0.3 Representations, units, and modeling investigations - Courses of Computing4All
Precalculus for High School Students 0.3 Representations, units, and modeling investigations
Precalculus for High School Students
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Learning goals: Connect a situation, table, graph, and equation; interpret units; build and critique a model.

Four ways to see the same relationship

A climbing gym charges a joining fee of $18 and $7 for each visit. Describe the total cost after \(v\) visits using words, a table, a graph, and an equation. The equation is \(C(v)=18+7v\). A table includes \((0,18),(1,25),(2,32)\). On a graph with visits horizontally and dollars vertically, the points rise seven dollars per visit. The intercept represents the joining fee, not a free visit.

Because visits are counted, the contextual domain consists of nonnegative integers. A continuous line is a useful extension of the model, but its point at half a visit is not an actual purchase. Distinguishing an algebraic domain from a contextual domain will matter throughout the course.

Words, an equation, a table, and a graph tell one storyWords, an equation, a table, and a graph tell one storyWORDSPay $18 to join, then $7 per visit.EQUATION: C(v) = 18 + 7vTABLEVisits vCost C ($)0181252323390123018253239vC ($)GRAPHCounted visits: v = 0, 1, 2, …Filled points represent possible purchases; the dashed line is an extension.The joining fee gives (0, 18). Each whole visit adds $7.
Read the same values in the situation, formula, table, and graph. The graph uses separate points because the contextual domain is the nonnegative integers. A point at half a visit would belong only to the continuous extension.

A disciplined modeling cycle

State the question, define variables and units, identify assumptions, choose a relationship, calculate, interpret, and check the model against the situation. Assumptions are part of the answer. The gym equation assumes no tax, discounts, or changes in price. If a promotion waives the fee, the intercept must change. A correct calculation using an unsuitable model is not a correct solution to the original problem.

Units are mathematical information

For constant motion, \(d=rt\). If rate is in miles per hour, time must be in hours to obtain miles. A trip lasting 90 minutes at 48 miles per hour covers \[d=48\frac{\text{mi}}{\text{h}}\cdot\frac{90}{60}\text{h}=72\text{ mi}.\] Multiplying 48 by 90 without converting would mix units. Conversion factors equal one: \(60\text{ min}/1\text{ h}\) changes units without changing the physical quantity.

Keep units attached: minutes must become hoursKeep units attached: minutes must become hoursTIME90 minCONVERSION1 h / 60 minSAME TIME1.5 h×=90 min × (1 h / 60 min) = 1.5 hThe minute units cancel; the hour unit remains.48 mi/h × 1.5 h = 72 miThe conversion factor equals 1 because 1 hour = 60 minutes.Rate × time gives distance only when the time units match.
Follow the units as well as the numbers. Dividing 90 minutes by 60 minutes per hour gives 1.5 hours; multiplying by 48 miles per hour gives 72 miles. Writing units helps detect the incorrect calculation 48 × 90.

Compare plans and explain a threshold

Plan A costs \(A(v)=18+7v\); Plan B costs \(B(v)=42+4v\). Set the costs equal: \(18+7v=42+4v\), so \(3v=24\) and \(v=8\). Both cost $74 at eight visits. The difference \(A(v)-B(v)=3v-24\) explains the entire comparison: Plan A is cheaper below eight visits, while Plan B is cheaper above eight. Checking one value on each side confirms the interpretation.

The intersection separates which plan is cheaperThe intersection separates which plan is cheaper048120184274100visitscost ($)(8, 74)A(v) = 18 + 7vB(v) = 42 + 4vAt 0–7 visits, Plan A costs less. At 8 visits, both cost $74.At 9 or more visits, Plan B costs less. Compare the vertical heights.
The lines intersect at eight visits and $74. Only whole-number visit counts represent actual purchases; the lines help reveal the trend between the marked points. Solving \(A(v)-B(v)=3v-24\) explains the same threshold algebraically.

Investigate, critique, and extend

Suppose a table contains \((1,3),(2,5),(3,7)\). The linear rule \(y=2x+1\) fits, but finitely many points do not uniquely determine a rule unless a model family is specified. A graphing tool can suggest structure, reveal an arithmetic error, or test a conjecture. It cannot justify an unspoken assumption about how the situation behaves between or beyond measured points.

Three matching data points do not determine one ruleThree matching data points do not determine one rule01234−2035712xyBoth curves pass through (1, 3), (2, 5), and (3, 7).Blue: y = 2x + 1Green: y = 2x + 1 + ½(x − 1)(x − 2)(x − 3)The added product is zero at x = 1, 2, and 3, but not everywhere.
Each red point lies on both graphs. At x = 4 the predictions differ: 9 for the line and 12 for the curve. To select the linear rule, state the assumption that the relationship is linear; matching a few data points alone is insufficient.

For a partner investigation, design two payment plans that agree at exactly six visits. Explain why your equations meet that requirement, identify when each is cheaper, and discuss whether a customer can make a fractional visit. Exchange plans and critique each other’s units, restrictions, and conclusions. A strong response combines computation with a clear account of what the result means.

Practice: attempt before reading the solutions

Show your reasoning, state restrictions, and check your result. For explanation questions, use complete mathematical sentences.

  1. A service charges $12 plus $3 per use. Write the cost for \(n\) uses and find the cost for nine uses.
  2. What does the coefficient 3 mean in \(C(n)=12+3n\)?
  3. Find the distance traveled in 40 minutes at 54 miles per hour.
  4. Compare \(A(n)=10+5n\) and \(B(n)=28+2n\). When are they equal?
  5. A rectangle has perimeter 30 cm. Express area using width \(w\) and state a physical domain.
  6. A model predicts 12.4 buses for a trip. What should a planner do?
  7. The table \((0,2),(1,6),(2,10)\) is assumed linear. Find its rule.
  8. A student uses a two-week plant-growth model to predict its height after 100 years. Critique the prediction.
Worked solutions and reasoning

  1. Separate the fixed cost from the cost that depends on use. The $12 fee is paid once, while n uses cost \(3n\) dollars. Adding them gives \(C(n)=12+3n\), with \(n=0,1,2,\ldots\) for counted uses.

    For nine uses, \(C(9)=12+3(9)=12+27=39\) dollars. At zero uses the model gives $12, confirming that the fee is fixed rather than a per-use charge. The model assumes the quoted charges do not change and that no additional fees or taxes are added.

  2. The coefficient 3 multiplies the number of uses, so its units are dollars per use. Increasing n by one changes the cost by \[C(n+1)-C(n)=[12+3(n+1)]-[12+3n]=3.\]

    Thus every additional use adds $3 to the total. On a graph, this is the slope: the output rises three dollars for an increase of one use. The number 12 plays a different role; it is the fixed starting cost, in dollars, even when the number of uses is zero.

  3. Use distance = rate × time, but first match the time units. Because the rate is 54 miles per hour, convert 40 minutes to hours: \[40\text{ min}\cdot\frac{1\text{ h}}{60\text{ min}}=\frac23\text{ h}.\]

    Then \(d=(54\text{ mi/h})(2/3\text{ h})=36\text{ mi}\). The hour units cancel, leaving miles. As a reasonableness check, forty minutes is two-thirds of an hour, so the distance should be two-thirds of 54 miles, not greater than 54 miles.

  4. Equal costs occur where the formulas have the same output for the same n. Set \(10+5n=28+2n\). Subtract 2n and then 10 to get \(3n=18\), so \(n=6\).

    Check both costs: \(A(6)=10+30=40\) and \(B(6)=28+12=40\). For the full comparison, \(A(n)-B(n)=3n-18=3(n-6)\). This difference is negative below six uses, so A is cheaper there, and positive above six, so B is cheaper there. Use whole-number n if uses are counted.

  5. Let the rectangle’s length be \(\ell\). Perimeter counts two widths and two lengths, so \(2w+2\ell=30\). Divide by two and isolate the length: \(w+\ell=15\), hence \(\ell=15-w\).

    Area is width × length, giving \(A(w)=w(15-w)=15w-w^2\) square centimeters. A nondegenerate rectangle needs both sides positive: \(w\gt0\) and \(15-w\gt0\). Together these give \(0\lt w\lt15\). The endpoints produce a zero side and zero area, so they are excluded from the physical domain.

  6. The model’s output is a required amount of bus capacity, but buses must be hired in whole numbers. If the buses are identical and twelve do not provide enough seats, the planner must round up: \(\lceil12.4\rceil=13\) buses.

    Ordinary rounding to the nearest integer would give twelve and could leave passengers without seats. Check the actual passenger count, seats available per bus, and any operating constraints before booking. The key is that “enough capacity” is a minimum requirement; the context determines the direction of rounding.

  7. For a linear rule, write \(y=mx+b\). Compute the slope from consecutive points: \((6-2)/(1-0)=4\) and \((10-6)/(2-1)=4\). The matching rates are consistent with the stated linear-model assumption.

    The point (0, 2) gives the intercept directly, so b = 2. Therefore \(y=4x+2\). Substitution returns 2, 6, and 10 at inputs 0, 1, and 2. Without the assumption of linearity, other curves could also fit these three points.

  8. This prediction extrapolates far beyond the two-week period used to build the model. Agreement with observed data over a short interval does not show that the same relationship remains valid for a century.

    A plant may grow rapidly at first, slow as it matures, face changing light or water, and eventually die. Those changes can invalidate a constant growth-rate assumption. A better response is to identify the model’s assumptions and supported time range, collect longer-term evidence, and use a biologically appropriate model before making a distant prediction.

Reflection

Choose one problem you initially found difficult. Explain the key idea, identify a possible mistake, and write a new problem that uses the same idea. Solve your new problem and check it independently.

Unit investigation: When evidence becomes an argument

For integers n, explore \(n^2+n\). Compute it for four values, including a negative integer. Conjecture whether the result is always even, then prove or refute your claim. A partner claims that \(n^2+n+1\) is always prime. Find a counterexample and explain why your proof for the first claim says nothing about primality. Finish by writing the domain of each claim explicitly.

Worked discussion — open after attempting

For example, n=−2,0,1,2 give 2,0,2,6. The product \(n(n+1)\) is even because consecutive integers have opposite parity, so one factor is even. This covers every integer, including negative ones. At n=4, \(n^2+n+1=21=3\cdot7\), disproving the prime claim. A parity argument establishes divisibility by two for one expression; it does not establish primality of a different expression.

Check your explanation: State assumptions, choose appropriate representations, show the reasoning, and interpret the result with units or restrictions.

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