Algebra 2

Rational functions in context

One quantity is divided by another that keeps changing. How do you write that as a function, and what does the formula tell you about the situation?

You need

Domain and zeros of a rational functionAlgebra 2 Function notation in a real situationAlgebra 1

Where you are 5 in Rational functions

Functions13

Rational functions5

Asymptotes3 Graphing rational functions Domain and zeros of a rational function Transforming the reciprocal graph

Rational functions in context

When you reach for it

Many quantities in the world are one amount per another amount. The average cost of an item is the total cost divided by the number of items. The concentration of a mixture is the amount of what is dissolved divided by the volume of liquid. The time a shared job takes comes from adding rates, and a rate is a job divided by a time. When the amount on the bottom changes with the input, the quantity is a rational function of that input: one polynomial divided by another, and in these models usually one straight-line expression divided by another.

Reach for this when a question asks what such a ratio is at some input, which input gives a chosen ratio, or what happens to the ratio as the input grows. Answering it uses three things you already have: function notation read in words, the domain of a rational function, and solving a rational equation.

The procedure

Take the lead model of this lesson. Theodric runs a pottery workshop. Setting up a batch of mugs (the kiln, the molds, the glaze) costs 2400 dollars however many mugs are made, and each mug then costs 9 dollars more in clay and glaze.

Build the ratio from the two amounts

Write the top and the bottom separately, each as a function of the input, before dividing. Let x be the number of mugs. The total cost is the fixed 2400 plus 9 for each mug, 2400 + 9x dollars, a linear function of x because the cost per mug is constant. The number of mugs is x. The average cost per mug is the total divided by the number:

C(x) = \frac{2400 + 9x}{x}

The units divide too: dollars divided by mugs gives dollars per mug. Checking that the units of the quotient are the units the question asks for is the quickest test that the right amounts were put on top and bottom.

Find the domain, then cut it down to the story

The implied domain of C is every real number except 0, the one input that makes the denominator zero. In the story x = 0 is excluded for a plain reason as well: no mugs means no average. But the story also removes far more. A number of mugs is a positive integer, so the domain in context is x = 1, 2, 3 and so on, and if the kiln held at most some number of mugs that would cap it too. The domain of the model is the part of the implied domain the story allows, never more.

Read values, and solve for inputs

A value of the function is read the way Function notation in a real situation reads any statement: \displaystyle C(100) = \frac{2400 + 900}{100} = 33 says that when 100 mugs are made, the average cost is 33 dollars a mug.

To find the input that gives a chosen output, set the formula equal to that output and solve the rational equation. Multiplying both sides by the denominator is allowed because, inside the domain, the denominator is never zero. Example 2 does this.

Rewrite to see the long run

Splitting the fraction into its two terms shows what the ratio does as the input grows:

C(x) = \frac{2400}{x} + \frac{9x}{x} = 9 + \frac{2400}{x}

This form says something about the workshop. Each mug carries its own 9 dollars of material, plus an equal share, \displaystyle \frac{2400}{x}, of the set-up cost. That share varies inversely with x: double the batch and each mug carries half as much of it. As x grows the share shrinks toward 0, so C(x) comes as close to 9 as you like. It never reaches 9, since \displaystyle \frac{2400}{x} > 0 for every positive x. On the graph the line y = 9 is a horizontal asymptote, and in words: however many mugs are made, a mug can never cost less on average than its own materials.

Figure 1The average cost per mug, \displaystyle C(x) = \frac{2400 + 9x}{x}. It falls as the set-up cost is shared over more mugs, and settles toward 9, the cost of the materials in one mug, without reaching it.

The same rewriting works on any linear-over-linear model. Write the numerator as a multiple of the denominator plus a constant, and divide: what is left is a constant, the long-run value, plus a leftover fraction that shrinks to zero. Its sign says whether the ratio approaches from above or from below.

Worked examples

Example 1

Reading the average cost

For Theodric’s mugs, \displaystyle C(x) = \frac{2400 + 9x}{x}. Find C(400) and say what it means. Then say how much cheaper each mug is on average in a batch of 400 than in a batch of 100.

Solution

Substitute 400 for x:

C(400) = \frac{2400 + 9(400)}{400} = \frac{6000}{400} = 15

In a batch of 400 mugs the average cost is 15 dollars a mug. From the procedure, C(100) = 33, so the larger batch saves 33 - 15 = 18 dollars a mug. The whole saving comes from the set-up cost: it is \displaystyle \frac{2400}{100} = 24 dollars a mug in the small batch and \displaystyle \frac{2400}{400} = 6 in the large one, and 24 - 6 = 18.

Example 2

A target average cost

How many mugs must Theodric make for the average cost to be 12 dollars? How many for it to be at most 10 dollars? Can the average ever be 9 dollars?

Solution

Set C(x) = 12 and use the rewritten form, which keeps the algebra short:

9 + \frac{2400}{x} = 12

Subtract 9: \displaystyle \frac{2400}{x} = 3. Multiply both sides by x, which is allowed because x > 0: 2400 = 3x, so x = 800. Check: \displaystyle \frac{2400 + 7200}{800} = \frac{9600}{800} = 12. A batch of 800 mugs averages 12 dollars a mug.

For “at most 10”, \displaystyle 9 + \frac{2400}{x} \le 10 means \displaystyle \frac{2400}{x} \le 1. Multiplying an inequality by a positive number keeps its direction, so 2400 \le x. Theodric needs at least 2400 mugs, and at x = 2400 exactly the average is 9 + 1 = 10.

For 9, the equation \displaystyle 9 + \frac{2400}{x} = 9 would need \displaystyle \frac{2400}{x} = 0, and a fraction is 0 only when its numerator is. The numerator is 2400, so there is no such x. This is the asymptote seen from the algebra side.

Example 3

A mixing tank

Ferdinanda’s tank holds 60 liters of water with 3 kg of salt dissolved in it. Brine containing 0.2 kg of salt per liter flows in at 4 liters a minute, and the tank is stirred so the salt is always spread evenly. The tank holds 300 liters. Write the concentration C(t) after t minutes, give its domain in this situation, and find C(10). When is the concentration 0.15 kg per liter?

Solution

Build the top and bottom separately. Each minute 4 liters arrive carrying 4 \times 0.2 = 0.8 kg of salt. So after t minutes the salt is 3 + 0.8t kg and the volume is 60 + 4t liters. Concentration is salt divided by volume:

C(t) = \frac{3 + 0.8t}{60 + 4t}

in kg per liter. The implied domain leaves out only t = -15, where 60 + 4t = 0. The story allows much less: time starts when the tap opens, so t \ge 0, and the tank is full when 60 + 4t = 300, at t = 60. The domain in context is 0 \le t \le 60.

After 10 minutes, \displaystyle C(10) = \frac{3 + 8}{60 + 40} = \frac{11}{100} = 0.11 kg per liter. At the start it was \displaystyle C(0) = \frac{3}{60} = 0.05, so the brine is making the water saltier.

For the target, set C(t) = 0.15 and multiply both sides by 60 + 4t, which is positive on the domain:

\begin{aligned} 3 + 0.8t &= 0.15(60 + 4t) \\ 3 + 0.8t &= 9 + 0.6t \\ 0.2t &= 6 \end{aligned}

so t = 30 minutes, which is inside the domain. Check: \displaystyle C(30) = \frac{3 + 24}{60 + 120} = \frac{27}{180} = 0.15.

Example 4

Where the mixture is heading

For the tank in Example 3, what concentration does C(t) approach as t grows? Does the tank ever get there, and what is the concentration when it fills?

Solution

Write the numerator as a multiple of the denominator plus a constant. Since 0.2(60 + 4t) = 12 + 0.8t, the numerator 3 + 0.8t is 0.2(60 + 4t) - 9. Dividing by 60 + 4t:

C(t) = 0.2 - \frac{9}{60 + 4t}

As t grows, \displaystyle \frac{9}{60 + 4t} shrinks toward 0, so C(t) approaches 0.2 kg per liter from below. That is the concentration of the brine flowing in, which makes sense: after a long time almost all the liquid in the tank is brine that came through the tap, and the first 60 liters are a smaller and smaller part of it.

The model never reaches 0.2, because \displaystyle \frac{9}{60 + 4t} is always positive. The tank stops even further short, since its domain ends at t = 60: \displaystyle C(60) = 0.2 - \frac{9}{300} = 0.2 - 0.03 = 0.17 kg per liter. The long-run value describes the formula as t grows without bound; the real tank only ever sees the part of the curve up to t = 60.

Example 5

Two painters on one fence

Wenzel paints a fence alone in 6 hours, and Zephyrine paints the same fence alone in 9 hours. How long do they take together? Then let Zephyrine’s time alone be h hours instead, and write the time together T as a function of h.

Solution

In one hour Wenzel paints \displaystyle \frac{1}{6} of the fence and Zephyrine \displaystyle \frac{1}{9}. Working side by side, their rates add, and together they paint \displaystyle \frac{1}{t} of the fence an hour, where t is the time together:

\frac{1}{6} + \frac{1}{9} = \frac{1}{t}

The left side is \displaystyle \frac{3}{18} + \frac{2}{18} = \frac{5}{18}, so \displaystyle \frac{1}{t} = \frac{5}{18} and \displaystyle t = \frac{18}{5} = 3.6 hours, which is 3 hours 36 minutes. It is less than Wenzel’s 6 hours, as it must be.

Now replace 9 by h, with h > 0. The same reasoning gives \displaystyle \frac{1}{6} + \frac{1}{h} = \frac{1}{T}. Multiply every term by 6hT, which is not zero:

\begin{aligned} hT + 6T &= 6h \\ T(h + 6) &= 6h \end{aligned}

and divide by h + 6, which is positive:

T(h) = \frac{6h}{h + 6}

Check: \displaystyle T(9) = \frac{54}{15} = \frac{18}{5}, the time found above. The domain in context is h > 0, since a time alone must be positive.

Example 6

What a slow helper is worth

For \displaystyle T(h) = \frac{6h}{h + 6}, find the value of h that gives a time together of 4 hours, then describe what T(h) does as h grows, and say what that means for Wenzel.

Solution

Set T(h) = 4 and multiply by h + 6, which is positive for h > 0: 6h = 4h + 24, so 2h = 24 and h = 12. A partner who would take 12 hours alone brings the job down to 4 hours. Check: \displaystyle \frac{1}{6} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}.

For the long run, write 6h as 6(h + 6) - 36:

T(h) = 6 - \frac{36}{h + 6}

As h grows, \displaystyle \frac{36}{h + 6} shrinks to 0, so T(h) rises toward 6 hours, staying below it. In words: a very slow partner adds almost nothing, and the job takes almost as long as Wenzel alone, but any partner at all, however slow, saves some time. The function also never exceeds h, since \displaystyle \frac{6h}{h + 6} < h is the same as 6 < h + 6 (divide by h and multiply by h + 6, both positive). Together they always beat each of them alone.

Figure 2The time together, \displaystyle T(h) = \frac{6h}{h + 6}, when one painter takes 6 hours alone and the other takes h. The slower the second painter, the closer the time gets to 6 hours, the first painter’s time alone.

Example 7

Choosing between two set-ups

A second supplier offers Theodric a cheaper set-up, 900 dollars a batch, but charges 12 dollars a mug for materials. For which batch sizes is the second supplier cheaper on average?

Solution

The second supplier’s average cost is \displaystyle B(x) = \frac{900 + 12x}{x} = 12 + \frac{900}{x}. Find where the two averages are equal:

\frac{2400 + 9x}{x} = \frac{900 + 12x}{x}

Both sides have the same denominator x, which is positive, so multiplying by it gives 2400 + 9x = 900 + 12x, then 1500 = 3x and x = 500. At 500 mugs both average \displaystyle \frac{6900}{500} = 13.8 dollars.

The difference is \displaystyle C(x) - B(x) = \frac{1500 - 3x}{x}. Its denominator is positive, so its sign is the sign of 1500 - 3x: positive below 500 mugs, negative above. So the second supplier is cheaper on average for batches of fewer than 500 mugs, and Theodric’s present supplier for more than 500. The long-run values agree: B heads to 12 and C to 9, so for large batches the cheaper materials win.

Comparing averages at the same x is the same as comparing total costs, since each average is its total divided by the same positive number. The rational model’s job here was to ask the question in the units the buyer thinks in, dollars a mug.

Boundaries

The story’s domain is smaller than the formula’s. The implied domain only removes the inputs that make the denominator zero. A count of items is a positive integer, a time starts at 0, a tank has a capacity, and every answer has to sit inside those limits as well. For a count, the real graph is a row of separate points; the smooth curve only shows their trend.
The model is only as good as its assumptions. The average cost formula assumes the cost per item never changes; the tank assumes instant, even mixing and a steady flow; the painters assume each keeps a steady rate and neither gets in the other’s way. Where an assumption fails, so does the formula.
A long-run value is approached, not reached. The ratio comes as close as you like to it as the input grows without bound, but a leftover term that is never zero keeps it from arriving. If the story caps the input, as the full tank does, the ratio stops even further away.
Rearranging can quietly widen the domain. Solving \displaystyle \frac{1}{6} + \frac{1}{h} = \frac{1}{T} for T gives \displaystyle \frac{6h}{h + 6}, which is defined at h = 0 even though \displaystyle \frac{1}{h} is not. The domain comes from the story and the original equation, not from the final formula.
Multiplying through needs a non-zero multiplier, and in an inequality a positive one, which is why each example states the sign of the denominator before clearing it.

Where this goes next

The long-run behavior seen here is the horizontal asymptote of the model, and the input the denominator forbids is where a vertical asymptote would sit if the story let the input go there. Graphing rational functions puts these pieces together into a full sketch, including the parts of the curve no story uses. The rewriting of a numerator as a multiple of the denominator plus a remainder is the first case of dividing one polynomial by another, which also finds slant asymptotes when the numerator’s degree is one higher. Concentration and shared work return wherever rates are combined, and average cost returns whenever a fixed cost is shared over a growing number of items.

Exercises

Build and read a model

Corvin’s club hires a hall for 600 dollars, and food costs 15 dollars a guest. The hall holds 120 guests. Write the average cost per guest A(n) for n guests, give its domain in this situation, and find A(40).
For the club in Exercise 1, how many guests make the average cost 20 dollars?
For Theodric’s mugs, find C(50) and say in a sentence what it means.
A 40-liter tank of water has 2 kg of sugar dissolved in it. Syrup with 0.3 kg of sugar per liter flows in at 5 liters a minute, and the tank is kept stirred. Write the concentration C(t) after t minutes and find C(8).
For the tank in Exercise 4, when is the concentration 0.2 kg per liter?

Solve for the input

A bucket holds 3 liters of cleaning solution containing 0.6 liters of bleach. Pure water is added. Write the concentration of bleach C(x) after x liters of water are added, and find how much water brings it down to 0.05.
A 10-liter mixture is 20\% acid. How many liters of pure acid must be added to make it 50\% acid?
How many mugs must Theodric make for the average cost to be 10.50 dollars?
For the club in Exercise 1, which numbers of guests make the average cost at most 25 dollars?

Shared work

A gardener weeds a plot alone in 4 hours. A helper alone would take h hours. Write the time together T(h) and find T(12).
For the gardener in Exercise 10, how long must the helper take alone for the two to finish in 2.4 hours?
For T(h) in Exercise 10, what value does T(h) approach as h grows, and what does that mean for the gardener?
One pipe fills a tank in 5 hours. A second pipe alone takes p hours. Write the time with both pipes open, T(p), and find p if both together take 2 hours.

The long run and comparisons

Write the club’s A(n) from Exercise 1 as a constant plus a fraction. What value does it approach as n grows, and why can it never reach it?
Show that the concentration in Exercise 4 is \displaystyle C(t) = 0.3 - \frac{10}{40 + 5t}, and say what value it approaches and why that value makes sense.
Plan P costs 1500 dollars to set up and 6 dollars an item; plan Q costs 300 to set up and 10 an item. At what number of items are the average costs equal, and which plan is cheaper on average for 500 items?

Explain

A firm has a fixed cost F > 0 and a cost v per item. Explain why its average cost \displaystyle \frac{F + vx}{x} is greater than v for every positive x, and why it gets closer to v as x grows.
Brine of concentration k flows at r liters a minute into a tank that starts with V liters of liquid containing s kg of salt. Explain why the concentration approaches k as time goes on, whatever V and s are.
Explain why the time together \displaystyle T(h) = \frac{ah}{a + h}, for workers who take a and h hours alone, is less than both a and h.
Someone says the domain of \displaystyle C(x) = \frac{2400 + 9x}{x} is every real number except 0. Say what is right about this and what the domain is for Theodric’s workshop.
The formula \displaystyle T(h) = \frac{6h}{h + 6} gives T(0) = 0. Explain why h = 0 is still not in the domain, and why the answer 0 makes no sense for the painters.

Answers

\displaystyle A(n) = \frac{600 + 15n}{n}, in dollars per guest. Domain: the integers n = 1, 2, \ldots, 120. \displaystyle A(40) = \frac{600 + 600}{40} = 30 dollars a guest.
120 guests. \displaystyle \frac{600 + 15n}{n} = 20 gives 600 + 15n = 20n, so 5n = 600.
\displaystyle C(50) = \frac{2400 + 450}{50} = 57: in a batch of 50 mugs the average cost is 57 dollars a mug.
\displaystyle C(t) = \frac{2 + 1.5t}{40 + 5t} kg per liter, since 5 \times 0.3 = 1.5 kg of sugar arrives each minute. \displaystyle C(8) = \frac{14}{80} = 0.175 kg per liter.
After 12 minutes. 2 + 1.5t = 0.2(40 + 5t) = 8 + t gives 0.5t = 6. Check: \displaystyle \frac{20}{100} = 0.2.
\displaystyle C(x) = \frac{0.6}{3 + x}, and 9 liters of water. 0.6 = 0.05(3 + x) gives 3 + x = 12.
6 liters. The mixture holds 2 liters of acid, so after adding x liters the concentration is \displaystyle \frac{2 + x}{10 + x} = \frac{1}{2}, which gives 4 + 2x = 10 + x. Check: \displaystyle \frac{8}{16}.
1600 mugs. \displaystyle 9 + \frac{2400}{x} = 10.5 gives \displaystyle \frac{2400}{x} = 1.5, so \displaystyle x = \frac{2400}{1.5}.
From 60 to 120 guests. \displaystyle A(n) = 15 + \frac{600}{n} \le 25 gives \displaystyle \frac{600}{n} \le 10, and since n > 0, n \ge 60; the hall caps n at 120.
\displaystyle T(h) = \frac{4h}{h + 4}, from \displaystyle \frac{1}{4} + \frac{1}{h} = \frac{1}{T}. \displaystyle T(12) = \frac{48}{16} = 3 hours.
h = 6 hours. 4h = 2.4(h + 4) = 2.4h + 9.6 gives 1.6h = 9.6. Check: \displaystyle \frac{1}{4} + \frac{1}{6} = \frac{5}{12}, and \displaystyle \frac{12}{5} = 2.4.
It approaches 4 hours, from below, since \displaystyle T(h) = 4 - \frac{16}{h + 4}. A very slow helper adds almost nothing, so the job takes nearly the gardener’s 4 hours alone, but always a little less.
\displaystyle T(p) = \frac{5p}{p + 5}, and \displaystyle p = \frac{10}{3} hours, which is 3 hours 20 minutes. 5p = 2(p + 5) gives 3p = 10.
\displaystyle A(n) = 15 + \frac{600}{n}, which approaches 15, the food cost per guest. It never reaches 15 because \displaystyle \frac{600}{n} > 0 for every positive n: each guest always carries a share of the hall.
0.3(40 + 5t) = 12 + 1.5t, so 2 + 1.5t = 0.3(40 + 5t) - 10, and dividing by 40 + 5t gives the form shown. It approaches 0.3 kg per liter from below, the concentration of the incoming syrup, because in the long run nearly all the liquid in the tank is syrup that came in.
Equal at 300 items, where both average \displaystyle \frac{3300}{300} = 11 dollars. For 500 items plan P averages \displaystyle \frac{4500}{500} = 9 and plan Q \displaystyle \frac{5300}{500} = 10.6, so P is cheaper.
\displaystyle \frac{F + vx}{x} = v + \frac{F}{x}. For x > 0 the term \displaystyle \frac{F}{x} is positive, so the average is above v. As x grows, \displaystyle \frac{F}{x} shrinks toward 0, so the average comes as close to v as you like: the fixed cost is spread over more and more items.
After t minutes the salt is s + krt and the volume V + rt, so \displaystyle C(t) = \frac{s + krt}{V + rt} = k + \frac{s - kV}{V + rt}. The numerator s - kV is a fixed number and the denominator grows without bound, so the second term shrinks to 0 and C(t) approaches k. In words, the original liquid becomes a smaller and smaller part of the tank.
Dividing \displaystyle \frac{ah}{a + h} < a by a > 0 gives \displaystyle \frac{h}{a + h} < 1, true since h < a + h. The same argument with h in place of a gives \displaystyle \frac{ah}{a + h} < h. Each worker adds work to the other’s, so together they finish before either would alone.
Right for the formula: 0 is the only input that makes the denominator zero, so the implied domain is every real number except 0. Wrong for the workshop: x counts mugs, so the domain is the positive integers 1, 2, 3, \ldots, and a negative or fractional batch means nothing.
The model comes from \displaystyle \frac{1}{6} + \frac{1}{h} = \frac{1}{T}, and \displaystyle \frac{1}{h} is undefined at h = 0; solving for T hid that exclusion. In the story, h = 0 would be a partner who paints the whole fence in no time, which is impossible, so the answer T = 0 describes no real pair of painters.