Constructing Functions from Context - Math 1

1. Which function $C(m)$ represents the total cost for a trip of $m$ miles?
   Answer: $C(m) = 2m + 8$ Let $C(m) = b + rm$ where $b$ is base fare and $r$ is rate per mile. From the given information: $b + 5r = 18$ and $b + 12r = 32$. Subtracting: $7r = 14$, so $r = 2$. Substituting back: $b + 5(2) = 18$, so $b = 8$. Therefore $C(m) = 8 + 2m = 2m + 8$. Choice B gives costs of $18 and $39 for the given trips. Choice C uses incorrect decimal approximations. Choice D reverses the base fare and rate per mile.
2. A phone plan charges a monthly fee plus a rate per gigabyte of data used. Sarah's bill was $45 for using 3 GB and $65 for using 8 GB. Which function represents the total cost $C(g)$ for using $g$ gigabytes of data?
   Answer: $C(g) = 4g + 33$ The rate per GB is (65 - 45)/(8 - 3) = 20/5 = $4 per GB. Using point-slope form with (3, 45): C(g) - 45 = 4(g - 3), so C(g) = 4g + 33. We can verify: C(3) = 4(3) + 33 = 45 ✓ and C(8) = 4(8) + 33 = 65 ✓. Choice B uses the cost at 3 GB as the y-intercept. Choice C uses the difference in costs (20) as the slope. Choice D reverses the calculation.
3. A rectangular garden has a perimeter of 80 feet. If the width is $w$ feet, which function represents the area $A(w)$ of the garden in terms of its width?
   Answer: $A(w) = w(40 - w)$ With perimeter 80, we have 2w + 2l = 80, so l = 40 - w. The area is A = wl = w(40 - w). Choice A uses the full perimeter instead of half. Choice C includes an extra factor of 2. Choice D uses 80 - 2w, which would give l = 80 - 2w, making the perimeter 2w + 2(80 - 2w) = 160, not 80.
4. A population of bacteria doubles every 4 hours. If there are initially 250 bacteria, which function represents the population $P(t)$ after $t$ hours?
   Answer: $P(t) = 250 \cdot 2^{t/4}$ When you encounter exponential growth problems, you're dealing with situations where a quantity multiplies by a constant factor at regular intervals. The key is translating the given information into the correct exponential function form: $P(t) = P_0 \cdot b^{t/d}$, where $P_0$ is the initial amount, $b$ is the growth factor, and $d$ is the time period for one complete cycle. Here, the bacteria population doubles (multiplies by 2) every 4 hours, starting with 250 bacteria. Since doubling means the growth factor is 2, and this happens every 4 hours, you need $P(t) = 250 \cdot 2^{t/4}$. The exponent $t/4$ ensures that when $t = 4$, you get $2^1 = 2$, doubling the population exactly once. Choice A ($250 \cdot 4^{t/2}$) incorrectly uses 4 as the base instead of 2, and has the wrong time scaling. Choice B ($250 \cdot 2^{4t}$) has the reciprocal time relationship—this would mean the population doubles every $1/4$ hour instead of every 4 hours, creating explosive growth. Choice C ($250 + 2t$) represents linear growth, not exponential growth, where the population would only increase by 2 bacteria per hour. For exponential growth problems, always identify three things: the initial value, the growth factor, and the time period. Then construct your function as $\text{initial} \cdot \text{factor}^{\text{time}/\text{period}}$. Double-check by substituting the given time period to verify you get exactly one growth cycle.
5. A population of bacteria doubles every 3 hours. If there are initially 200 bacteria, which function $P(t)$ models the population after $t$ hours?
   Answer: $P(t) = 200 \cdot 2^{t/3}$ This question tests exponential growth models, where a quantity changes by a constant factor over regular time intervals. When you see "doubles every [time period]," you're dealing with exponential growth with base 2. The general form for exponential growth is $P(t) = P_0 \cdot b^{t/k}$, where $P_0$ is the initial amount, $b$ is the growth factor, and $k$ is the time period for one complete growth cycle. Here, the bacteria population starts at 200, doubles (factor of 2), and completes this doubling every 3 hours. So we need $P(t) = 200 \cdot 2^{t/3}$. Let's verify: at $t = 0$, $P(0) = 200 \cdot 2^0 = 200$. At $t = 3$, $P(3) = 200 \cdot 2^{3/3} = 200 \cdot 2^1 = 400$. The population correctly doubles from 200 to 400 after 3 hours. Choice A ($200 + 2^{t/3}$) represents linear growth plus an exponential term, not pure exponential growth. Choice B ($200 \cdot 2^{3t}$) has the exponent backwards—this would mean the population doubles every $\frac{1}{3}$ hour instead of every 3 hours, causing explosive growth. Choice C ($200 \cdot 3^{t/2}$) uses the wrong base (3 instead of 2) and wrong time period (2 instead of 3). The correct answer is D. Study tip: For exponential growth problems, always check your function at $t = 0$ (initial value) and at one complete growth period. This quick verification catches most common mistakes with the base and exponent structure.
6. A spring's length depends on the weight attached to it. With no weight, the spring is 8 inches long. Each pound of weight stretches the spring an additional 0.5 inches, but the spring cannot stretch beyond 15 inches total. Which function $L(w)$ models the spring's length for weight $w$ pounds?
   Answer: $L(w) = \begin{cases} 8 + 0.5w & \text{if } 0 \leq w \leq 14 \\ 15 & \text{if } w > 14 \end{cases}$ When you encounter a piecewise function modeling a real-world constraint, you need to identify the natural behavior and where limitations kick in. This spring follows a linear relationship until it hits a physical maximum. The spring starts at 8 inches with no weight, then stretches 0.5 inches per pound: $L = 8 + 0.5w$. However, it can't exceed 15 inches total length. To find when this limit is reached, solve: $8 + 0.5w = 15$, which gives $w = 14$ pounds. So the spring follows the linear formula $8 + 0.5w$ when $0 \leq w \leq 14$, and stays constant at 15 inches when $w > 14$. This matches choice C perfectly. Choice A has the right idea but poor execution—the condition "if $8 + 0.5w > 15$" in the second piece should specify the weight range, not repeat the length formula. Choice B incorrectly uses $w \leq 15$ and $w > 15$ as the boundary conditions. The weight limit isn't 15 pounds; that's the maximum length in inches. The actual weight limit is 14 pounds. Choice D uses a minimum function, but $\min(8 + 0.5w, 15)$ would give the smaller value, which doesn't model this situation correctly—we want the spring length, not the minimum of length and 15. Study tip: For piecewise functions with physical constraints, always solve for the exact point where the limitation kicks in. Don't assume the constraint value appears directly as a boundary in the domain.
7. A ball is thrown upward from a building. Its height in feet after $t$ seconds is given by $h(t) = -16t^2 + 64t + 80$. A second ball is thrown from ground level 2 seconds later with initial velocity 48 ft/s. Which function $d(t)$ represents the distance between the two balls for $t \geq 2$?
   Answer: $d(t) = |-16t^2 + 64t + 80 - (-16(t-2)^2 + 48(t-2))|$ The first ball's height is $h_1(t) = -16t^2 + 64t + 80$. The second ball starts at $t = 2$ from ground level with velocity 48 ft/s, so its height is $h_2(t) = -16(t-2)^2 + 48(t-2)$ for $t \geq 2$. The distance between them is $d(t) = |h_1(t) - h_2(t)|$. Choice B ignores the 2-second delay. Choice C adds the heights instead of finding their difference. Choice D has an error in the first ball's velocity coefficient.
8. A company's profit depends on the number of items sold. The profit is $800 when 20 items are sold, and the profit decreases by $15 for each additional item sold beyond 20 due to bulk discounts. Which function represents the profit $P(x)$ when $x$ items are sold, where $x ≥ 20$?
   Answer: $P(x) = -15(x - 20) + 800$ The profit decreases by $15 for each item beyond 20, so the function is P(x) = 800 - 15(x - 20). This can also be written as P(x) = -15(x - 20) + 800. Choice A incorrectly assumes the decrease starts from x = 0. Choice B has the wrong sign and wrong y-intercept. Choice D doesn't account for the baseline of 20 items.
9. A cylindrical water tank has radius 4 feet. Water is pumped in at 8 cubic feet per minute. Which function $h(t)$ represents the height of water in the tank after $t$ minutes, assuming the tank starts empty?
   Answer: $h(t) = \frac{8t}{16\pi}$ This is a related rates problem involving the volume of a cylinder. When you see questions about filling containers, think about the relationship between volume, rate of change, and the container's dimensions. To find the height function, start with the volume formula for a cylinder: $V = \pi r^2 h$. Since the radius is 4 feet, we have $V = \pi (4^2) h = 16\pi h$. Water enters at 8 cubic feet per minute, so after $t$ minutes, the total volume is $V = 8t$. Setting these equal: $8t = 16\pi h$. Solving for height: $h = \frac{8t}{16\pi}$, which matches answer D. Let's examine why the other answers are incorrect. Answer A, $h(t) = \frac{2t}{\pi}$, appears to use an incorrect radius calculation—possibly confusing radius with diameter or making an arithmetic error with the area formula. Answer B, $h(t) = \frac{8t}{4\pi}$, uses $4\pi$ in the denominator, suggesting the student used $r = 2$ instead of $r = 4$ when calculating the area. Answer C, $h(t) = \frac{8t}{8\pi}$, uses $8\pi$ in the denominator, which might result from using $r^2 = 8$ instead of $r^2 = 16$. Study tip: For cylinder filling problems, always write out $V = \pi r^2 h$ first, then substitute the given radius carefully. Double-check that $r^2$ calculation—it's the most common source of error in these problems.
10. A store offers a discount structure where customers get 5% off their total if they spend at least $100, and an additional 3% off the discounted amount if they spend at least $200. Which function $D(x)$ gives the amount a customer pays for a purchase of $x$ dollars?
    Answer: $D(x) = \begin{cases} x & \text{if } x < 100 \\ 0.95x & \text{if } 100 \leq x < 200 \\ 0.9215x & \text{if } x \geq 200 \end{cases}$ For $x \geq 200$: First apply 5% discount: $0.95x$. Then apply additional 3% discount to the already discounted amount: $0.95x \times 0.97 = 0.9215x$. Choice A incorrectly calculates the second discount as 8% total instead of applying 3% to the discounted amount. Choice C uses fixed dollar amounts instead of percentages. Choice D shows the calculation incorrectly as $0.95x - 0.03x = 0.92x$, applying the 3% to the original amount rather than the discounted amount.
11. A company's profit depends on the number of units sold. The company breaks even (zero profit) when selling 20 units or 80 units. The maximum profit occurs when selling 50 units, where the profit is $9000. Assuming the profit function is quadratic, which function models the company's profit $P(x)$ based on $x$ units sold?
    Answer: $P(x) = -10(x - 50)^2 + 9000$ The quadratic has zeros at $x = 20$ and $x = 80$, with vertex at $(50, 9000)$. Using vertex form $P(x) = a(x - 50)^2 + 9000$, we find $a$ using a zero: $0 = a(20 - 50)^2 + 9000 = 900a + 9000$, so $a = -10$. Therefore $P(x) = -10(x - 50)^2 + 9000$. Choice A uses the wrong coefficient. Choice C has the correct zeros but wrong vertex value (it gives $P(50) = -5400$). Choice D opens upward, contradicting the maximum at the vertex.
12. A manufacturer's cost to produce $x$ items includes a setup cost of $500 plus $12 per item. Each item sells for $20. However, if more than 100 items are produced, the per-item production cost decreases to $10 for all items. Which function $P(x)$ represents the profit from producing and selling $x$ items?
    Answer: $P(x) = \begin{cases} 8x - 500 & \text{if } 0 \leq x \leq 100 \\ 10x - 500 & \text{if } x > 100 \end{cases}$ When you encounter profit problems with changing cost structures, remember that profit equals revenue minus total cost, and you need to account for different pricing tiers separately. Let's build the profit function step by step. Revenue is always $20x$ since each item sells for $20. For costs, you have a $500 setup cost plus variable costs that change at 100 items. For $0 \leq x \leq 100$: Total cost = $500 + 12x$, so profit = $20x - (500 + 12x) = 8x - 500$ For $x > 100$: Total cost = $500 + 10x$ (the $10 per item applies to ALL items when production exceeds 100), so profit = $20x - (500 + 10x) = 10x - 500$ This matches answer choice C. Answer A incorrectly calculates the second piece. It uses $8(100) + 10(x-100) - 500$, which assumes the first 100 items still cost $12 each and only additional items cost $10. The problem states that when production exceeds 100, the per-item cost "decreases to $10 for all items." Answer B shows the correct setup ($20x - 12x - 500$ and $20x - 10x - 500$) but fails to simplify, leaving it as separate terms rather than the simplified forms $8x - 500$ and $10x - 500$. Answer D uses the same logic as the correct answer but presents it in unsimplified form with parentheses around $(20-12)$ and $(20-10)$, making it unnecessarily complex. Study tip: In piecewise cost problems, pay careful attention to whether cost changes apply to all units or only additional units beyond the threshold.
13. A rectangular garden has a perimeter of 100 feet. If the length is $x$ feet, which function $A(x)$ represents the area of the garden in terms of its length?
    Answer: $A(x) = x(50 - x)$ Given perimeter = 100, we have $2x + 2w = 100$, so $w = 50 - x$. The area is $A(x) = x \cdot w = x(50 - x)$. Choice A incorrectly uses the full perimeter instead of solving for width. Choice C uses $100 - 2x$ which would be correct for width if the perimeter equation were $x + w = 100$. Choice D includes an extra factor of 2, confusing area with perimeter calculations.
14. The cost to produce $x$ units of a product includes a fixed cost of $500 and a variable cost of $8 per unit. However, due to bulk purchasing of materials, the variable cost decreases by $0.02 for each unit produced. Which function represents the total cost $C(x)$ to produce $x$ units?
    Answer: $C(x) = 500 + (8 - 0.02x)x$ The variable cost per unit is (8 - 0.02x), so the total variable cost for x units is x(8 - 0.02x). The total cost is C(x) = 500 + x(8 - 0.02x) = 500 + 8x - 0.02x². Choice A incorrectly subtracts 0.02x instead of 0.02x². Choice B has the correct expanded form but doesn't show the reasoning clearly. Choice D adds instead of subtracts the bulk discount term.
15. A taxi company charges a base fare plus a rate per mile. The total cost for a 5-mile trip is $12, and for a 12-mile trip is $23.50. If a customer wants to spend exactly $30, which equation can be used to find the maximum distance $d$ they can travel?
    Answer: $30 = \frac{23}{14}d + \frac{17}{14}$ The rate per mile is (23.50 - 12)/(12 - 5) = 11.5/7 = 23/14 dollars per mile. Using point (5, 12): C(d) - 12 = (23/14)(d - 5), so C(d) = (23/14)d - 115/14 + 12 = (23/14)d + 17/14. To find distance for $30: 30 = (23/14)d + 17/14. Choice A uses an incorrect rate. Choice B uses wrong calculations. Choice D uses 11.5 as the rate instead of 11.5/7.
16. A projectile is launched from the ground with initial velocity $v_0$ feet per second. Its height after $t$ seconds is $h(t) = v_0 t - 16t^2$. If the projectile reaches its maximum height of 64 feet, which function correctly models this specific projectile's height?
    Answer: $h(t) = 64t - 16t^2$ When you encounter projectile motion problems, you're working with quadratic functions where the maximum or minimum occurs at the vertex. The general form $h(t) = v_0 t - 16t^2$ represents height over time, where $v_0$ is the initial velocity and $-16t^2$ accounts for gravity's effect. To find the maximum height, you need to locate the vertex of this parabola. For a quadratic $h(t) = at^2 + bt + c$, the vertex occurs at $t = -\frac{b}{2a}$. In our case, $h(t) = -16t^2 + v_0 t$, so $a = -16$ and $b = v_0$. The time at maximum height is $t = \frac{v_0}{32}$. Substituting this back into the height equation: $h_{max} = v_0 \cdot \frac{v_0}{32} - 16 \cdot \left(\frac{v_0}{32}\right)^2 = \frac{v_0^2}{32} - \frac{v_0^2}{64} = \frac{v_0^2}{64}$ Since the maximum height is 64 feet: $\frac{v_0^2}{64} = 64$, which gives us $v_0^2 = 4096$, so $v_0 = 64$. Therefore, the correct function is $h(t) = 64t - 16t^2$, which is choice C. Choice A ($v_0 = 128$) would give a maximum height of 256 feet. Choice B ($v_0 = 32$) yields a maximum of 16 feet. Choice D ($v_0 = 16$) produces a maximum of only 4 feet. Remember: for projectile problems, always find the vertex to determine maximum height, then work backward to find the initial velocity that produces your given maximum.
17. Which function best represents the volume $V(T)$ in liters as a function of temperature $T$ in Kelvin?
    Answer: $V(T) = 0.008T - 0.6$ The slope is (2.8 - 2.4)/(350 - 300) = 0.4/50 = 0.008 L/K. Using point-slope form with (300, 2.4): V(T) - 2.4 = 0.008(T - 300), so V(T) = 0.008T - 2.4 + 2.4 = 0.008T - 0.6. Choice A assumes the line passes through origin. Choice C uses 0.8 instead of 0.008 for slope. Choice D uses 0.08 instead of 0.008 for slope.
18. A ball is dropped from a building. The height of the ball above ground is given by $h(t) = -16t^2 + h_0$, where $t$ is time in seconds and $h_0$ is the initial height. If the ball hits the ground after 3.5 seconds, which function represents the height of the ball at time $t$?
    Answer: $h(t) = -16t^2 + 196$ When the ball hits the ground, h = 0. So 0 = -16(3.5)² + h₀. Solving: h₀ = 16(3.5)² = 16(12.25) = 196. Therefore h(t) = -16t² + 196. Choice B uses 16 × 3.5 instead of 16 × (3.5)². Choice C uses just the time value. Choice D uses (3.5)² without the coefficient 16.
19. Which function represents the height $h(t)$ of the projectile $t$ seconds after launch, and what does the coefficient of $t$ represent in this context?
    Answer: $h(t) = -16t^2 + 48t + 64$; the coefficient 48 represents the initial upward velocity Substituting the given values: h(t) = -16t² + 48t + 64. The coefficient of t (which is 48) represents the initial upward velocity in ft/s. Choice A has the correct function but wrong interpretation of the coefficient. Choice C switches the initial velocity and height values. Choice D switches the gravity and velocity coefficients.
20. A water tank is being filled at a constant rate. After 5 minutes, the tank contains 120 gallons. After 12 minutes, it contains 204 gallons. If the tank was not empty when filling began, which function represents the total amount of water $W(t)$ in gallons after $t$ minutes of filling?
    Answer: $W(t) = 12t + 60$ The rate of filling is (204 - 120)/(12 - 5) = 84/7 = 12 gallons per minute. Using point-slope form with (5, 120): W(t) - 120 = 12(t - 5), so W(t) = 12t - 60 + 120 = 12t + 60. Choice B uses an incorrect initial amount. Choice C doubles the rate and uses wrong initial amount. Choice D uses the total change (84) as the rate.

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