Intermediate Algebra: Functions and Graphs

The common factor for numerator and denominator is $2x^2y\text{.}$ We factor $2x^2y$ from the numerator and denominator, then divide by the common factor.

\begin{equation*} \dfrac{8x^3y}{6x^2y^3} = \dfrac{4x \cdot \cancel{2x^2y}}{3y^2 \cdot \cancel{2x^2y}} = \dfrac{4x}{3y^2} \end{equation*}

Checkpoint 8.2.3. Practice 1.

Reduce $\displaystyle\frac{-15ab^4}{25a^2b}\text{:}$

$-3b^{3}$

$5a$

$\dfrac{-3b^3}{5a}$

Caution 8.2.4.

When we cancel common factors, we are dividing. Because division is the inverse or opposite operation for multiplication, we can cancel common factors, but we cannot cancel common terms.

Example 8.2.5.

Use your calculator to decide which calculation is correct.

$\displaystyle \dfrac{12}{8} = \dfrac{4 \cdot 3}{4 \cdot 2} \rightarrow \dfrac{3}{2}$
$\displaystyle \dfrac{7}{6} = \dfrac{4 + 3}{4 + 2} \rightarrow \dfrac{3}{2}$

We can cancel the 4's in part (a), because they are factors of numerator and denominator.

We cannot cancel the 4's in part (b), because they are terms.

You can verify that $\dfrac{12}{8} = \dfrac{3}{2}\text{,}$ but $\dfrac{7}{6} \ne \dfrac{3}{2}\text{.}$

Checkpoint 8.2.6. QuickCheck 1.

Fill in the blanks.

An algebraic fraction is undefined when the
- denominator
- index
- numerator
- radicand
is
- even
- negatove
- positive
- zero
.
“Canceling” common factors uses the operation of
- addition
- division
- subtraction
- squaring
.
Terms are expressions that are
- added
- exponentiated
- multiplied
- reduced
or
- divided
- expended
- simplified
- subtracted
.
When we reduce a fraction, the new fraction is
- equivalent
- inverse
- superior
- unrelated
to the old one.

$\text{denominator}$

$\text{zero}$

$\text{division}$

$\text{added}$

$\text{subtracted}$

$\text{equivalent}$

denominator, zero
division
added, subtracted
equivalent

Checkpoint 8.2.7. Practice 2.

Which calculation is correct? Choose a value for $x$ and use your calculator to verify your answer.

(a) $\quad \displaystyle\frac{5x}{8x}=\frac{5}{8} $
(b) $\quad \displaystyle\frac{x+5}{x+8}=\frac{5}{8} $

$\text{Choice 1}$

(a) $~\dfrac{5x}{8x}=\dfrac{5}{8}$

If the numerator or denominator of the fraction contains more than one term, it is especially important to factor before reducing, so that numerator and denominator are written as products of factors, instead of sums of terms.

Example 8.2.8.

Reduce each fraction.

$\displaystyle \dfrac{4x+2}{4}$
$\displaystyle \dfrac{9x^2+3}{6x+3}$

First we factor the numerator and denominator. Then we divide numerator and denominator by any common factors.

$\displaystyle \dfrac{4x+2}{4} = \dfrac{\cancel{2}(2x+1)}{\cancel{2}(2)} = \dfrac{2x+1}{2}$
$\displaystyle \dfrac{9x^2+3}{6x+3} = \dfrac{\cancel{3}(3x^2+1)}{\cancel{3}(2x+1)} = \dfrac{3x^2+1}{2x+1}$

Caution 8.2.9.

We cannot cancel the 4's in part (a) of the Example above, because 4 is not a factor of the entire numerator. Thus,

\begin{equation*} \dfrac{4x+2}{4} \ne x+2 \end{equation*}

In the Example part (b), we cannot cancel the 3's, because they are terms and not factors. Thus,

\begin{equation*} \dfrac{9x^2+3}{6x+3} \ne \dfrac{9x^2}{6x} \end{equation*}

Checkpoint 8.2.10. Practice 3.

Explain why each calculation is incorrect.

$\dfrac{x-6}{x-9} \rightarrow \dfrac{x-2}{x-3}$
- A) 3 is not a factor of numerator or denominator.
- B) More cancelling is required.
$\dfrac{2x+5}{2} \rightarrow x+5$
- A) 2 is not a factor of the numerator.
- B) 2 should be subtracted from 5.

$\text{A) ... denominator.}$

$\text{A) 2 ... the numerator.}$

3 is not a factor of numerator or denominator
2 is not a factor of the numerator

We summarize the procedure for reducing algebraic fractions as follows.

To reduce an algebraic fraction.

Factor the numerator and the denominator.
Divide the numerator and denominator by any common factors.

Example 8.2.11.

Reduce each fraction.

$\displaystyle \dfrac{3x+12}{6x+24}$
$\displaystyle \dfrac{27x^3-1}{9x^2-1}$

We first factor the numerator and denominator completely.

\begin{equation*} \dfrac{3x+12}{6x+24} = \dfrac{3(x+4)}{2 \cdot 3(x+4)} \end{equation*}

Then we divide numerator and denominator by the common factors 3 and $(x+4)\text{.}$ We must cancel the entire expression $(x+4)$ from numerator and denominator (we cannot cancel the $x$'s or the 4's separately!).

\begin{equation*} \dfrac{3(x+4)}{2 \cdot 3(x+4)} = \dfrac{\cancel{3}\cancel{(x+4)}}{2 \cdot \cancel{3}\cancel{(x+4)}}= \dfrac{1}{2} \end{equation*}

All the factors are canceled from the numerator, so we replace them by 1, because any expression divided by itself is 1.
The numerator of the fraction is a difference of two cubes, and the denominator is a difference of two squares. We factor each to obtain

\begin{equation*} \dfrac{27x^3-1}{9x^2-1} = \dfrac{(3x-1)(9x^2+3x+1)}{(3x-1)(3x+1)} \end{equation*}

We cancel the factor $(3x-1)$ from top and bottom, to get

\begin{equation*} \dfrac{\cancel{(3x-1)}(9x^2+3x+1)}{\cancel{(3x-1)}(3x+1)} = \dfrac{9x^2+3x+1}{3x+1} \end{equation*}

Because we cannot factor any further, we cannot reduce the fraction any further.

Checkpoint 8.2.12. QuickCheck 2.

Fill in the blanks.

If we want to cancel an expression from a fraction, we must be able to
- exponentiate
- factor
- invalidate
- subtract
it from numerator and denominator.
When we factor an expression, we write it as a
- difference
- product
- quotient
- sum
.
If a factor appears in both numerator and denominator, it is called a
- common
- fallacious
- missing
- zero
factor.
If all the factors in the numerator or denominator cancel out, we replace it by .

$\text{factor}$

$\text{product}$

$\text{common}$

$1$

factor
product
common
1

Checkpoint 8.2.13. Practice 4.

Reduce each fraction.

$\dfrac{x^2-x-6}{x^2-9}$
- $\displaystyle \quad \displaystyle\frac{x+6}{9} $
- $\displaystyle \quad \displaystyle\frac{x+2}{x+3} $
- $\displaystyle \quad \displaystyle\frac{x-2}{x-3} $
$\dfrac{16t^2-4}{8t+4}$
- $\displaystyle \quad 2t-1 $
- $\displaystyle \quad \displaystyle\frac{t-2}{2} $
- $\displaystyle \quad3t^2 $

$\text{Choice 2}$

$\text{Choice 1}$

$\displaystyle \dfrac{x+2}{x+3}$
$\displaystyle 2t-1$

Keep in mind that the reduced form is equivalent to the original form of the fraction. If we evaluate the original form and the reduced form at the same value of the variable, the results are equal.

Example 8.2.14.

Verify that $\dfrac{x^2-x-6}{x^2-9}$ is equal to $\dfrac{x+2}{x+3}$ for $x=2\text{.}$

We evaluate each fraction at $x=\alert{2}\text{.}$

\begin{equation*} \dfrac{x^2-x-6}{x^2-9} = \dfrac{(\alert{2})^2-\alert{2}-6}{(\alert{2})^2-9} = \dfrac{-4}{-5} = \dfrac{4}{5} \end{equation*}

and

\begin{equation*} \dfrac{x+2}{x+3} = \dfrac{\alert{2}+2}{\alert{2}+3} = \dfrac{4}{5} \end{equation*}

Checkpoint 8.2.15. Practice 5.

If you evaluate $\dfrac{3x+12}{6x+24}$ for $x=2$ and for $x=-5\text{,}$ what answer do you expect to get? (Recall the earlier Example with the same algebraic fraction in part a.)

$\frac{1}{2}$

$\dfrac{1}{2}$

Subsection 8.2.3 Opposite of a Binomial

Any number (except zero) divided by itself is 1, and any number divided by its opposite is $-1\text{.}$ For example,

\begin{equation*} \dfrac{5}{5} = 1 ~~~~~\text{and}~~~~~\dfrac{-5}{5} = -1 \end{equation*}

The same is true for binomials and other algebraic expressions. The opposite of an expression can be found by multiplying it by $-1\text{.}$ Thus, the opposite of $a-b$ is

\begin{equation*} -(a-b) = -a + b= b-a \end{equation*}

and so

\begin{equation*} \dfrac{b-a}{a-b} = \dfrac{-\cancel{(a-b)}}{\cancel{(a-b)}} = -1 \end{equation*}

Here are some examples of opposites.

\begin{align*} 2a-3b~~~~\amp \text{and}~~~~3b-2a\\ 2a+3b~~~~\amp \text{and}~~~~-2a-3b\\ -x-1~~~~\amp \text{and}~~~~x+1\\ -x+1~~~~\amp \text{and}~~~~x-1 \end{align*}

We can cancel opposites when we reduce fractions.

Example 8.2.16.

Reduce $~\dfrac{2x-4y}{6y-3x}$

We first factor the numerator and denominator.

\begin{equation*} \dfrac{2x-4y}{6y-3x} = \dfrac{2(x-2y)}{3(2y-x)} \end{equation*}

We see that $x-2y$ is the opposite of $2y-x\text{,}$ that is, $x-2y = -(2y-x)\text{.}$ Thus,

\begin{equation*} \dfrac{2(x-2y)}{3(2y-x)} = \dfrac{-2\cancel{(2y-x)}}{3\cancel{(2y-x)}} = \dfrac{-2}{3} \end{equation*}

Checkpoint 8.2.17. Practice 6.

Reduce if possible.

$\dfrac{-x+1}{1-x}$
$\dfrac{1+x}{1-x}$
$\dfrac{2a-3b}{3b-2a}$
$\dfrac{2a-3b}{2b-3a}$

$1$

$\text{cannot be reduced}\hbox{ or }\text{cbr}$

Answer 3.

$\text{-1}$

Answer 4.

$\text{cannot be reduced}\hbox{ or }\text{cbr}$

1
cannot be reduced
$\displaystyle -1$
cannot be reduced

Subsection 8.2.4 Rational Functions

A rational function is a function defined by an algebraic fraction. That is, it has the form

\begin{equation*} f(x) = \dfrac{P(x)}{Q(x)} \end{equation*}

where $P(x)$ and $Q(x)$ are polynomials. A rational function is undefined at any $x$-values where $Q(x) = 0\text{.}$

Example 8.2.18.

Francine is planning a 60-mile training flight through the desert on her cycle-plane, a pedal-driven aircraft. If there is no wind, she can pedal at an average speed of 15 miles per hour, so she can complete the flight in 4 hours.

If there is a headwind of $x$ miles per hour, it will take Francine longer to fly 60 miles. Express the time it will take to complete the training flight as a function of $x\text{.}$
Make a table of values for the function.
Graph the function and explain what it tells you about the time Francine should allot for the flight.

If there is a headwind of $x$ miles per hour, Francine's ground speed will be $15 - x$ miles per hour. Using the fact that time $\text{time} = \dfrac{\text{distance}}{\text{rate}}\text{,}$ we find that the time needed for the flight will be

\begin{equation*} t = f(x) = \frac{60}{15 - x} \end{equation*}
We evaluate the function for several values of $x\text{,}$ as shown in the table below.

$x$ $0$ $3$ $5$ $7$ $9$ $10$

$t$ $4$ $5$ $6$ $7.5$ $10$ $12$

For example, if the headwind is $\alert{5}$ miles per hour, then

\begin{equation*} t=\frac{60}{15-\alert{5}}=\frac{60}{10}=6 \end{equation*}

Francine's effective speed is only 10 miles per hour, and it will take her 6 hours to fly the 60 miles. The table shows that as the speed of the headwind increases, the time required for the flight increases also.
The graph of the function is shown below. You can use your calculator with the window

\begin{align*} \text{Xmin} \amp = -8.5 \amp\amp \text{Xmax} = 15\\ \text{Ymin} \amp = 0 \amp\amp \text{Ymax} = 30 \end{align*}

to verify the graph. In particular, the point $(0,4)$ lies on the graph. This point tells us that if there is no wind, Francine can fly 60 miles in 4 hours, as we calculated earlier.

The graph is increasing, as indicated by the table of values. In fact, as the speed of the wind gets close to 15 miles per hour, Francine's flying time becomes extremely large. In theory, if the wind speed were exactly 15 miles per hour, Francine would never complete her flight. On the graph, the time becomes infinite at $x = 15\text{.}$

What about negative values for $x\text{?}$ If we interpret a negative headwind as a tailwind, Francine's flying time should decrease for negative $x$-values. For example, if $x = -5\text{,}$ there is a tailwind of 5 miles per hour, so Francine's effective speed is 20 miles per hour, and she can complete the flight in 3 hours. As the tailwind gets stronger (that is, as we move farther to the left in the $x$-direction), Francine's flying time continues to decrease, and the graph approaches the $x$-axis.

The vertical dashed line at $x=15$ on the graph of $t=\dfrac{60}{15-x}$ is a vertical asymptote for the graph. We first encountered asymptotes in Section 5.3 when we studied the graph of $y=\dfrac{1}{x}\text{.}$ Locating the vertical asymptotes of a rational function is an important part of determining the shape of the graph.

Checkpoint 8.2.19. Practice 7.

EarthCare decides to sell T-shirts to raise money. The company makes an initial investment of $100 to pay for the design of the T-shirt and to set up the printing process. After that, the T-shirts cost $5 each for labor and materials.

Express the average cost, $C\text{,}$ per T-shirt as a function of the number of T-shirts EarthCare produces.

$C=$
Make a table of values for the function.

$x$ $1$ $2$ $4$ $5$ $10$ $20$

$C$
Graph the function and explain what it tells you about the cost of the T-shirts.

$\frac{100+5x}{x}$

$105$

$55$

$30$

$25$

$15$

$10$

$\displaystyle C=g(x) = \dfrac{100+5x}{x}$
$x$ $1$ $2$ $4$ $5$ $10$ $20$

$C$ $105$ $55$ $30$ $25$ $15$ $10$
A graph is below.

As the number of T-shirts increases, the average cost per shirt decreases, but eventually levels off and approaches $5 per T-shirt.

Graph for part (c):

$graph of average cost of production$

In Practice 7, the horizontal line $C=5$ is a horizontal asymptote for the graph of the function. As $x$ increases, the graph approaches the line $C=5$ but will never actually meet it. The average price per T-shirt will always be slightly more than $5. Horizontal asymptotes are also important in sketching the graphs of rational functions.

Checkpoint 8.2.20. QuickCheck 3.

True or False.

A rational function is a quotient of polynomials.
- True
- False
A vertical asymptote occurs where a rational function is undefined.
- True
- False
A horizontal asymptote is a line that the graph of a rational function approaches as $x$ increases.
- True
- False
The algebraic fraction $\dfrac{60}{x-15}$ is never equal to zero.
- True
- False

$\text{True}$

$\text{True}$

$\text{True}$

$\text{True}$

True
True
True
True

Exercises 8.2.5 Problem Set 8.2

Warm Up

Exercise Group.

For Problems 1–2,

Evaluate the fraction for the given values of the variable.
For what values of the variable is the fraction undefined?
Find a value of $x$ for which the fraction is equal to zero.

1.

$\dfrac{x+1}{x-3},~~~~~x=\dfrac{1}{2},~{-4}$

$\dfrac{-3}{5} \text{,}$ $\dfrac{3}{7} $
$\displaystyle 3 $
$\displaystyle -1$

2.

$\dfrac{2a-a^2}{a^2+1},~~~~~a=3,~{-1}$

3.

$\dfrac{2x}{x^2-1},~~~~~x=-2,~20$

$\dfrac{-4}{3} \text{,}$ $\dfrac{40}{399} $
$\displaystyle 1,~-1 $
$\displaystyle 0$

4.

$\dfrac{x-2}{x^2-2x+1},~~~~~x=-1,~\dfrac{1}{2}$

Exercise Group.

For Problems 5–8, reduce the fraction.

5.

$\displaystyle \dfrac{15}{3x}$
$\displaystyle \dfrac{24b}{14} $

$\displaystyle \dfrac{5}{x}$
$\displaystyle \dfrac{12b}{7} $

6.

$\displaystyle \dfrac{-5z}{6z}$
$\displaystyle \dfrac{5u}{120uv}$

7.

$\displaystyle \dfrac{16ab}{-10ab}$
$\displaystyle \dfrac{3a^2}{27a}$

$\displaystyle \dfrac{-8}{5} $
$\displaystyle \dfrac{a}{9}$

8.

$\displaystyle \dfrac{-9y^3z}{42yz}$
$\displaystyle \dfrac{8u^3v^2}{12v^2w}$

Skills Practice

9.

Which of the following is a correct application of the fundamental principle of fractions? Explain why or why not in each case.

$\displaystyle \dfrac{3x+5}{3x} \rightarrow 5$
$\displaystyle \dfrac{4x+3}{2y} \rightarrow \dfrac{2x+3}{y}$
$\displaystyle \dfrac{2x^2+x-12}{x-12} \rightarrow 2x^2$
$\displaystyle \dfrac{8x^2-9}{2x-3} \rightarrow 4x+3$

None are correct

Exercise Group.

For Problems 10 and 11, decide whether the fraction is equivalent to 1, to $-1\text{,}$ or cannot be reduced.

10.

$\displaystyle \dfrac{x+4}{x-4}$
$\displaystyle \dfrac{t+5w}{5w-t}$
$\displaystyle \dfrac{x+3z}{z+3x}$
$\displaystyle \dfrac{-(m-1)}{1-m}$

11.

$\displaystyle \dfrac{2a+b}{2a-b}$
$\displaystyle \dfrac{-(a-b)}{b-a}$
$\displaystyle \dfrac{2a^2-1}{2a^2}$
$\displaystyle \dfrac{-a^2+3)}{a^2+3}$

cannot be reduced
$\displaystyle 1$
cannot be reduced
cannot be reduced

Exercise Group.

For Problems 12-15, reduce if possible, and select the correct reduced form.

12.

$\dfrac{2x+3}{2x}$

$\displaystyle \dfrac{x+3}{x}$
$\displaystyle 3$
neither of these

13.

$\dfrac{3a+a^2}{3a}$

$\displaystyle a^2$
$\displaystyle \dfrac{3+a}{3}$
neither of these

(b)

14.

$\dfrac{y+3}{2y^2+6y}$

$\displaystyle \dfrac{1}{2y}$
$\displaystyle y$
neither of these

15.

$\dfrac{a^3}{a^4-a^3}$

$\displaystyle \dfrac{1}{a-1}$
$\displaystyle \dfrac{1}{a^4}$
neither of these

(a)

16.

Reduce each fraction. Which of the fractions are equivalent to $3b\text{?}$

$\displaystyle \dfrac{9b^2-3b}{3b}$
$\displaystyle \dfrac{b+2}{3b^2+6b}$
$\displaystyle \dfrac{3b-9}{9}$
$\displaystyle \dfrac{9b^2-3b)}{3b-1}$

Exercise Group.

For Problems 17-28, reduce if possible.

17.

$\dfrac{2a^2}{2a^2-6a}$

$\dfrac{a}{a-3}$

18.

$\dfrac{b-2}{4-2b}$

19.

$\dfrac{a-b}{a^2-b^2}$

$\dfrac{1}{a+b}$

20.

$\dfrac{(3x+2y)^2}{4y^2-9x^2}$

21.

$\dfrac{y^2-9x^2}{(3x-y)^2}$

$\dfrac{y+3x}{y-3x}$

22.

$\dfrac{6-6t^2}{(t-1)^2}$

23.

$\dfrac{2y^2-8}{2y+4}$

$y-2$

24.

$\dfrac{4x^3-36x}{6x^2-18x}$

25.

$\dfrac{3a-a^2}{a^2-2a-3}$

$\dfrac{-a}{a+1}$

26.

$\dfrac{2x^2+x-6}{x^2+x-2}$

27.

$\dfrac{8z^3-27}{4z^2-9}$

$\dfrac{4z^2+6z+9}{2z+3}$

28.

$\dfrac{6-2v}{v^3-27}$

Applications

29.

The crew team can row at a steady pace of 10 miles per hour in still water. Every afternoon, their training includes a five-mile row upstream on the river. If the current in the river on a given day is $v$ miles per hour, then the time required for this exercise, in minutes, is given by

\begin{equation*} t=\dfrac{300}{10-v} \end{equation*}

Use the graph of this equation shown in the figure to answer questions (a) and (b).

$curve$

How long does the exercise take if there is no current in the river?
How long will it take if the current is 4 miles per hour?
Find an exact answer for part (b) by using the equation.
If the exercise took 2 hours, what was the current in the river?
As the speed of the current increases, what happens to the time needed for the exercise?

30 min
50 min
50 min
6 mph
The time increases. If the current is 10 mph, the team will not be able to row upstream

30.

The eider duck, one of the world's fastest flying birds, can exceed an airspeed of 65 miles per hour. A flock of eider ducks is migrating south at an average airspeed of 50 miles per hour against a moderate headwind. Their next feeding grounds are 150 miles away.

Express the ducks' travel time, $t\text{,}$ as a function of the windspeed, $v\text{.}$
Complete the table showing the travel time for various windspeeds.

$v$ $0$ $10$ $20$ $30$ $40$ $50$

$t$ $\hphantom{00}$ $\hphantom{00}$    

What happens to the travel time as the headwind increases?
Use the table to choose an appropriate window and graph your function $t(v)\text{.}$ Label the scales on the axes.
Estimate the wind speed if the travel time was 12 hours. Illustrate your result on the graph.
Give the equations of any horizontal or vertical asymptotes. What does the vertical asymptote tell us about the problem?

31.

The cost, in thousands of dollars, for extracting $p$ percent of a precious ore from a mine is given by the equation

\begin{equation*} C(p)=\frac{360p}{100-p} \end{equation*}

What input values make sense for $C\text{?}$
Complete the table showing the cost of extracting various percentages of the ore. (Note: do not convert percents to decimals.)

$p$ $0$ $25$ $50$ $75$ $90$ $100$

$C$ $\hphantom{00}$ $\hphantom{00}$    
Graph the function $C$ in an appropriate window. What percentage of the ore can be extracted if $540,000 can be spent on the extraction?
For what values of $p$ is the total cost less than $1,440,000?
The graph has a vertical asymptote. What is it? What is its significance in the context of this problem?

$\displaystyle 0\le p \lt 100$
$p$ $0$ $25$ $50$ $75$ $90$ $100$

$C$ $0$ $120$ $360$ $1080$ $3240$ $-- $
$curve$

60%
$\displaystyle p\lt 80\%$
$p=100\text{:}$ The cost of extracting more ore grows without bound as the amount extracted approaches 100%.

32.

The total cost in dollars of producing $n$ calculators is approximately $20,000+8n\text{.}$

Express the cost per calculator, $C\text{,}$ as a function of the number $n$ of calculators produced.
Complete the table showing the cost per calculator for various production levels.

$n$ $100$ $400$ $500$ $1000$ $4000$ $5000$

$C$      
Graph the function $C(n)$ for the cost per calculator. Use the window

\begin{align*} {\text{Xmin}} \amp = 0 \amp\amp {\text{Xmax}} = 9400\\ {\text{Ymin}} \amp = 0 \amp\amp {\text{Ymax}} = 50 \end{align*}
How many calculators should be produced so that the cost per calculator is $18?
For what values of $n$ is the cost less than $12 per calculator?
Find the horizontal asymptote of the graph. What does it represent in this context?

33.

The volume of a test tube is given by its height times the area of its cross-section. A test tube that holds 200 cubic centimeters is $2x-1$ centimeters long.

What is the area of its cross-section?
Evaluate your fraction for $x=13\text{.}$ What does your answer mean in the context of the problem?

$\dfrac{200}{2x-1} $ square centimeters
8; If $x=13\text{,}$ the area of the cross-section is 8 $\text{cm}^2\text{.}$

34.

Delbert prepares a 25% glucose solution of by mixing 2 ml of glucose with 8 ml of water. If he adds $x$ ml of glucose to the solution, its concentration is given by

\begin{equation*} C(x)=\dfrac{2+x}{8+x} \end{equation*}

How many ml of glucose should Delbert add to increase the concentration to 50%?
Graph the function for $0 \le x \le 100$
What is the horizontal asymptote of the graph? What does it tell you about the solution?

35.

A computer store sells approximately 300 of its most popular model per year. The manager would like to minimize her annual inventory cost by ordering the optimal number of computers, $x\text{,}$ at regular intervals. If she orders $x$ computers in each shipment, the cost of storage will be $6x$ dollars, and the cost of reordering will be $\dfrac{300}{x} (15x + 10)$ dollars. The inventory cost is the sum of the storage cost and the reordering cost.

Use the distributive law to simplify the expression for the reordering cost. Then express the inventory cost, $C\text{,}$ as a function of $x\text{.}$
Complete the table of values for the inventory cost for various reorder sizes.

$x$ $20$ $40$ $60$ $80$ $100$

$C$     
Graph the function $C$ for the cost per calculator. Use the window

\begin{align*} {\text{Xmin}} \amp = 0 \amp\amp {\text{Xmax}} = 150\\ {\text{Ymin}} \amp = 4500 \amp\amp {\text{Ymax}} = 5500 \end{align*}

Estimate the minimum possible value for $C\text{.}$
How many computers should the manager order in each shipment so as to minimize the inventory cost? How many orders will she make during the year?
Graph the function $y = 6x + 4500$ in the same window with the function $C\text{.}$ What do you observe?