Rational Functions · applications

Applications: Rational Functions

10 question types · Secondary 3 · each worked step by step with a figure that follows the steps

H2

01

The Average Cost of a Printed Yearbook: The Cost That a Long Print Run Approaches and the Run That Meets a Target

methodDivide the Total Cost by the Number of Copies, Compare the Degrees of the Top and the Bottom to Read the Horizontal Asymptote, Then Solve an Inequality for the Target

A school prints its yearbook. Setting up the press costs $600, and after that each copy costs $4 to print. When x copies are printed, the average cost of a copy is A(x) = 600 + 4xx dollars. (a) Find the horizontal asymptote of the graph of A, and say what it means for the school. (b) The school wants the average cost to be no more than $6 a copy. Find the least number of copies it must print.

04812162002004006008001000copies printed, xaverage cost of a copy ($)A(x) = (600 + 4x)/x = 600/x + 410 copies: A(10) = 60 + 4 = 64 dollars each
The total cost 600 + 4x is shared among x copies, so A(x) = 600x + 4. A short run is expensive: A(10) = 64.
The total cost of x copies is 600 + 4x dollars, and it is shared among x copies. Divide each term of the top by x: A(x) = 600x + 4. The number of copies is positive, so the domain is x > 0. For a short run the set-up cost dominates: A(10) = 60 + 4 = 64.
step 1 of 5

The average cost is the total cost shared equally among the copies, so it is a rational function of the number of copies. The set-up cost is shared among more and more copies as the print run grows, and its share fades. What the average cost closes on is the horizontal asymptote, and the target in part (b) is an inequality in x.

  1. The total cost of x copies is 600 + 4x dollars, and it is shared among x copies. Divide each term of the top by x: A(x) = 600x + 4. The number of copies is positive, so the domain is x > 0. For a short run the set-up cost dominates: A(10) = 60 + 4 = 64.
  2. The top and the bottom both have degree 1, so the horizontal asymptote is the ratio of the leading coefficients: y = 41 = 4. The term 600x shows why: A(1000) = 0.6 + 4 = 4.6, and the longer the run, the smaller that term becomes.
  3. (a) The horizontal asymptote is y = 4. With a long print run the average cost closes on $4 a copy, which is the printing cost alone. It never reaches $4, because 600x is always more than 0.
  4. For the target, 600x + 4 ≤ 6, so 600x ≤ 2. Since x is positive, multiplying both sides by x keeps the inequality sign: 600 ≤ 2x, so x ≥ 300.
  5. (b) The school must print at least 300 copies. Check: A(300) = 600 + 1200300 = 1800300 = 6, and A(299) = 1796299 ≈ 6.007, which is just over the target.

answer(a) The horizontal asymptote is y = 4: with a long print run the average cost closes on $4 a copy, the printing cost alone, and never reaches it; (b) 300 copies

techniqueHorizontal Asymptotes by Degree · Zeros and Vertical Asymptotes

examsH2

Common pitfalls

  • Reading the horizontal asymptote as y = 600 or as y = 0. When the top and the bottom have the same degree, the asymptote is the ratio of the leading coefficients, the coefficients of x: 41 = 4. The 600 is a constant term, and its share 600x is the part that fades.
  • Saying that the average cost becomes $4 when enough copies are printed. An asymptote is a value the curve approaches and does not reach: A(x) = 4 would need 600x = 0, and no number of copies makes that true.
02

The Average Cost of a Chair When Overtime Makes Large Outputs Expensive: A Slant Asymptote

methodDivide the Total Cost by the Output to Get a Line Plus a Remainder Term, Read the Line as the Slant Asymptote, Then Multiply Through by x to Solve for a Given Average

A workshop makes x chairs in a day. Wood, wages and overtime bring the total cost of the day to x2 + 50x + 400 dollars, so the average cost of a chair is A(x) = x2 + 50x + 400x dollars. (a) Divide to write A(x) as a quotient plus a remainder term, and give the slant asymptote of the graph. Say what the asymptote tells the workshop. (b) Find the two outputs at which the average cost is $100 a chair.

0501001500102030405060chairs made in a day, xaverage cost of a chair ($)divide each term of the top by xA(x) = x + 50 + 400/x
Divide each term of the top by x: A(x) = x + 50 + 400x. The quotient is x + 50, and the remainder term is 400x.
Divide each term of the top by x: x2x = x, 50xx = 50, and 400 is left over as the remainder. So A(x) = x + 50 + 400x. The quotient is x + 50 and the remainder term is 400x.
step 1 of 6

The top of this fraction has degree 2 and the bottom has degree 1. When the top is one degree higher than the bottom, the graph has no horizontal asymptote. Instead, division gives a straight line plus a remainder term that fades, and the line is the slant asymptote. Part (b) is an equation that becomes a quadratic once both sides are multiplied by x.

  1. Divide each term of the top by x: x2x = x, 50xx = 50, and 400 is left over as the remainder. So A(x) = x + 50 + 400x. The quotient is x + 50 and the remainder term is 400x.
  2. As x grows, the remainder term 400x fades toward 0, and the curve closes on the line y = x + 50. For example A(400) = 400 + 50 + 1 = 451, and the line gives 450.
  3. (a) A(x) = x + 50 + 400x, and the slant asymptote is y = x + 50. For large outputs the average cost follows this line: it rises by about $1 for every extra chair, and it is always a little above the line, by 400x dollars.
  4. Put A(x) = 100: x + 50 + 400x = 100. Multiply both sides by x: x2 + 50x + 400 = 100x, so x2 − 50x + 400 = 0.
  5. Factorize: (x − 10)(x − 40) = 0, so x = 10 or x = 40. Both are positive whole numbers of chairs, so both fit the situation.
  6. (b) The average cost is $100 at 10 chairs and at 40 chairs. Check: A(10) = 10 + 50 + 40 = 100 and A(40) = 40 + 50 + 10 = 100. Between the two outputs the average cost is lower: A(20) = 20 + 50 + 20 = 90.

answer(a) A(x) = x + 50 + 400x, and the slant asymptote is y = x + 50: for large outputs the average cost rises by about $1 for every extra chair; (b) 10 chairs or 40 chairs

techniqueSlant Asymptotes by Long Division · Long Division of Polynomials

examsH2

Common pitfalls

  • Looking for a horizontal asymptote by comparing leading coefficients. That rule is for a top and a bottom of the same degree. Here the top is one degree higher, so the average cost does not level off: it follows the line y = x + 50, which keeps rising.
  • Stopping at x = 10. A quadratic equation can have two solutions, and the curve falls to a lowest point and rises again, so it meets the level 100 twice. The second output, 40 chairs, is just as real as the first.
03

Brine Poured into a Tank of Fresh Water: The Concentration That Is Never Passed

methodWrite the Concentration as the Mass of Salt Divided by the Volume of Liquid, Read the Horizontal Asymptote from the Leading Coefficients, Then Solve for the Volume That Gives a Stated Concentration

A large tank holds 200 liters of fresh water. Brine that contains 30 grams of salt in every liter is poured in and stirred. After x liters of brine have been poured in, the concentration of salt in the tank is C(x) grams per liter. (a) Find a formula for C(x), and find the concentration that the tank approaches but never reaches. (b) Find how much brine has been poured in when the concentration is 10 grams per liter.

010203002004006008001000liters of brine poured in, xsalt (g per liter)30x grams of salt in x + 200 litersC(x) = 30x/(x + 200)
After x liters of brine the tank holds 30x grams of salt in x + 200 liters, so C(x) = 30xx + 200.
After x liters of brine, the tank holds 30x grams of salt in x + 200 liters of liquid. So C(x) = 30xx + 200, with domain x ≥ 0.
step 1 of 6

A concentration is the mass of salt divided by the volume of liquid that holds it. Both of these grow as the brine is poured in, which makes the concentration a rational function of x. The top and the bottom have the same degree, so the graph levels off at a horizontal asymptote.

  1. After x liters of brine, the tank holds 30x grams of salt in x + 200 liters of liquid. So C(x) = 30xx + 200, with domain x ≥ 0.
  2. The top is zero at x = 0: before any brine is poured in, there is no salt. The bottom is zero at x = −200, which is outside the domain, so the graph has no vertical asymptote in this situation.
  3. The top and the bottom both have degree 1, so the horizontal asymptote is the ratio of the leading coefficients: y = 301 = 30. For example C(800) = 240001000 = 24, and the value keeps rising toward 30.
  4. (a) C(x) = 30xx + 200, and the concentration approaches 30 grams per liter, which is the concentration of the brine itself. It never reaches 30, because the 200 liters of fresh water are still in the tank and always dilute the brine a little.
  5. Put C(x) = 10: 30xx + 200 = 10. Multiply both sides by x + 200: 30x = 10x + 2000, so 20x = 2000 and x = 100.
  6. (b) The concentration is 10 grams per liter after 100 liters of brine. Check: there are 30 × 100 = 3000 grams of salt in 100 + 200 = 300 liters, and 3000300 = 10.

answer(a) C(x) = 30xx + 200, and the concentration approaches 30 grams per liter, the concentration of the brine itself, without ever reaching it; (b) 100 liters

techniqueHorizontal Asymptotes by Degree · Zeros and Vertical Asymptotes

examsH2

Common pitfalls

  • Dividing the salt by the volume of brine only, which gives 30xx = 30 at every moment. The salt is spread through all the liquid in the tank, so the bottom of the fraction is x + 200 liters and not x liters.
  • Expecting the concentration to reach 30 grams per liter once enough brine has been added. 30xx + 200 = 30 would need 30x = 30x + 6000, which is impossible. The value 30 is an asymptote: the curve closes on it and never meets it.
04

Caffeine in the Blood After a Cup of Coffee: The Concentration Over the Hours That Follow

methodCompare the Degrees of the Top and the Bottom for the Horizontal Asymptote, Then Turn a Stated Concentration into a Quadratic Equation and Read the Time Between Its Two Solutions

After a cup of coffee, the concentration of caffeine in a person's blood x hours later is modeled by C(x) = 20xx2 + 4 milligrams per liter, for x ≥ 0. (a) Find C(2), and find the horizontal asymptote of the graph. Say what the asymptote means. (b) For how long is the concentration at least 4 milligrams per liter?

0246014812hours after the coffee, xcaffeine (mg per liter)(2, 5)C(2) = 20 × 2/(22+ 4)= 40/8 = 5 mg per liter
C(2) = 408 = 5 milligrams per liter.
C(2) = 20 × 222 + 4 = 408 = 5 milligrams per liter.
step 1 of 6

Here the bottom of the fraction has a higher degree than the top. For large x the bottom grows much faster than the top, so the value of the fraction returns toward zero. The model rises while the caffeine is absorbed and falls as the body clears it. The times at which it passes a given level come from a quadratic equation.

  1. C(2) = 20 × 222 + 4 = 408 = 5 milligrams per liter.
  2. The top has degree 1 and the bottom has degree 2. When the bottom has the higher degree, the horizontal asymptote is y = 0. For example C(20) = 400404 ≈ 0.99 and C(100) = 200010004 ≈ 0.2.
  3. (a) C(2) = 5, and the horizontal asymptote is y = 0: the caffeine leaves the blood, and the concentration returns toward zero without becoming negative. The bottom x2 + 4 is never zero, so there is no vertical asymptote, and the only zero of the function is at x = 0, the moment the coffee is drunk.
  4. Put C(x) = 4: 20xx2 + 4 = 4. Multiply both sides by x2 + 4: 20x = 4x2 + 16. Divide every term by 4 and collect the terms on one side: x2 − 5x + 4 = 0.
  5. Factorize: (x − 1)(x − 4) = 0, so x = 1 or x = 4. The curve rises through the level 4 at x = 1 and falls back through it at x = 4.
  6. (b) The concentration is at least 4 milligrams per liter from x = 1 to x = 4, which is 4 − 1 = 3 hours. Check: C(1) = 205 = 4 and C(4) = 8020 = 4, and between them C(2) = 5 is above the level.

answer(a) C(2) = 5 milligrams per liter, and the horizontal asymptote is y = 0: the caffeine leaves the blood and the concentration returns toward zero; (b) for 3 hours, from x = 1 to x = 4

techniqueHorizontal Asymptotes by Degree · Zeros and Vertical Asymptotes

examsH2

Common pitfalls

  • Reading the horizontal asymptote as y = 20 from the leading coefficient of the top. The ratio of leading coefficients is the asymptote only when the two degrees are equal. Here the bottom has the higher degree, so the fraction fades to 0.
  • Giving 4 hours as the answer because the concentration falls back to the level at x = 4. The concentration first reaches the level at x = 1, so the time spent at or above the level is 4 − 1 = 3 hours.
05

A Lens, an Object and a Screen: What the Two Asymptotes Mean for Where the Image Forms

methodSet the Bottom Equal to Zero for the Vertical Asymptote and Check That the Top Is Not Zero There, Compare Leading Coefficients for the Horizontal Asymptote, and Read Each One as a Statement About the Distances

A lens forms a sharp image of a candle on a screen. When the candle is x cm from the lens, the screen has to be v(x) = 10xx − 10 cm from the lens on the other side, for x > 10. (a) Find the vertical asymptote and the horizontal asymptote of the graph of v, and say what each one means for the candle and the screen. (b) The screen is 35 cm from the lens. How far from the lens is the candle?

010203040506001020304050candle to lens (cm), xlens to screen (cm)x = 10x − 10 = 0 at x = 10, where the top is 100vertical asymptote x = 10; v(11) = 110
The bottom is zero at x = 10 and the top is 100 there, so the vertical asymptote is x = 10. Just past it the values are very large: v(11) = 110.
The bottom x − 10 is zero at x = 10, and the top there is 10 × 10 = 100, which is not zero. So the vertical asymptote is x = 10. Just beyond it the values are very large: v(11) = 1101 = 110 and v(10.5) = 1050.5 = 210.
step 1 of 5

A rational function has a vertical asymptote where its bottom is zero and its top is not, and a horizontal asymptote that depends on the two degrees. In this situation both asymptotes can be seen on the bench: one is a position of the candle that gives no image, and the other is the closest that the screen ever comes to the lens.

  1. The bottom x − 10 is zero at x = 10, and the top there is 10 × 10 = 100, which is not zero. So the vertical asymptote is x = 10. Just beyond it the values are very large: v(11) = 1101 = 110 and v(10.5) = 1050.5 = 210.
  2. The top and the bottom both have degree 1, so the horizontal asymptote is the ratio of the leading coefficients: y = 101 = 10. For example v(1010) = 101001000 = 10.1.
  3. (a) The vertical asymptote is x = 10: as the candle is moved in toward 10 cm from the lens, the screen has to be moved farther and farther away, and at 10 cm no screen position gives an image. The horizontal asymptote is y = 10: for a candle far from the lens, the screen sits just beyond 10 cm from the lens and never closer.
  4. Put v(x) = 35: 10xx − 10 = 35. Multiply both sides by x − 10: 10x = 35x − 350, so 25x = 350 and x = 14.
  5. (b) The candle is 14 cm from the lens. Check: v(14) = 1404 = 35, and 14 > 10, so the value is in the domain.

answer(a) The vertical asymptote is x = 10: a candle 10 cm from the lens gives no image, and near that distance the screen must be very far away. The horizontal asymptote is y = 10: for a distant candle the screen sits just beyond 10 cm from the lens; (b) 14 cm

techniqueZeros and Vertical Asymptotes · Horizontal Asymptotes by Degree

examsH2

Common pitfalls

  • Mixing up the two asymptotes. The vertical asymptote comes from the bottom of the fraction and is a value of x, the distance of the candle. The horizontal asymptote comes from the leading coefficients and is a value of y, the distance of the screen. Here both happen to be 10, but they say different things.
  • Multiplying out 35(x − 10) as 35x − 10. The 35 multiplies both terms of the bracket, so the right side is 35x − 350. With the slip the equation gives x = 0.4, which is not even in the domain x > 10.
06

A Coach Driving Faster or Slower Than Its Timetable: A Vertical Asymptote and a Domain Cut by the Situation

methodWrite the Time as the Distance Divided by the Speed, Find the Vertical Asymptote Where the Speed Would Be Zero, Let the Situation Limit the Domain, Then Solve for a Stated Time

A coach has 240 km to drive, and its timetable assumes an average speed of 60 km/h. If the coach averages x km/h more than that, the journey takes T(x) = 24060 + x hours. A negative value of x means that the coach is slower than the timetable. The coach may not drive faster than 100 km/h. (a) Find the vertical asymptote of the graph of T and say what it means. Give the domain that the situation allows. (b) The driver wants to arrive 1 hour sooner than the timetable says. Find x.

36912−60−40−2002040km/h faster than the timetable, xhours, T(0, 4)the speed is 60 + x, so T(x) = 240/(60 + x)on the timetable: T(0) = 240/60 = 4 hours
The speed is 60 + x km/h, so T(x) = 24060 + x. On the timetable x = 0 and the journey takes T(0) = 4 hours.
The speed of the coach is 60 + x km/h, and the time is the distance divided by the speed: T(x) = 24060 + x. On the timetable x = 0, so the planned time is T(0) = 24060 = 4 hours.
step 1 of 5

The time for a journey is the distance divided by the speed, and the speed here is 60 + x. The function is undefined where the speed would be zero, and that is the vertical asymptote. The formula accepts many values of x that make no sense for a coach, so the situation decides the domain.

  1. The speed of the coach is 60 + x km/h, and the time is the distance divided by the speed: T(x) = 24060 + x. On the timetable x = 0, so the planned time is T(0) = 24060 = 4 hours.
  2. The bottom 60 + x is zero at x = −60, and the top is 240, which is never zero. So the vertical asymptote is x = −60. Close to it the journey is very long: T(−50) = 24010 = 24 hours.
  3. (a) The vertical asymptote is x = −60: a coach that is 60 km/h slower than the timetable is not moving, so it never arrives. The speed must be more than 0 and at most 100 km/h, so 0 < 60 + x ≤ 100, and the domain is −60 < x ≤ 40. The shortest possible journey is T(40) = 240100 = 2.4 hours.
  4. One hour sooner than 4 hours is 3 hours. Put T(x) = 3: 24060 + x = 3, so 240 = 180 + 3x, 3x = 60 and x = 20.
  5. (b) x = 20: the coach must average 20 km/h more than the timetable, a speed of 80 km/h. Check: 24080 = 3 hours, and 20 ≤ 40, so the value is in the domain.

answer(a) The vertical asymptote is x = −60: a coach 60 km/h slower than its timetable is standing still and never arrives. The domain is −60 < x ≤ 40; (b) x = 20, a speed of 80 km/h

techniqueZeros and Vertical Asymptotes

examsH2

Common pitfalls

  • Giving the vertical asymptote as x = 60. The bottom is 60 + x, which is zero when x = −60. In the situation that is a coach slowed by the whole of its planned 60 km/h, which is a coach standing still.
  • Reasoning that 1 hour out of 4 is a quarter, so the speed must rise by a quarter, to 75 km/h. Time and speed are inversely proportional: the time is multiplied by 34, so the speed is multiplied by 43, and 60 × 43 = 80 km/h.
07

The Width of a Vegetable Plot from Its Area and Its Length: A Hole Where the Length Would Be Zero

methodFactorize the Top, Cancel the Common Factor While Recording the Value of x It Excludes, and Mark That Value as a Hole on the Simplified Graph

A rectangular vegetable plot is planned with an area of x2 − 9 square meters and a length of x − 3 meters, where x > 3. Its width is W(x) = x2 − 9x − 3 meters. (a) Simplify W(x). Describe what happens on the graph at x = 3, and say what x = 3 would mean for the plot. (b) The width is to be 10 meters. Find x, and the length and the area of the plot.

area x2− 9length x − 3width ?x2− 9 = (x − 3)(x + 3)a difference of two squares
The width is the area divided by the length. The top is a difference of two squares: x2 − 9 = (x − 3)(x + 3).
The top is a difference of two squares: x2 − 9 = (x − 3)(x + 3).
step 1 of 6

When the top and the bottom of a rational function share a factor, the factor cancels, and the graph is the graph of the simpler expression with one point missing. The missing point is a hole. It is not a vertical asymptote, because near that value of x the function stays close to an ordinary number.

  1. The top is a difference of two squares: x2 − 9 = (x − 3)(x + 3).
  2. So W(x) = (x − 3)(x + 3)x − 3 = x + 3, as long as x ≠ 3. Canceling x − 3 is a division by x − 3, and that is allowed only when x − 3 is not zero.
  3. At x = 3 the top is 9 − 9 = 0 and the bottom is 3 − 3 = 0, and 00 is undefined. Because the factor cancels, the graph has a hole there and not a vertical asymptote. The line y = x + 3 would pass through (3, 6), so the hole is an open circle at (3, 6).
  4. (a) W(x) = x + 3 for x ≠ 3, and the graph is the line y = x + 3 with a hole at (3, 6). At x = 3 the length would be 0 meters and the area 0 square meters: there is no plot, so there is no width to find, even though x + 3 on its own would give 6.
  5. Put W(x) = 10: x + 3 = 10, so x = 7. The length is 7 − 3 = 4 meters, and the area is 72 − 9 = 40 square meters.
  6. (b) x = 7: the plot is 4 meters long and 10 meters wide, with an area of 40 square meters. Check: 404 = 10.

answer(a) W(x) = x + 3 for x ≠ 3: the graph is the line y = x + 3 with a hole at (3, 6), because at x = 3 the length and the area would both be 0 and there is no plot; (b) x = 7, so the plot is 4 meters long and 10 meters wide, with an area of 40 square meters

techniqueHoles Where a Factor Cancels · Zeros and Vertical Asymptotes

examsH2

Common pitfalls

  • Calling x = 3 a vertical asymptote because the bottom is zero there. A vertical asymptote needs the top to be non-zero at that value. Here the top is zero as well and the factor x − 3 cancels, so the graph has a single missing point and the values near it stay close to 6.
  • Writing W(x) = x + 3 and forgetting the condition x ≠ 3. The simplified expression and the original fraction agree everywhere except at x = 3, where the original is undefined. Without the condition the two are not the same function.
08

The Average Rate of Climb of a Hot-Air Balloon Measured from the 2-Minute Mark: A Hole at the Instant Itself

methodDivide the Top by the Bottom with Long Division, Take a Remainder of Zero as the Sign of a Common Factor and a Hole, and Read the Value the Hole Leaves Out

For the first 10 minutes after it lifts off, a hot-air balloon is x2 + 3x meters above the ground after x minutes. After 2 minutes its height is 10 meters. For any other time x, the average rate of climb between the 2-minute mark and minute x is R(x) = x2 + 3x − 10x − 2 meters per minute. (a) Use long division to simplify R(x), and explain why the graph has a hole and where the hole is. (b) Find the average rate of climb between the 2-minute mark and minute 6, and find the rate that the hole leaves out.

x − 2x2+ 3x− 10xx2− 2x5x− 10x2divided by x is x, and x(x − 2) = x2− 2xsubtract: 3x + 2x = 5x, then bring down −10
The balloon climbs (x2 + 3x) − 10 meters in x − 2 minutes. Divide: x2 ÷ x = x, and x(x − 2) = x2 − 2x. Subtracting leaves 5x, and −10 comes down.
The balloon climbs (x2 + 3x) − 10 meters in x − 2 minutes, which gives the formula for R(x). Start the long division of x2 + 3x − 10 by x − 2: x2 ÷ x = x, and x(x − 2) = x2 − 2x. Subtract: (x2 + 3x) − (x2 − 2x) = 5x. Bring down the −10 to get 5x − 10.
step 1 of 6

An average rate is a change in height divided by the time it took. Measured from the 2-minute mark, both the change and the time are zero at x = 2, so the function is undefined there. Long division shows that the bottom divides the top exactly, which means the two share a factor and the graph has a hole. The value that the hole leaves out is the rate of climb at the 2-minute mark itself.

  1. The balloon climbs (x2 + 3x) − 10 meters in x − 2 minutes, which gives the formula for R(x). Start the long division of x2 + 3x − 10 by x − 2: x2 ÷ x = x, and x(x − 2) = x2 − 2x. Subtract: (x2 + 3x) − (x2 − 2x) = 5x. Bring down the −10 to get 5x − 10.
  2. Next, 5x ÷ x = 5, and 5(x − 2) = 5x − 10. Subtract: the remainder is 0. So x2 + 3x − 10 = (x − 2)(x + 5) exactly.
  3. So R(x) = (x − 2)(x + 5)x − 2 = x + 5 for x ≠ 2. At x = 2 the balloon has climbed 0 meters in 0 minutes, and 00 is undefined.
  4. (a) R(x) = x + 5 for x ≠ 2. The graph is a straight line with a hole at (2, 7). It is a hole and not a vertical asymptote because the factor x − 2 cancels, and the values near x = 2 stay close to 7.
  5. Between the 2-minute mark and minute 6, R(6) = 6 + 5 = 11. Check with the heights: at minute 6 the height is 36 + 18 = 54 meters, and 54 − 106 − 2 = 444 = 11.
  6. (b) The average rate of climb is 11 meters per minute. The hole leaves out 7 meters per minute: R(1.9) = 6.9 and R(2.1) = 7.1, so the averages close on 7 from both sides, and 7 meters per minute is the rate of climb at the 2-minute mark.

answer(a) R(x) = x + 5 for x ≠ 2: a straight line with a hole at (2, 7), because the factor x − 2 cancels and at x = 2 no time has passed; (b) 11 meters per minute, and the hole leaves out 7 meters per minute, the rate of climb at the 2-minute mark

techniqueHoles Where a Factor Cancels · Long Division of Polynomials

examsH2

Common pitfalls

  • Substituting x = 2 into x + 5 and reporting R(2) = 7. The function R is not defined at x = 2, because no time has passed and no average can be taken. The value 7 is what the averages approach, and it has to be described that way.
  • Subtracting x2 − 2x from x2 + 3x and getting x. Subtracting −2x adds 2x, so the difference is 3x + 2x = 5x. A sign slip in this line changes the quotient, and the remainder no longer comes to 0.
09

Fencing a Plot That Must Hold a Lawn of Fixed Area Beside a Vegetable Bed: A Vertical and a Slant Asymptote

methodExpress the Length from the Fixed Area, Add Up the Perimeter as a Line Plus a Fraction, Read the Vertical Asymptote from the Fraction and the Slant Asymptote from the Line, Then Clear the Fraction to Solve

A rectangular plot is x meters wide. A vegetable bed 4 meters wide runs along the whole of one long side, and the rest of the plot is a lawn, so the lawn is x − 4 meters wide. The lawn must have an area of 48 square meters, and a fence goes round the whole plot. (a) Find a formula for the length of fence P(x) in meters, and find the vertical asymptote and the slant asymptote of its graph. Say what each one means. (b) There are 40 meters of fence. Find the two possible widths of the plot.

the fence goes round the whole plotlawn 48 m2bedx − 44xlength 48/(x − 4)the lawn is x − 4 wide, so it is 48/(x − 4) longP(x) = 2x + 2 × 48/(x − 4) = 2x + 96/(x − 4)
The lawn is x − 4 meters wide and covers 48 square meters, so the plot is 48x − 4 meters long. The fence goes round the whole plot: P(x) = 2x + 96x − 4.
The lawn is x − 4 meters wide and has an area of 48 square meters, so the length of the lawn, and of the plot, is 48x − 4 meters. The fence goes round a plot that is x meters by 48x − 4 meters: P(x) = 2x + 96x − 4.
step 1 of 6

The fixed area of the lawn ties the length of the plot to its width, so the perimeter becomes a function of the width alone. The function is a line plus a fraction. The fraction is very large when the lawn is only a sliver, which gives a vertical asymptote, and it fades for a very wide plot, which leaves the line as a slant asymptote.

  1. The lawn is x − 4 meters wide and has an area of 48 square meters, so the length of the lawn, and of the plot, is 48x − 4 meters. The fence goes round a plot that is x meters by 48x − 4 meters: P(x) = 2x + 96x − 4.
  2. The bottom x − 4 is zero at x = 4, and the top 96 is never zero, so the vertical asymptote is x = 4. The lawn needs some width, so the domain is x > 4. Close to 4 the fence is very long: P(5) = 10 + 96 = 106.
  3. As one fraction, P(x) = 2x2 − 8x + 96x − 4. The top is one degree higher than the bottom, so there is a slant asymptote, and it is the quotient y = 2x. As x grows, the remainder term 96x − 4 fades: P(100) = 200 + 1 = 201.
  4. (a) P(x) = 2x + 96x − 4. The vertical asymptote is x = 4: a plot only just wider than the bed leaves a sliver of lawn, which must be very long to cover 48 square meters. The slant asymptote is y = 2x: a very wide plot is very short, so its fence is little more than its two long sides, 2x meters.
  5. Put P(x) = 40 and multiply both sides by x − 4: 2x(x − 4) + 96 = 40(x − 4), so 2x2 − 8x + 96 = 40x − 160. Collect the terms and divide by 2: x2 − 24x + 128 = 0.
  6. (b) (x − 8)(x − 16) = 0, so the plot is 8 meters or 16 meters wide. Check: at 8 meters the lawn is 4 by 12 and the plot is 8 by 12, with a perimeter of 40. At 16 meters the lawn is 12 by 4 and the plot is 16 by 4, with a perimeter of 40.

answer(a) P(x) = 2x + 96x − 4. The vertical asymptote is x = 4: a plot only just wider than the bed needs a very long sliver of lawn. The slant asymptote is y = 2x: a very wide plot is short, so its fence is little more than its two long sides; (b) 8 meters or 16 meters

techniqueSlant Asymptotes by Long Division · Zeros and Vertical Asymptotes

examsH2

Common pitfalls

  • Using the width of the plot for the lawn and writing the length as 48x. The bed takes 4 meters of the width, so the lawn is x − 4 meters wide, and the length is 48x − 4. With the slip the vertical asymptote moves to x = 0 and the meaning of the bed is lost.
  • Multiplying only the fraction by x − 4 when clearing it. Every term on both sides must be multiplied: 2x becomes 2x(x − 4) and 40 becomes 40(x − 4). Leaving 2x alone gives a linear equation and only one width.
10

The Base of a Box from Its Volume and Its Height: Long Division with a Remainder of Zero as the Check

methodDivide the Volume by the Height Term by Term with Long Division, Expect a Remainder of Zero Because the Height Is a Factor of the Volume, Then Factorize the Quotient into the Two Sides of the Base

A packaging firm makes a family of boxes from one design. For a size setting x, the box has a volume of x3 + 6x2 + 11x + 6 cubic centimeters and a height of x + 1 centimeters. (a) Use long division to find the area of the base as a polynomial in x, and state the remainder. (b) Factorize the area of the base to find the two sides of the base. Find the three dimensions, the area of the base and the volume of the box when x = 4.

x + 1x3+ 6x2+ 11x+ 6x2x3+ x25x2+ 11xx3divided by x is x2, and x2(x + 1) = x3+ x2subtract: 5x2, then bring down 11x
The area of the base is the volume divided by the height. x3 ÷ x = x2, and x2(x + 1) = x3 + x2. Subtracting leaves 5x2, and 11x comes down.
Divide the leading terms: x3 ÷ x = x2. Multiply back: x2(x + 1) = x3 + x2. Subtract from x3 + 6x2 to get 5x2, and bring down 11x: the working line is 5x2 + 11x.
step 1 of 6

The volume of a box is the area of its base multiplied by its height, so the area of the base is the volume divided by the height. Dividing one polynomial by another follows the same four moves as long division with numbers: divide the leading terms, multiply back, subtract, and bring down the next term. Because the height is a factor of the volume, the remainder must come to zero.

  1. Divide the leading terms: x3 ÷ x = x2. Multiply back: x2(x + 1) = x3 + x2. Subtract from x3 + 6x2 to get 5x2, and bring down 11x: the working line is 5x2 + 11x.
  2. Divide again: 5x2 ÷ x = 5x. Multiply back: 5x(x + 1) = 5x2 + 5x. Subtract to get 6x, and bring down 6: the working line is 6x + 6.
  3. Divide once more: 6x ÷ x = 6. Multiply back: 6(x + 1) = 6x + 6. Subtract: the remainder is 0.
  4. (a) The area of the base is x2 + 5x + 6 square centimeters, and the remainder is 0, as it must be when the height is a factor of the volume. Check by multiplying back: (x + 1)(x2 + 5x + 6) = x3 + 5x2 + 6x + x2 + 5x + 6 = x3 + 6x2 + 11x + 6.
  5. Factorize the area of the base: x2 + 5x + 6 = (x + 2)(x + 3), because 2 + 3 = 5 and 2 × 3 = 6. The sides of the base are x + 2 and x + 3 centimeters.
  6. (b) When x = 4 the base is 6 cm by 7 cm, with an area of 42 square centimeters, and the height is 5 cm. The volume is 42 × 5 = 210 cubic centimeters. Check with the formula: 64 + 96 + 44 + 6 = 210.

answer(a) The area of the base is x2 + 5x + 6 square centimeters, with remainder 0; (b) the sides of the base are x + 2 and x + 3 centimeters, and when x = 4 the box is 6 cm by 7 cm by 5 cm, with a base of 42 square centimeters and a volume of 210 cubic centimeters

techniqueLong Division of Polynomials

examsO-Level · SAT

Common pitfalls

  • Subtracting only the first term of each product. After multiplying back, the whole of x3 + x2 is subtracted, so 6x2 − x2 = 5x2 is carried forward. Carrying 6x2 instead gives a quotient of x2 + 6x + 5 and a remainder of 1, which warns that something has gone wrong.
  • Dividing each term of the volume by x alone and ignoring the + 1 in the height. The divisor is the whole of x + 1. Each new term of the quotient comes from the leading terms only, but it is multiplied by both terms of the divisor before subtracting.
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