Building Function Models - Math 2

1. A radioactive substance has a half-life of 8 years, meaning that every 8 years, exactly half of the remaining substance decays. Starting with 100 grams, a scientist wants to model the amount remaining after t years. Which function correctly models this situation and what key property distinguishes it from other function types?
   Answer: Exponential function $A(t) = 100(0.5)^{t/8}$, because the amount decreases by a constant percentage over equal intervals When you encounter radioactive decay problems, you're dealing with exponential functions because the substance decreases by a constant percentage (not a fixed amount) over equal time intervals. Let's build the correct model step by step. You start with 100 grams, and every 8 years, half remains. After 8 years: 50 grams. After 16 years: 25 grams. After 24 years: 12.5 grams. Notice that in each 8-year period, the amount is multiplied by 0.5. For exponential decay, the general form is $A(t) = A_0 \cdot b^{t/h}$, where $A_0$ is the initial amount, $b$ is the decay factor, and $h$ is the time for one complete cycle. Here: $A(t) = 100(0.5)^{t/8}$. When $t = 8$, you get $100(0.5)^1 = 50$ grams, confirming our model. Option A is wrong because linear functions show constant amount decrease, but radioactive decay involves constant percentage decrease. The calculation is also incorrect—after 8 years you'd have 50 grams, not 100 - 6.25(8) = 50 grams (coincidentally right for one point, but wrong pattern). Option B incorrectly suggests quadratic decay. While decay rates do change over time, they follow exponential, not quadratic patterns. Option C has the wrong exponent structure. The form $(0.5)^{8t}$ would mean the substance halves every 1/8 year, not every 8 years. Remember: exponential decay problems always involve constant percentage changes over equal intervals. Look for the pattern where the decay factor is raised to $t$ divided by the period length.
2. If the marketing team's assumption is correct, which function would best model the total number of users as a function of time in months, and what characteristic of the data supports this choice?
   Answer: Exponential function, because the monthly user gains double each month, indicating exponential total growth When analyzing growth patterns, you need to distinguish between the rate of change and the type of function that models the total. The key is examining what's happening to the monthly gains: 1000, 2000, 4000 users. Notice that each month's user gain doubles the previous month's gain: $2000 = 2 \times 1000$ and $4000 = 2 \times 2000$. This doubling pattern in the monthly additions is the hallmark of exponential growth in the total. When the incremental changes themselves grow exponentially, the cumulative total follows an exponential function. Let's trace the total users: Month 1: 1000, Month 2: 3000, Month 3: 7000. The pattern $1000, 3000, 7000$ can be expressed as $1000(2^1 - 1), 1000(2^2 - 1), 1000(2^3 - 1)$, confirming exponential behavior. Answer choice A incorrectly confuses "increasing rate" with linear growth—linear functions have constant rates of change. Choice B misidentifies the pattern as quadratic; while the gains do increase, they're doubling (exponential), not following a quadratic sequence like 1000, 4000, 9000. Choice C gives the right function type but wrong reasoning—not all social media growth is exponential, and network effects alone don't guarantee this pattern. Choice D correctly identifies both the exponential function and the supporting evidence: the monthly gains double each month. Study tip: When identifying growth patterns, always examine the differences between consecutive terms first. Constant differences suggest linear, constant ratios suggest exponential, and increasing differences that aren't in constant ratio suggest quadratic or polynomial growth.
3. Which function model best describes this user engagement pattern, and what key business assumption does this model make?
   Answer: Linear model $U(d) = 240d$ assuming steady user acquisition rate of 240 users per day without market saturation The users increase by exactly 2,400 every 10 days, giving a constant rate of 240 users per day. This linear pattern $U(d) = 240d$ assumes the company can maintain steady user acquisition indefinitely without hitting market limits. Choice A incorrectly identifies this as exponential (users are adding, not multiplying). Choice C incorrectly identifies quadratic growth. Choice D incorrectly applies logarithmic growth to this linear pattern.
4. What function type best fits this revenue pattern, and what assumption makes this model most questionable for long-term forecasting?
   Answer: Linear model $R(q) = 120q + 280$ with the assumption that market conditions and competitive landscape remain constant indefinitely The revenue increases by exactly $120k each quarter, showing a constant rate of change characteristic of linear growth. The model $R(q) = 120q + 280$ fits perfectly. The main assumption is that this steady growth rate continues indefinitely, ignoring market saturation, competition, and economic cycles. Choice B incorrectly identifies exponential growth. Choice C incorrectly identifies quadratic growth. Choice D incorrectly applies logarithmic growth to constant increases.
5. A manufacturing company finds that production cost per unit decreases with volume. At 100 units: $50 per unit, at 200 units: $35 per unit, at 300 units: $30 per unit, at 400 units: $27.50 per unit. Which model type fits this cost behavior?
   Answer: Inverse/rational model $C(x) = 20 + \frac{3000}{x}$ representing fixed costs spread over more units plus variable costs The cost structure shows fixed costs distributed over increasing volume plus variable costs. Testing $C(x) = 20 + \frac{3000}{x}$: at 100 units: $20 + $30 = $50 ✓, at 200 units: $20 + $15 = $35 ✓. This rational function models economies of scale where fixed costs are spread over more units. Choice A incorrectly assumes linear cost reduction. Choice B incorrectly models exponential decay. Choice C incorrectly suggests costs will eventually increase quadratically.
6. A population of rabbits in a controlled environment shows the following counts: Month 0: 50 rabbits, Month 1: 65 rabbits, Month 2: 84.5 rabbits, Month 3: 109.85 rabbits. What model best represents this population growth and what does it predict for Month 6?
   Answer: Exponential growth $P(t) = 50(1.3)^t$ predicting approximately 214 rabbits, assuming constant proportional growth rate each month Each month the population multiplies by 1.3: 65/50 = 1.3, 84.5/65 = 1.3, 109.85/84.5 = 1.3. This constant growth factor indicates exponential growth with $P(t) = 50(1.3)^t$. For Month 6: $50(1.3)^6 \approx 214$ rabbits. Choice A incorrectly assumes linear growth. Choice C incorrectly assumes quadratic growth. Choice D uses an incorrect exponential model with the wrong growth factor.
7. The startup wants to model their revenue for the entire time period described. Which approach would be most appropriate for capturing all three distinct phases of growth?
   Answer: Piecewise function combining linear, exponential, and quadratic portions to match each distinct growth phase When you encounter a problem describing multiple distinct phases of behavior, you need to think about whether a single function can capture all the complexity or if you need different functions for different phases. The passage describes three very different growth patterns: steady linear growth (months 1-6), rapid exponential acceleration (months 7-12), and continued growth at a decreasing rate (after month 12). Each phase has fundamentally different mathematical characteristics that require different types of functions to model accurately. Option A is correct because a piecewise function allows you to use the most appropriate mathematical model for each distinct phase. You can use a linear function for the steady growth period, an exponential function for the viral marketing acceleration, and a logarithmic or power function for the saturated market phase. This approach captures the real behavior described in each time period. Option B fails because a single exponential function cannot model the initial linear phase or the eventual slowdown - exponential functions maintain constant percentage growth rates. Option C is incorrect because while quadratic functions can show changing rates of growth, they cannot capture the distinct behavioral differences between linear, exponential, and saturation phases described. Option D misses the complexity entirely - a linear function shows constant growth rate and cannot model the acceleration or deceleration phases. Remember: when you see problems describing multiple distinct behavioral phases, look for piecewise functions. Single functions work best when the underlying process remains consistent, but real-world scenarios often involve phase changes that require different mathematical models.
8. A rocket's altitude during the first 10 seconds of flight is recorded: $t = 2$s: 156 ft, $t = 4$s: 384 ft, $t = 6$s: 684 ft, $t = 8$s: 1056 ft. Which function family models this motion and why?
   Answer: Quadratic function because rocket motion involves constant acceleration, and the second differences in altitude are approximately constant Rocket motion with constant acceleration follows $h(t) = \frac{1}{2}at^2 + v_0t + h_0$, a quadratic function. Checking second differences: first differences are 228, 300, 372 (increasing by 72 each time), confirming constant acceleration. The pattern fits $h(t) = 18t^2 + 12t$ approximately. Choice A incorrectly assumes linear motion despite accelerating pattern. Choice B incorrectly applies exponential growth to physics motion. Choice D unnecessarily assumes variable acceleration without evidence.
9. For the first 4 hours, which function type best models the engagement, and how should the model be modified to account for the plateau effect?
   Answer: Linear growth $E(t) = 45t + 5$ for hours 1-4, then piecewise function with constant $E(t) = K$ after Hour 6 The engagement increases by exactly 45 interactions each hour (95-50=45, 140-95=45, 185-140=45), indicating linear growth $E(t) = 45t + 5$. To model the plateau, a piecewise function works best: linear growth initially, then constant value after Hour 6. Choice A incorrectly identifies exponential growth. Choice B correctly identifies linear growth but unnecessarily complicates the plateau modeling. Choice C incorrectly identifies quadratic growth and suggests the parabola naturally plateaus (it doesn't).
10. A bank offers two investment options: Option A pays 3% simple interest annually, while Option B pays 2.8% interest compounded annually. An investor wants to model the value of a $10,000 investment over time for each option. What function types should be used, and which investment performs better after 20 years?
    Answer: Option A linear, Option B exponential; Option A: $V(t) = 10000 + 300t$, Option B: $V(t) = 10000(1.028)^t$; Option B is better When you encounter investment problems, the key is identifying whether interest is simple or compound, as this determines the function type you'll need. Simple interest grows linearly (adds the same amount each year), while compound interest grows exponentially (multiplies by the same factor each year). Option A uses simple interest: 3% of the original $10,000 = $300 added each year. This creates the linear function $V(t) = 10000 + 300t$. Option B uses compound interest: the investment is multiplied by 1.028 each year, creating the exponential function $V(t) = 10000(1.028)^t$. To determine which performs better after 20 years: - Option A: $V(20) = 10000 + 300(20) = 16,000$ - Option B: $V(20) = 10000(1.028)^{20} ≈ 10000(1.756) ≈ 17,560$ Option B performs better despite the lower interest rate because compound interest allows earnings to generate their own earnings over time. Choice B incorrectly concludes Option A is better. Choice C treats both as linear functions, which misrepresents compound interest—compound interest can never be linear because it involves multiplying the growing balance by the interest factor each period. Choice D incorrectly models simple interest as exponential by using $V(t) = 10000(1.03)^t$, which would mean the original investment multiplies by 1.03 each year rather than adding a fixed $300. Remember: Simple interest = linear growth (addition), compound interest = exponential growth (multiplication). The compound effect often overcomes initially lower rates over longer time periods.
11. Which function type best models this user growth pattern, and what is the primary limitation of extending this model beyond the observed data?
    Answer: Exponential growth with rate $r = 1.5$; the limitation is that exponential growth is unrealistic for long-term user adoption due to market saturation The users increase by a factor of 1.5 each month (750/500 = 1.5, 1125/750 = 1.5), indicating exponential growth with $f(t) = 500(1.5)^{t-1}$. However, exponential growth models become unrealistic over time due to market saturation and resource constraints. Choice B incorrectly identifies this as linear growth. Choice C misidentifies the pattern as quadratic and incorrectly states that quadratic functions always decrease. Choice D incorrectly identifies this as exponential decay.
12. What function type models this account balance, and what is the annual interest rate?
    Answer: Exponential function $B(t) = 5000(1.05)^t$ representing 5% annual compound interest, since $(1.05)^2 \approx 1.1025$ Compound interest follows exponential growth $B(t) = P(1+r)^t$. The balance grows by factor 1.1025 every 2 years ($5,512.50/$5,000 = 1.1025), so annually it grows by $\sqrt{1.1025} = 1.05$, indicating 5% annual interest. Choice A incorrectly models compound interest as linear. Choice C incorrectly assumes quadratic growth. Choice D correctly identifies the 2-year factor but incorrectly suggests biennial rather than annual compounding.
13. Based on this data, what type of function best models the profit, and what does the model predict about selling 100 specialty drinks?
    Answer: Linear function $P(x) = 5x + 80$; predicting $580 profit, assuming fixed costs and constant marginal profit per drink The profit increases by exactly $100 for every 20 additional drinks: from 20 to 40 drinks (+$100), from 40 to 60 drinks (+$100). This constant rate of change ($5 per drink) indicates a linear relationship. The function $P(x) = 5x + 80$ fits all three points. Choice B assumes diminishing returns without evidence. Choice C incorrectly models this as exponential growth. Choice D incorrectly applies a square root model.
14. A water tank drains through a hole at the bottom. The drainage rate depends on water pressure (height). Data shows: Hour 0: 1000 gallons, Hour 1: 900 gallons, Hour 2: 810 gallons, Hour 3: 729 gallons. What model fits this drainage pattern?
    Answer: Exponential decay $V(t) = 1000(0.9)^t$ representing drainage rate proportional to current water volume and pressure The volume multiplies by 0.9 each hour: 900/1000 = 0.9, 810/900 = 0.9, 729/810 = 0.9. This exponential decay reflects that drainage rate is proportional to water pressure (which depends on height/volume). As volume decreases, pressure decreases, so drainage rate slows. Choice A incorrectly assumes constant drainage rate. Choice C incorrectly suggests accelerating drainage. Choice D incorrectly applies Torricelli's law (which involves height, not volume directly).
15. A pharmaceutical company studies drug concentration in blood over time. Measurements show: Hour 0: 100 mg/L, Hour 2: 60 mg/L, Hour 4: 36 mg/L, Hour 6: 21.6 mg/L. What model best represents this and what is the half-life?
    Answer: Exponential decay $C(t) = 100(0.6)^{t/2}$ with half-life of approximately 1.44 hours, assuming first-order elimination kinetics Drug elimination typically follows first-order kinetics (exponential decay). The concentration multiplies by 0.6 every 2 hours: 60/100 = 0.6, 36/60 = 0.6, 21.6/36 = 0.6. So $C(t) = 100(0.6)^{t/2}$. Half-life occurs when $(0.6)^{t/2} = 0.5$, solving gives $t \approx 1.44$ hours. Choice A incorrectly assumes linear elimination. Choice C incorrectly assumes quadratic decay. Choice D uses wrong decay factor and incorrect half-life calculation.
16. A ball is thrown upward from a building. The height data shows: at $t = 1$ second, height is 84 feet; at $t = 2$ seconds, height is 100 feet; at $t = 3$ seconds, height is 84 feet. Which model best represents this situation and why?
    Answer: Quadratic model $h(t) = -16t^2 + 84t + 16$ because projectile motion follows parabolic paths due to constant acceleration Projectile motion is governed by gravity, which creates constant downward acceleration, resulting in a quadratic (parabolic) path. The data points (1,84), (2,100), (3,84) form a parabola opening downward. Choice A incorrectly assumes linear motion despite the curved trajectory. Choice B misapplies exponential functions to projectile motion. Choice D incorrectly suggests piecewise linear motion, but gravity acts continuously, creating smooth parabolic motion.
17. A coffee shop notices that their daily profit follows a predictable pattern based on the price they charge per cup. When they charge too little, they don't make enough per cup despite high volume. When they charge too much, they lose too many customers. The data shows profit starts at $0 when the price is$2, reaches a maximum at some intermediate price, and returns to $0 when the price is$8. Which function model best represents this relationship?
    Answer: Quadratic function $P(x) = a(x-2)(x-8)$ where $a < 0$, because profit has two zeros and one maximum The scenario describes profit that starts at zero, increases to a maximum, then decreases back to zero. This is characteristic of a quadratic function with a negative leading coefficient. The factored form P(x) = a(x-2)(x-8) where a < 0 gives zeros at x = 2 and x = 8 (the break-even prices) and a maximum between them. Choice A is wrong because linear functions don't have maximums. Choices B and D are wrong because exponential functions don't return to zero or have interior maximums.
18. A bacterial culture starts with 1,000 bacteria. After 2 hours, there are 1,440 bacteria. After 4 hours, there are 2,074 bacteria. Which function family best models this growth, and what is the growth factor per hour?
    Answer: Exponential growth with factor $r = 1.44$ per 2-hour period, equivalent to $r \approx 1.2$ per hour From 1,000 to 1,440 bacteria in 2 hours gives a factor of 1.44 per 2-hour period. From 1,440 to 2,074 in the next 2 hours gives approximately 1.44 again (2074/1440 ≈ 1.44). This indicates exponential growth with $r = \sqrt{1.44} \approx 1.2$ per hour. Choice A uses the correct hourly rate but incorrect reasoning. Choice B incorrectly assumes linear growth. Choice C incorrectly assumes quadratic growth without checking second differences.
19. A farmer wants to build a rectangular pen against the side of her barn. She has 200 feet of fencing and will use the barn wall as one side, so she only needs to fence three sides. She wants to maximize the area enclosed. What type of function should she use to model area as a function of the pen's width, and what is the domain restriction?
    Answer: Quadratic function $A(w) = w(200-2w)$, with domain $0 < w < 100$, because area depends on two varying dimensions Let w = width (perpendicular to barn). The farmer needs fencing for: two widths + one length = 2w + l = 200, so l = 200 - 2w. Area = w × l = w(200 - 2w), which is quadratic. Domain: w > 0 and l > 0, so 200 - 2w > 0, giving w < 100. Choice A has the wrong function form. Choice C uses the wrong constraint equation. Choice D incorrectly calls a quadratic function linear.

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