draw the graph by starting near an asymptote and drawing a curve through points ending near another asymptote. Do not cross a vertical asymptote, but you may cross the horizontal asymptote. Repeat for each section of the graph. Finally, make sure the graph matches the predicted behavior. Both sides of the vertical asymptote x = 1 go the same direction because it has even multiplicity. Near the vertical asymptote x = −2, one side goes up and the other side goes down due to odd multiplicity. The graph crosses the x-axis at both x-intercepts because they each have odd multiplicity (1).

   

Write a Rational Function from Intercepts and Asymptotes

Write an equation for the rational function shown below in figure 6.

Solution

1. The x-intercepts are −4 and 3. Write those as factors.

   (x + 4)(x − 3)

   Notice that the graph does not cross the x-axis at 3, so make the factor (x − 3) squared.

   (x + 4)(x − 3)²

2. The vertical asymptotes are x = −2 and x = 1. Write those as factors.

   (x + 2)(x − 1)

   Notice that the graph goes up on both sides of the asymptotes x = 1, so make that factor squared.

   (x + 2)(x − 1)²

3. Write the function in the form \(f(x) = a\frac{\text{factors from x-intercepts}}{\text{factors from vertical asymptotes}}\).

   $$ f(x) = a\frac{(x+4)(x-3)^2}{(x+2)(x-1)^2} $$

4. Substitute any clear point to find a. (−1, 3) works in this case.

   $$ 3 = a\frac{(-1+4)(-1-3)^2}{(-1+2)(-1-1)^2} $$

   3 = a(12)

   $$ a = \frac{1}{4} $$

5. Write the final function by substituting for a.

   $$ f(x) = \frac{(x + 4)(x - 3)^2}{4(x + 2)(x - 1)^2} $$

Solve an Applied Problem Involving a Rational Function

A large mixing tank is used to prepare water for a salt-water aquarium. It currently contains 100 gallons of water into which 5 pounds of salt have been mixed. A tap will open pouring 10 gallons per minute of water into the tank at the same time salt is poured into the tank at a rate of 1 pound per minute. Find the concentration (pounds per gallon) of salt in the tank after 12 minutes. Is that a greater concentration than at the beginning?

Solution

Let t be the number of minutes since the tap opened. Since the water starts at 100 gallons and increases at 10 gallons per minute, its volume is given by

W(t) = 100 + 10t = 10t + 100.

The salt starts at 5 pounds and increases at 1 pounds per minute, so its quantity is given by

S(t) = 5 + 1t = t + 5.

The concentration in pounds per gallon can be found by taking the pounds of salt and dividing by the gallons of water.

$$ C(t) = \frac{t + 5}{10t+100} $$

The problem asks for the concentration at t = 12 minutes. Plug in a 12 and evaluate the function.

$$ C(12) = \frac{12 + 5}{10(12) + 100} = 0.077 \text{ lbs/gal} $$

The last part asks if this concentration is higher than the beginning concentration. Answer that by finding the concentration when t = 0.

$$ C(0) = \frac{0 + 5}{10(0) + 100} = \frac{1}{20} = 0.05 \text{ lbs/gal} $$

C(12) is higher than C(0).

Analysis

Find the horizontal asymptote. Since the degree of the numerator and denominator are the same, the horizontal asymptote is the ratio of the leading coefficients.

$$ y = \frac{1}{10} = 0.1 $$

This means the concentration of salt to water, will approach 0.1 lbs/gal in the long term.