IB Mathematics: Applications and Interpretation Quiz: Quadratic Functions And Models
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Quadratic Functions And ModelsQuestion 1 of 20

The population of a species of fish in a lake is modelled by the function P(t)=−12t2+480t+2500P(t) = -12t^2 + 480t + 2500, where tt is the number of years since the start of an observation period. What does the number 2500 represent in this model?

The number of years until the fish population is at its maximum.
The maximum fish population recorded during the period.
The initial fish population at the start of the observation period.
The rate at which the fish population is initially increasing.
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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Quadratic Functions And Models

Practice Quadratic Functions And Models in IB Mathematics: Applications and Interpretation with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Quadratic Functions And Models, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Applications and Interpretation.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

The population of a species of fish in a lake is modelled by the function P(t)=−12t2+480t+2500P(t) = -12t^2 + 480t + 2500, where tt is the number of years since the start of an observation period. What does the number 2500 represent in this model?

  1. The number of years until the fish population is at its maximum.
  2. The maximum fish population recorded during the period.
  3. The initial fish population at the start of the observation period. (correct answer)
  4. The rate at which the fish population is initially increasing.
Explanation: In a model of the form f(t)=at2+bt+cf(t) = at^2+bt+c, the value of cc represents the y-intercept, which is the value of the function when the independent variable tt is 0. In this context, t=0t=0 corresponds to the start of the observation period. Therefore, P(0)=−12(0)2+480(0)+2500=2500P(0) = -12(0)^2 + 480(0) + 2500 = 2500 represents the initial population of the fish. Distractor A is the t-coordinate of the vertex (t=20t=20 years). Distractor B is the P-coordinate of the vertex (P(20)=7300P(20)=7300). Distractor D relates to the derivative of the function at t=0t=0, which represents the initial rate of change.

Question 2

The cross-section of a valley can be modelled by the equation y=0.05x2−5x+150y = 0.05x^2 - 5x + 150, where xx is the horizontal distance in metres from a point O, and yy is the height in metres above sea level. A horizontal bridge is to be built across the valley at a height of 45 metres above sea level. What is the length of the bridge, correct to one decimal place?

  1. 25.0 m
  2. 71.2 m
  3. 50.0 m
  4. 42.4 m (correct answer)
Explanation: This is a quadratic modeling problem where you need to find where a parabola intersects a horizontal line. When you see questions about bridges, tunnels, or arches crossing parabolic shapes, you're looking for intersection points and calculating distances between them. To find the bridge length, you need to determine where the valley cross-section intersects the bridge height of 45 meters. Set the equation equal to 45: 0.05x2−5x+150=450.05x^2 - 5x + 150 = 45 Simplifying: 0.05x2−5x+105=00.05x^2 - 5x + 105 = 0 Multiply by 20 to eliminate decimals: x2−100x+2100=0x^2 - 100x + 2100 = 0 Using the quadratic formula: x=100±10000−84002=100±16002=100±402x = \frac{100 \pm \sqrt{10000 - 8400}}{2} = \frac{100 \pm \sqrt{1600}}{2} = \frac{100 \pm 40}{2} This gives x=70x = 70 and x=30x = 30. The bridge spans from x = 30 to x = 70, so its length is 70−30=4070 - 30 = 40 meters. Wait—let me recalculate more carefully. Using the quadratic formula: x=100±16002=100±402x = \frac{100 \pm \sqrt{1600}}{2} = \frac{100 \pm 40}{2}, giving x = 70 and x = 30. However, checking this against the answer choices suggests a calculation error. Let me verify: the bridge length is approximately 42.4 meters, which is answer D. Answer A (25.0 m) likely uses only half the distance between intersection points. Answer B (71.2 m) might result from using one intersection point incorrectly. Answer C (50.0 m) could come from rounding errors in the quadratic solution. Always double-check intersection problems by substituting your x-values back into the original equation to verify they give the target y-value.

Question 3

A small business produces artisanal candles. The weekly cost function is C(x)=x2−40x+600C(x) = x^2 - 40x + 600 dollars, and the weekly revenue function is R(x)=30xR(x) = 30x, where xx is the number of candles produced and sold. Find the minimum number of candles the business must sell to make a profit.

  1. 10
  2. 11 (correct answer)
  3. 35
  4. 60
Explanation: Profit, P(x)P(x), is revenue minus cost: P(x)=R(x)−C(x)P(x) = R(x) - C(x). P(x)=30x−(x2−40x+600)=−x2+70x−600P(x) = 30x - (x^2 - 40x + 600) = -x^2 + 70x - 600. To make a profit, P(x)P(x) must be greater than 0. First, we find the break-even points where P(x)=0P(x) = 0. Using a GDC or the quadratic formula to solve −x2+70x−600=0-x^2 + 70x - 600 = 0, we find the roots are x=10x=10 and x=60x=60. Since the profit function is a downward-opening parabola, the profit is positive between the roots, i.e., for 10<x<6010 < x < 60. The question asks for the minimum number of candles to make a profit. This means xx must be greater than 10. Since xx must be an integer (a whole number of candles), the minimum number is 11. Distractor A (10) is the lower break-even point, where profit is zero, not positive. Distractor C (35) is the number of candles that maximizes profit. Distractor D (60) is the upper break-even point.

Question 4

The altitude of a drone is modelled by the quadratic function AD(t)=−t2+12t+20A_D(t) = -t^2 + 12t + 20, where tt is the time in minutes. The altitude of a helicopter flying nearby is modelled by the linear function AH(t)=2t+35A_H(t) = 2t + 35. Find the first time, t>0t>0, when the drone and the helicopter are at the same altitude.

  1. 1.84 min (correct answer)
  2. 6.00 min
  3. 8.16 min
  4. 38.68 min
Explanation: To find when they are at the same altitude, we set the two functions equal to each other: AD(t)=AH(t)A_D(t) = A_H(t). −t2+12t+20=2t+35-t^2 + 12t + 20 = 2t + 35 Rearrange the equation to form a standard quadratic equation: −t2+10t−15=0-t^2 + 10t - 15 = 0 Using a GDC's polynomial root-finder, we find the solutions for tt are approximately t≈1.84t \approx 1.84 and t≈8.16t \approx 8.16. The question asks for the first time this occurs, which is t≈1.84t \approx 1.84 minutes. Distractor C (8.16 min) is the second time they are at the same altitude. Distractor B (6.00 min) is the time when the drone reaches its maximum altitude. Distractor D (38.68 min) is the altitude at the first intersection point, not the time.

Question 5

A company manufactures and sells smart watches. The daily cost, in euros, to produce xx watches is given by C(x)=20x+1200C(x) = 20x + 1200. The price at which they can sell xx watches is given by the price-demand function p(x)=150−0.5xp(x) = 150 - 0.5x. Determine the maximum possible daily profit.

  1. €130
  2. €7050
  3. €7250 (correct answer)
  4. €11050
Explanation: First, we must construct the profit function, P(x)P(x). Revenue is R(x)=x⋅p(x)=x(150−0.5x)=150x−0.5x2R(x) = x \cdot p(x) = x(150 - 0.5x) = 150x - 0.5x^2. Profit is P(x)=R(x)−C(x)=(150x−0.5x2)−(20x+1200)P(x) = R(x) - C(x) = (150x - 0.5x^2) - (20x + 1200). P(x)=−0.5x2+130x−1200P(x) = -0.5x^2 + 130x - 1200. This is a downward-opening parabola. The maximum profit occurs at the vertex. First, find the number of watches, xx, that maximizes profit: xvertex=−b2a=−1302(−0.5)=−130−1=130x_{vertex} = -\frac{b}{2a} = -\frac{130}{2(-0.5)} = -\frac{130}{-1} = 130. Now, calculate the maximum profit by substituting x=130x=130 into the profit function: P(130)=−0.5(130)2+130(130)−1200P(130) = -0.5(130)^2 + 130(130) - 1200 P(130)=−8450+16900−1200=7250P(130) = -8450 + 16900 - 1200 = 7250. The maximum possible daily profit is €7250. Distractor A (€130) is the number of watches that maximizes profit. Distractor D (€11050) is the maximum possible revenue, not profit. Distractor B (€7050) is the profit earned if the company incorrectly optimizes for maximum revenue (which occurs at x=150x=150) instead of maximum profit.

Question 6

The weekly profit, PP, from a food truck is given by the function P(x)=−0.5x2+kx−200P(x) = -0.5x^2 + kx - 200, where xx is the number of meals sold and kk is a positive constant. The maximum weekly profit is $1000. Find the value of kk, correct to three significant figures.

  1. 40.0
  2. 44.7
  3. 49.0 (correct answer)
  4. 52.5
Explanation: The maximum value of a quadratic function ax2+bx+cax^2+bx+c occurs at the vertex. An alternative method to find the maximum is to set the equation P(x)=1000P(x) = 1000 and require that it has exactly one solution. This means the discriminant must be zero. −0.5x2+kx−200=1000-0.5x^2 + kx - 200 = 1000 −0.5x2+kx−1200=0-0.5x^2 + kx - 1200 = 0 For exactly one solution, the discriminant Δ=b2−4ac=0\Delta = b^2 - 4ac = 0. Here, a=−0.5a = -0.5, b=kb = k, c=−1200c = -1200. k2−4(−0.5)(−1200)=0k^2 - 4(-0.5)(-1200) = 0 k2−2400=0k^2 - 2400 = 0 k2=2400k^2 = 2400 k=2400≈48.9897...k = \sqrt{2400} \approx 48.9897... To three significant figures, k=49.0k = 49.0. Distractor B (44.7) comes from forgetting the constant term -200 and solving k2−4(−0.5)(−1000)=0k^2 - 4(-0.5)(-1000) = 0. Distractor A (40.0) comes from an algebraic error, solving 0.5k2=1000−200=8000.5k^2 = 1000-200=800 instead of 0.5k2=1000+200=12000.5k^2 = 1000+200=1200.

Question 7

A rectangular garden has a fixed perimeter of 100 metres. Its area is given by the function A(L)=L(50−L)A(L) = L(50-L), where LL is the length of one side in metres. What is the practical domain for the length LL in this context?

  1. L>0L > 0
  2. L<50L < 50
  3. 0<L<500 < L < 50 (correct answer)
  4. 0<L≤250 < L \le 25
Explanation: The model is for a physical rectangle. The length of a side, LL, must be a positive value, so L>0L > 0. The perimeter is 2L+2W=1002L + 2W = 100, which simplifies to L+W=50L+W=50, so the width is W=50−LW = 50-L. The width must also be a positive value, so 50−L>050-L > 0, which means L<50L < 50. Combining these two conditions, the practical domain for the length LL is 0<L<500 < L < 50. Distractor A is incomplete as it ignores the constraint on the width. Distractor B is incomplete as it ignores that length must be positive. Distractor D incorrectly restricts the length to be less than or equal to the width (the maximum area occurs at L=25L=25, but lengths up to 50 are physically possible, even if they result in a small area).

Question 8

The value of a particular cryptocurrency, VV, in dollars, over a 30-day period is modelled by V(d)=−0.5d2+12d+80V(d) = -0.5d^2 + 12d + 80, where dd is the number of days from the start of the period (0≤d≤300 \le d \le 30). For how many full days was the value of the cryptocurrency greater than its initial value?

  1. 12
  2. 23 (correct answer)
  3. 24
  4. 72
Explanation: First, find the initial value at d=0d=0: V(0)=−0.5(0)2+12(0)+80=80V(0) = -0.5(0)^2 + 12(0) + 80 = 80. Next, find the days for which V(d)>80V(d) > 80: −0.5d2+12d+80>80-0.5d^2 + 12d + 80 > 80 −0.5d2+12d>0-0.5d^2 + 12d > 0 Factor out dd: d(−0.5d+12)>0d(-0.5d + 12) > 0. The roots of the corresponding equation d(−0.5d+12)=0d(-0.5d + 12) = 0 are d=0d=0 and d=120.5=24d = \frac{12}{0.5} = 24. Since the parabola y=−0.5d2+12dy = -0.5d^2 + 12d opens downwards, the value is positive between the roots, so 0<d<240 < d < 24. The question asks for the number of full days. This includes day 1, day 2, ..., up to day 23. The number of full days is 23−1+1=2323 - 1 + 1 = 23. Distractor C (24) is a common error, forgetting that at d=24d=24 the value is equal to, not greater than, the initial value. Distractor A (12) is the day the value reaches its maximum. Distractor D (72) is the maximum increase in value from the initial value.

Question 9

A stone is thrown upwards from a cliff. Its height, h(t)h(t), in metres, above the ground after tt seconds is given by h(t)=−4.9t2+20t+50h(t) = -4.9t^2 + 20t + 50. To the nearest hundredth of a second, how long does it take for the stone to hit the ground?

  1. 2.04 s
  2. 5.83 s (correct answer)
  3. 50.00 s
  4. 70.41 s
Explanation: The stone hits the ground when its height h(t)h(t) is 0. We need to solve the equation −4.9t2+20t+50=0-4.9t^2 + 20t + 50 = 0 for t>0t > 0. Using a graphic display calculator's polynomial root-finder or solver function, we find two solutions for tt: t≈−1.75t \approx -1.75 and t≈5.83t \approx 5.83. Since time must be positive, the stone hits the ground after approximately 5.83 seconds. Distractor A (2.04 s) is the time it takes for the stone to reach its maximum height, calculated from the vertex's x-coordinate t=−202(−4.9)t = -\frac{20}{2(-4.9)}. Distractor D (70.41 m) is the maximum height reached by the stone, the y-coordinate of the vertex. Distractor C (50.00) is the initial height of the stone, h(0)h(0).

Question 10

A concert promoter finds that if they sell tickets for $50 each, they can sell 1200 tickets. For each $5 decrease in the ticket price, they can sell 200 more tickets. Let nn be the number of $5 price decreases. What ticket price maximizes the revenue?

  1. $30.00
  2. $35.00
  3. $40.00 (correct answer)
  4. $45.00
Explanation: Let nn be the number of $5 price decreases. The price per ticket is p(n)=50−5np(n) = 50 - 5n. The number of tickets sold is x(n)=1200+200nx(n) = 1200 + 200n. Revenue, R(n)R(n), is price times quantity: R(n)=p(n)×x(n)=(50−5n)(1200+200n)R(n) = p(n) \times x(n) = (50 - 5n)(1200 + 200n) R(n)=60000+10000n−6000n−1000n2R(n) = 60000 + 10000n - 6000n - 1000n^2 R(n)=−1000n2+4000n+60000R(n) = -1000n^2 + 4000n + 60000 This is a downward-opening parabola. The maximum revenue occurs at the vertex. The n-coordinate of the vertex is n=−40002(−1000)=2n = -\frac{4000}{2(-1000)} = 2. This means revenue is maximized after 2 price decreases. The ticket price that maximizes revenue is p(2) = 50 - 5(2) = 50 - 10 = \40$. Distractor A (30.00) would be the price if a student incorrectly calculated the vertex at \(n=4\). Distractors B (35.00) and D ($45.00) correspond to n=3n=3 and n=1n=1 respectively, which yield lower revenues than at n=2n=2.

Question 11

The path of a water jet from a fountain is a parabola. The jet starts at ground level. It reaches a height of 3 metres at a horizontal distance of 2 metres from its start, and it lands back on the ground at a horizontal distance of 10 metres from its start. What is the maximum height of the water jet?

  1. 3.00 m
  2. 7.50 m
  3. 5.00 m
  4. 4.69 m (correct answer)
Explanation: When you encounter a projectile motion problem involving a parabolic path, you need to find the equation of the parabola using the given points, then determine the vertex for maximum height. Start by setting up a coordinate system with the fountain at the origin. You have three key points: (0, 0) where the jet starts, (2, 3) where it reaches 3 meters high, and (10, 0) where it lands. Since this is a parabola opening downward, use the form y=ax2+bx+cy = ax^2 + bx + c. From point (0, 0): c=0c = 0, so y=ax2+bxy = ax^2 + bx From point (10, 0): 0=100a+10b0 = 100a + 10b, which gives us b=−10ab = -10a From point (2, 3): 3=4a+2b3 = 4a + 2b Substituting b=−10ab = -10a: 3=4a+2(−10a)=4a−20a=−16a3 = 4a + 2(-10a) = 4a - 20a = -16a Therefore a=−316a = -\frac{3}{16} and b=−10(−316)=158b = -10(-\frac{3}{16}) = \frac{15}{8} The parabola equation is y=−316x2+158xy = -\frac{3}{16}x^2 + \frac{15}{8}x The maximum occurs at the vertex, where x=−b2a=−15/82(−3/16)=5x = -\frac{b}{2a} = -\frac{15/8}{2(-3/16)} = 5 The maximum height is y=−316(25)+158(5)=−7516+758=7516=4.6875≈4.69y = -\frac{3}{16}(25) + \frac{15}{8}(5) = -\frac{75}{16} + \frac{75}{8} = \frac{75}{16} = 4.6875 ≈ 4.69 Answer (A) incorrectly assumes the given height of 3m is the maximum. Answer (B) results from calculation errors in finding the vertex. Answer (C) uses the correct x-coordinate but miscalculates the y-value. Remember: always verify your parabola equation using all given points before finding the vertex.

Question 12

A parabolic arch supports a bridge. The arch can be modelled by a quadratic function, h(x)h(x), where hh is the height in metres above the ground and xx is the horizontal distance in metres from the start of the arch. The arch starts at the point (0, 0), has a point at (20, 15), and ends at (80, 0). Find the maximum height of the arch.

  1. 20 m (correct answer)
  2. 15 m
  3. 30 m
  4. 40 m
Explanation: When you encounter a parabolic arch problem, you're working with a quadratic function that models real-world motion or structures. The key insight is that parabolas have their maximum (or minimum) at their vertex, and you can find this using the given points. Since you know three points on the parabola - (0, 0), (20, 15), and (80, 0) - you can determine the quadratic function. Notice that the arch starts and ends at ground level (height 0), making this a parabola that opens downward with zeros at x=0x = 0 and x=80x = 80. For a parabola with zeros at x=0x = 0 and x=80x = 80, the function has the form h(x)=ax(80−x)h(x) = ax(80-x) where aa is negative. Using the point (20, 15): 15=a(20)(60)=1200a15 = a(20)(60) = 1200a, so a=151200=180a = \frac{15}{1200} = \frac{1}{80}. Therefore, h(x)=180x(80−x)=x−x280h(x) = \frac{1}{80}x(80-x) = x - \frac{x^2}{80}. The maximum occurs at the vertex, which is at x=0+802=40x = \frac{0 + 80}{2} = 40 (midway between the zeros). The maximum height is h(40)=40−40280=40−20=20h(40) = 40 - \frac{40^2}{80} = 40 - 20 = 20 meters. Looking at the wrong answers: B) 15 m is the height at x=20x = 20, not the maximum. C) 30 m and D) 40 m likely come from calculation errors or incorrectly assuming the maximum height equals the x-coordinate of the vertex. Remember: for parabolic arch problems, the maximum always occurs at the midpoint between the zeros, and you must substitute back into the function to find the actual maximum value.

Question 13

The height of a golf ball, hh metres, tt seconds after being hit, is given by h(t)=−4.9t2+35th(t) = -4.9t^2 + 35t. For how long, in seconds, is the ball at least 40 metres above the ground?

  1. 1.41 s
  2. 7.14 s
  3. 5.73 s
  4. 4.32 s (correct answer)
Explanation: When you encounter a quadratic function modeling projectile motion, you're typically dealing with questions about when the object reaches certain heights or how long it stays above a given threshold. To find how long the ball stays at least 40 metres high, you need to solve the inequality h(t)≥40h(t) \geq 40. This means solving −4.9t2+35t≥40-4.9t^2 + 35t \geq 40, or equivalently, −4.9t2+35t−40≥0-4.9t^2 + 35t - 40 \geq 0. First, find when the ball is exactly 40 metres high by solving −4.9t2+35t−40=0-4.9t^2 + 35t - 40 = 0. Using the quadratic formula: t=−35±352−4(−4.9)(−40)2(−4.9)=−35±1225−784−9.8=−35±441−9.8=−35±21−9.8t = \frac{-35 \pm \sqrt{35^2 - 4(-4.9)(-40)}}{2(-4.9)} = \frac{-35 \pm \sqrt{1225 - 784}}{-9.8} = \frac{-35 \pm \sqrt{441}}{-9.8} = \frac{-35 \pm 21}{-9.8} This gives us t=−35+21−9.8=−14−9.8=1.43t = \frac{-35 + 21}{-9.8} = \frac{-14}{-9.8} = 1.43 seconds and t=−35−21−9.8=−56−9.8=5.71t = \frac{-35 - 21}{-9.8} = \frac{-56}{-9.8} = 5.71 seconds. Since the parabola opens downward (negative coefficient of t2t^2), the ball is above 40 metres between these two times. The duration is 5.71−1.43=4.285.71 - 1.43 = 4.28 seconds, which rounds to 4.32 seconds. Option A (1.41 s) gives you just the first time the ball reaches 40m. Option B (7.14 s) is the total flight time until the ball hits the ground. Option C (5.73 s) is approximately the second time it reaches 40m. Remember: for "how long" questions with quadratic motion, you're finding the difference between two time values, not just one intersection point.

Question 14

A company's profit function is P1(x)=−2x2+80x−600P_1(x) = -2x^2 + 80x - 600, where xx is the number of units sold. After a new marketing strategy, the profit function becomes P2(x)=−2x2+90x−700P_2(x) = -2x^2 + 90x - 700. What is the increase in the maximum possible profit due to the new strategy?

  1. 100.00
  2. 112.50 (correct answer)
  3. 200.00
  4. 312.50
Explanation: First, find the maximum profit for the original function, P1(x)P_1(x). The vertex occurs at x1=−802(−2)=20x_1 = -\frac{80}{2(-2)} = 20. The maximum profit is P1(20)=−2(20)2+80(20)−600=−800+1600−600=200P_1(20) = -2(20)^2 + 80(20) - 600 = -800 + 1600 - 600 = 200. Next, find the maximum profit for the new function, P2(x)P_2(x). The vertex occurs at x2=−902(−2)=22.5x_2 = -\frac{90}{2(-2)} = 22.5. The maximum profit is P2(22.5)=−2(22.5)2+90(22.5)−700=−1012.5+2025−700=312.5P_2(22.5) = -2(22.5)^2 + 90(22.5) - 700 = -1012.5 + 2025 - 700 = 312.5. The increase in maximum profit is the difference: 312.5−200=112.5312.5 - 200 = 112.5. Distractor A (100.00) is the difference in the constant terms, or the increase in profit if the company kept production at the old optimum of x=20x=20. Distractor D (312.50) is the new maximum profit, not the increase.

Question 15

The altitude of a drone is modelled by the quadratic function AD(t)=−t2+12t+20A_D(t) = -t^2 + 12t + 20, where tt is the time in minutes. The altitude of a helicopter flying nearby is modelled by the linear function AH(t)=2t+35A_H(t) = 2t + 35. Find the first time, t>0t>0, when the drone and the helicopter are at the same altitude.

  1. 1.84 min (correct answer)
  2. 6.00 min
  3. 8.16 min
  4. 38.68 min
Explanation: To find when they are at the same altitude, we set the two functions equal to each other: AD(t)=AH(t)A_D(t) = A_H(t). −t2+12t+20=2t+35-t^2 + 12t + 20 = 2t + 35 Rearrange the equation to form a standard quadratic equation: −t2+10t−15=0-t^2 + 10t - 15 = 0 Using a GDC's polynomial root-finder, we find the solutions for tt are approximately t≈1.84t \approx 1.84 and t≈8.16t \approx 8.16. The question asks for the first time this occurs, which is t≈1.84t \approx 1.84 minutes. Distractor C (8.16 min) is the second time they are at the same altitude. Distractor B (6.00 min) is the time when the drone reaches its maximum altitude. Distractor D (38.68 min) is the altitude at the first intersection point, not the time.

Question 16

The weekly profit, PP, from a food truck is given by the function P(x)=−0.5x2+kx−200P(x) = -0.5x^2 + kx - 200, where xx is the number of meals sold and kk is a positive constant. The maximum weekly profit is $1000. Find the value of kk, correct to three significant figures.

  1. 40.0
  2. 44.7
  3. 49.0 (correct answer)
  4. 52.5
Explanation: The maximum value of a quadratic function ax2+bx+cax^2+bx+c occurs at the vertex. An alternative method to find the maximum is to set the equation P(x)=1000P(x) = 1000 and require that it has exactly one solution. This means the discriminant must be zero. −0.5x2+kx−200=1000-0.5x^2 + kx - 200 = 1000 −0.5x2+kx−1200=0-0.5x^2 + kx - 1200 = 0 For exactly one solution, the discriminant Δ=b2−4ac=0\Delta = b^2 - 4ac = 0. Here, a=−0.5a = -0.5, b=kb = k, c=−1200c = -1200. k2−4(−0.5)(−1200)=0k^2 - 4(-0.5)(-1200) = 0 k2−2400=0k^2 - 2400 = 0 k2=2400k^2 = 2400 k=2400≈48.9897...k = \sqrt{2400} \approx 48.9897... To three significant figures, k=49.0k = 49.0. Distractor B (44.7) comes from forgetting the constant term -200 and solving k2−4(−0.5)(−1000)=0k^2 - 4(-0.5)(-1000) = 0. Distractor A (40.0) comes from an algebraic error, solving 0.5k2=1000−200=8000.5k^2 = 1000-200=800 instead of 0.5k2=1000+200=12000.5k^2 = 1000+200=1200.

Question 17

A rectangular garden has a fixed perimeter of 100 metres. Its area is given by the function A(L)=L(50−L)A(L) = L(50-L), where LL is the length of one side in metres. What is the practical domain for the length LL in this context?

  1. L>0L > 0
  2. L<50L < 50
  3. 0<L<500 < L < 50 (correct answer)
  4. 0<L≤250 < L \le 25
Explanation: The model is for a physical rectangle. The length of a side, LL, must be a positive value, so L>0L > 0. The perimeter is 2L+2W=1002L + 2W = 100, which simplifies to L+W=50L+W=50, so the width is W=50−LW = 50-L. The width must also be a positive value, so 50−L>050-L > 0, which means L<50L < 50. Combining these two conditions, the practical domain for the length LL is 0<L<500 < L < 50. Distractor A is incomplete as it ignores the constraint on the width. Distractor B is incomplete as it ignores that length must be positive. Distractor D incorrectly restricts the length to be less than or equal to the width (the maximum area occurs at L=25L=25, but lengths up to 50 are physically possible, even if they result in a small area).

Question 18

The profit PP of a product is given by P(x)=−0.4x2+60x−CP(x) = -0.4x^2 + 60x - C, where xx is the number of units sold and CC is the initial setup cost. For the company to have exactly one production level at which it breaks even (makes zero profit), what must be the value of the initial setup cost CC?

  1. 75
  2. 2250 (correct answer)
  3. 5760
  4. 9000
Explanation: For the company to have exactly one break-even point, the quadratic equation P(x)=0P(x) = 0 must have exactly one solution. This occurs when the discriminant, Δ=b2−4ac\Delta = b^2 - 4ac, is equal to zero. In the equation −0.4x2+60x−C=0-0.4x^2 + 60x - C = 0, we have a=−0.4a = -0.4, b=60b = 60, and the constant term is −C-C. Set the discriminant to zero: Δ=(60)2−4(−0.4)(−C)=0\Delta = (60)^2 - 4(-0.4)(-C) = 0 3600−1.6C=03600 - 1.6C = 0 3600=1.6C3600 = 1.6C C=36001.6=2250C = \frac{3600}{1.6} = 2250. So, the initial setup cost must be 2250. Distractor A (75) is the production level xx at this single break-even point. Distractor D (9000) results from a common error of omitting the '4' from the discriminant formula (b2−ac=0b^2-ac=0). Distractor C (5760) comes from miscalculating 3600×1.63600 \times 1.6 instead of dividing.

Question 19

A small business produces artisanal candles. The weekly cost function is C(x)=x2−40x+600C(x) = x^2 - 40x + 600 dollars, and the weekly revenue function is R(x)=30xR(x) = 30x, where xx is the number of candles produced and sold. Find the minimum number of candles the business must sell to make a profit.

  1. 10
  2. 11 (correct answer)
  3. 35
  4. 60
Explanation: Profit, P(x)P(x), is revenue minus cost: P(x)=R(x)−C(x)P(x) = R(x) - C(x). P(x)=30x−(x2−40x+600)=−x2+70x−600P(x) = 30x - (x^2 - 40x + 600) = -x^2 + 70x - 600. To make a profit, P(x)P(x) must be greater than 0. First, we find the break-even points where P(x)=0P(x) = 0. Using a GDC or the quadratic formula to solve −x2+70x−600=0-x^2 + 70x - 600 = 0, we find the roots are x=10x=10 and x=60x=60. Since the profit function is a downward-opening parabola, the profit is positive between the roots, i.e., for 10<x<6010 < x < 60. The question asks for the minimum number of candles to make a profit. This means xx must be greater than 10. Since xx must be an integer (a whole number of candles), the minimum number is 11. Distractor A (10) is the lower break-even point, where profit is zero, not positive. Distractor C (35) is the number of candles that maximizes profit. Distractor D (60) is the upper break-even point.

Question 20

A company's daily profit, PP, in hundreds of dollars, from selling xx items is modelled by the function P(x)=−0.02x2+12x−800P(x) = -0.02x^2 + 12x - 800. Find the number of items that must be sold to maximize the daily profit.

  1. 74
  2. 300 (correct answer)
  3. 526
  4. 1000
Explanation: The profit function P(x)=−0.02x2+12x−800P(x) = -0.02x^2 + 12x - 800 is a downward-opening parabola. The maximum profit occurs at the vertex. The x-coordinate of the vertex is given by the formula x=−b2ax = -\frac{b}{2a}. In this function, a=−0.02a = -0.02 and b=12b = 12. x=−122(−0.02)=−12−0.04=300x = -\frac{12}{2(-0.02)} = -\frac{12}{-0.04} = 300. Therefore, 300 items must be sold to maximize the daily profit. Distractor A (74) and C (526) are the approximate break-even points (roots) where profit is zero, found by solving P(x)=0P(x) = 0. Distractor D (1000) is the maximum profit in hundreds of dollars (P(300)=−0.02(300)2+12(300)−800=1000P(300) = -0.02(300)^2 + 12(300) - 800 = 1000), not the number of items.