Precalculus by Richard Wright
Are you not my student and
has this helped you?
This book is available
to download as an epub.
Precalculus by Richard Wright
Are you not my student and
has this helped you?
This book is available
to download as an epub.
So the last will be first, and the first will be last. Matthew 20:16 NIV
Summary: In this section, you will:
SDA NAD Content Standards (2018): PC.5.3
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. What is the concentration, in pounds per gallon, of salt in the tank after a given amount of time? This problem can be solved using a rational function and will be analyzed later in this lesson.
Start by factoring and canceling factors common to the numerator and denominator.
y-intercepts occur when x = 0.
x-intercepts occur when y = 0. A fraction only equals zero when the numerator equals zero.
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).
Find the intercepts of \(t(x) = \frac{3x - 4}{x+5}\).
Answer
y-intercept: \(\left(0, -\frac{4}{5}\right)\); x-intercept: \(\left(\frac{4}{3}, 0\right)\)
The numerator of a rational function gives the x-intercepts of the graph. When the x-intercept has an even multiplicity because the corresponding factor is present an even number of times in the numerator, the graph touches the x-axis without crossing it at that point. When the x-intercept has an odd multiplicity, the graph crosses the x-intercept at that point. For example, then graph of \(f(x) = \frac{(x-2)^2(x+3)}{(x-3)^2(x+1)}\) is shown in figure 3 below. The x-intercepts are (2, 0) with multiplicity 2 (from the factor (x − 2)2) and (−3, 0) (from the factor (x + 3)). Notice that the graph does not cross the x-intercept at (2, 0) which has even multiplicity, but does cross it at (−3, 0) which has odd multiplicity.
The denominator gives the vertical asymptotes of the graph. When the factor giving a vertical asymptote has even multiplicity because it is present an even number of times, both sides of the graph by the asymptote go the same direction. When the factor giving a vertical asymptote has an odd multiplicity, one side of the graph goes up and the other side goes down near the asymptote. This can be seen in figure 3 above where the vertical asymptote x = 3 has an even multiplicity ((x − 3) is present twice) and the vertical asymptote x = −2 has odd multiplicity ((x + 2) is present once).
Sketch a graph of \(k(x) = \frac{x^2 - 9}{x^3 - 3x + 2}\).
Solution
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 \(\left(0, -\frac{9}{2}\right)\).
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
$$ k(x) = \frac{(x - 3)(x + 3)}{(x - 1)^2(x + 2)} $$
There are no factors common to the numerator and denominator to cancel, so there are also no removable discontinuities.
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).
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.
There are no removable discontinuities because no factors canceled in step 3.
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.
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 |
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).
Graph the function \(f(x) = \frac{x - 2}{x^2 - 3x + 2}\).
Answer
y-intercept at (0, −1); no x-intercepts; V.A. at x = 1; Removable discontinuity at (2, 1); H.A. at y = 0
Sometimes an equation is needed to match a graph. The process is somewhat like graphing in reverse. The x-intercepts give the factors of the numerator. The vertical asymptotes give the factors of the denominator. The behavior of the graph near the x-intercepts and vertical asymptotes gives the multiplicity of the factors. A stretch factor, a, can be used to scale the graph vertically for the y-intercept.
Write an equation for the rational function shown below in figure 6.
Solution
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
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
Write the function in the form \(f(x) = a\frac{\text{factors from x-intercepts}}{\text{factors from vertical asymptotes}}\).
$$ f(x) = a\frac{(x+4)(x-3)^2}{(x+2)(x-1)^2} $$
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)
$$ a = \frac{1}{4} $$
Write the final function by substituting for a.
$$ f(x) = \frac{(x + 4)(x - 3)^2}{4(x + 2)(x - 1)^2} $$
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.
There are 150 mosquitoes and 25 ants at picnic lunch at noon. After 12 p.m., 10 mosquitoes arrive at the picnic every five minutes while only 2 ants come. Find the ratio of mosquitoes to ants at 1 p.m.
Answer
\(\frac{270}{49}\)
Start by factoring and canceling factors common to the numerator and denominator.
y-intercepts occur when x = 0.
x-intercepts occur when y = 0. A fraction only equals zero when the numerator equals zero.
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