Math 2 Quiz: Interpreting Quadratic Model Features
18 questions · exam conditions
0:00
Interpreting Quadratic Model FeaturesQuestion 1 of 18

A ball is thrown upward from a platform. Its height hh (in meters) after tt seconds follows h(t)=−5t2+20t+15h(t) = -5t^2 + 20t + 15. An observer wants to know the initial height of the platform and when the ball reaches its peak. Which interpretation is correct?

Platform height is 2020 meters, and the ball reaches peak height of 3535 meters at t=2t = 2 seconds.
Platform height is 1515 meters, and the ball reaches peak height of 3535 meters at t=2t = 2 seconds.
Platform height is 1515 meters, and the ball reaches peak height of 2020 meters at t=4t = 4 seconds.
Platform height is 55 meters, and the ball reaches peak height of 3535 meters at t=2t = 2 seconds.
← Back to quizzes

Math 2 Quiz

Math 2 Quiz: Interpreting Quadratic Model Features

Practice Interpreting Quadratic Model Features in Math 2 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 Interpreting Quadratic Model Features, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

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.

All questions

Question 1

A ball is thrown upward from a platform. Its height hh (in meters) after tt seconds follows h(t)=−5t2+20t+15h(t) = -5t^2 + 20t + 15. An observer wants to know the initial height of the platform and when the ball reaches its peak. Which interpretation is correct?

  1. Platform height is 2020 meters, and the ball reaches peak height of 3535 meters at t=2t = 2 seconds.
  2. Platform height is 1515 meters, and the ball reaches peak height of 3535 meters at t=2t = 2 seconds. (correct answer)
  3. Platform height is 1515 meters, and the ball reaches peak height of 2020 meters at t=4t = 4 seconds.
  4. Platform height is 55 meters, and the ball reaches peak height of 3535 meters at t=2t = 2 seconds.
Explanation: The initial height (platform height) is h(0) = 15 meters. The vertex occurs at t = -20/(2(-5)) = 2 seconds. The peak height is h(2) = -5(4) + 20(2) + 15 = -20 + 40 + 15 = 35 meters. Choice A confuses the initial velocity coefficient (20) with initial height. Choice C miscalculates both the peak height and time. Choice D confuses the acceleration coefficient with initial height.

Question 2

A farmer models the relationship between fertilizer amount xx (in pounds per acre) and crop yield YY (in bushels per acre) using Y(x)=−0.1x2+8x+200Y(x) = -0.1x^2 + 8x + 200. The farmer currently uses no fertilizer but is considering optimization. Based on the vertex of this model, what strategy should the farmer adopt?

  1. Use exactly 4040 pounds per acre to achieve the maximum possible yield of 360360 bushels per acre. (correct answer)
  2. Use exactly 8080 pounds per acre to achieve the maximum possible yield of 520520 bushels per acre.
  3. Use exactly 200200 pounds per acre since this matches the current yield without fertilizer.
  4. Use exactly 4040 pounds per acre to achieve the maximum possible yield of 320320 bushels per acre.
Explanation: The vertex occurs at x = -b/(2a) = -8/(2(-0.1)) = 40 pounds per acre. The maximum yield is Y(40) = -0.1(1600) + 8(40) + 200 = -160 + 320 + 200 = 360 bushels per acre. Choice B uses the wrong vertex x-coordinate (confusing -b/2a with -b/a). Choice C misinterprets the y-intercept value. Choice D has the correct fertilizer amount but miscalculates the maximum yield.

Question 3

A company's quarterly profit QQ (in thousands of dollars) is given by Q(x)=−x2+10x−16Q(x) = -x^2 + 10x - 16, where xx represents thousands of units produced. The financial analyst needs to determine the break-even production levels and optimal production for a board presentation. What key findings should be reported?

  1. Break-even at 22 and 88 thousand units, maximum profit of 99 thousand dollars at 1010 thousand units.
  2. Break-even at 44 and 66 thousand units, maximum profit of 99 thousand dollars at 55 thousand units.
  3. Break-even at 22 and 88 thousand units, maximum profit of 2525 thousand dollars at 55 thousand units.
  4. Break-even at 22 and 88 thousand units, maximum profit of 99 thousand dollars at 55 thousand units. (correct answer)
Explanation: When analyzing a quadratic profit function, you need to find two key business metrics: break-even points (where profit equals zero) and the production level that maximizes profit. To find break-even points, set Q(x)=0Q(x) = 0: −x2+10x−16=0-x^2 + 10x - 16 = 0. Multiplying by −1-1: x2−10x+16=0x^2 - 10x + 16 = 0. This factors as (x−2)(x−8)=0(x-2)(x-8) = 0, giving break-even points at x=2x = 2 and x=8x = 8 thousand units. For maximum profit, since this is a downward-opening parabola (coefficient of x2x^2 is negative), the vertex represents the maximum. The vertex occurs at x=−b2a=−102(−1)=5x = -\frac{b}{2a} = -\frac{10}{2(-1)} = 5 thousand units. Substituting: Q(5)=−(5)2+10(5)−16=−25+50−16=9Q(5) = -(5)^2 + 10(5) - 16 = -25 + 50 - 16 = 9 thousand dollars. Choice A incorrectly places the maximum at x=10x = 10, which would be far outside the profitable range and gives the wrong profit value. Choice B has incorrect break-even points of 4 and 6, likely from factoring errors or misreading the quadratic. Choice C correctly identifies break-even points but claims maximum profit is 25 thousand dollars, probably from calculating 10(5)−25=2510(5) - 25 = 25 instead of the correct substitution. Choice D correctly identifies break-even at 2 and 8 thousand units and maximum profit of 9 thousand dollars at 5 thousand units. Remember: for business quadratics, always check that your break-even points make the function equal zero and that your maximum/minimum occurs at the vertex using the formula x=−b2ax = -\frac{b}{2a}.

Question 4

A soccer ball is kicked from ground level, and its height hh (in meters) after tt seconds is given by h(t)=−4.9t2+19.6th(t) = -4.9t^2 + 19.6t. A coach wants to determine both the maximum height achieved and the total time the ball spends in the air for training purposes. What should the coach calculate?

  1. Maximum height of 19.619.6 meters at t=2t = 2 seconds, total air time of 44 seconds. (correct answer)
  2. Maximum height of 19.619.6 meters at t=4t = 4 seconds, total air time of 22 seconds.
  3. Maximum height of 39.239.2 meters at t=2t = 2 seconds, total air time of 44 seconds.
  4. Maximum height of 9.89.8 meters at t=1t = 1 seconds, total air time of 44 seconds.
Explanation: The vertex occurs at t = 19.6/(2×4.9) = 2 seconds. Maximum height is h(2) = -4.9(4) + 19.6(2) = -19.6 + 39.2 = 19.6 meters. For landing time, solve -4.9t² + 19.6t = 0: t(-4.9t + 19.6) = 0, giving t = 0 (kick) or t = 4 seconds (landing). Total air time is 4 seconds. Choice B confuses the maximum height timing with air time. Choice C miscalculates the maximum height by doubling it. Choice D uses wrong vertex timing and miscalculates maximum height.

Question 5

The number of customers NN at a restaurant tt hours after opening is modeled by N(t)=−3t2+24t+12N(t) = -3t^2 + 24t + 12. The manager needs to schedule staff based on peak customer times and when the restaurant should expect to be empty. What should the manager conclude?

  1. Peak customers occur at t=4t = 4 hours with 6060 customers, restaurant is empty at t=−0.5t = -0.5 and t=8.5t = 8.5 hours.
  2. Peak customers occur at t=8t = 8 hours with 6060 customers, restaurant is empty only at t=12t = 12 hours.
  3. Peak customers occur at t=4t = 4 hours with 6060 customers, restaurant is empty only at closing time. (correct answer)
  4. Peak customers occur at t=4t = 4 hours with 4848 customers, restaurant is empty at t=−0.5t = -0.5 and t=8.5t = 8.5 hours.
Explanation: The vertex is at t = -24/(2(-3)) = 4 hours with N(4) = -3(16) + 24(4) + 12 = 60 customers. For zeros: -3t² + 24t + 12 = 0 gives t ≈ -0.5 and t ≈ 8.5 hours. Since negative time and time beyond operating hours aren't meaningful in this context, the restaurant is never actually empty during operating hours. Choice A includes mathematically correct but contextually irrelevant times. Choice B miscalculates the vertex time. Choice D miscalculates the peak number of customers.

Question 6

The temperature TT (in degrees Celsius) in a greenhouse hh hours after midnight is modeled by T(h)=−h2+14h+15T(h) = -h^2 + 14h + 15. The greenhouse manager needs to know the initial temperature, peak temperature timing, and when the temperature drops to 15°C15°C again. What information describes this temperature cycle?

  1. Initial temperature 15°C15°C, peak of 49°C49°C at 77 hours, returns to 15°C15°C at 1414 hours.
  2. Initial temperature 14°C14°C, peak of 64°C64°C at 77 hours, returns to 15°C15°C at 1515 hours.
  3. Initial temperature 15°C15°C, peak of 64°C64°C at 1414 hours, returns to 15°C15°C at 77 hours.
  4. Initial temperature 15°C15°C, peak of 64°C64°C at 77 hours, returns to 15°C15°C at 1414 hours. (correct answer)
Explanation: When analyzing quadratic temperature models, you need to extract three key pieces of information: the initial value (when h = 0), the vertex (maximum or minimum point), and when the function returns to a specific value. For T(h)=−h2+14h+15T(h) = -h^2 + 14h + 15, let's find each component systematically. The initial temperature occurs at midnight (h = 0): T(0)=−(0)2+14(0)+15=15°CT(0) = -(0)^2 + 14(0) + 15 = 15°C. Since the coefficient of h2h^2 is negative, this parabola opens downward, so the vertex represents the maximum temperature. The vertex occurs at h=−b2a=−142(−1)=7h = -\frac{b}{2a} = -\frac{14}{2(-1)} = 7 hours. The peak temperature is T(7)=−(7)2+14(7)+15=−49+98+15=64°CT(7) = -(7)^2 + 14(7) + 15 = -49 + 98 + 15 = 64°C. To find when the temperature returns to 15°C, solve −h2+14h+15=15-h^2 + 14h + 15 = 15. This simplifies to −h2+14h=0-h^2 + 14h = 0, or h(−h+14)=0h(-h + 14) = 0. So h = 0 or h = 14, meaning the temperature is 15°C at midnight and again at 14 hours. Choice A incorrectly calculates the peak temperature as 49°C instead of 64°C. Choice B gives the wrong initial temperature (14°C instead of 15°C) and wrong return time (15 hours instead of 14 hours). Choice C reverses the peak time and return time, placing the maximum at 14 hours instead of 7 hours. Therefore, choice D correctly identifies all three values: initial temperature 15°C, peak of 64°C at 7 hours, and return to 15°C at 14 hours. Remember: for quadratic word problems, always verify your vertex calculation and check your algebra when solving equations.

Question 7

The daily attendance AA at a theme park depends on the ticket price pp (in dollars) according to A(p)=−50p2+2000p−15000A(p) = -50p^2 + 2000p - 15000. Park management wants to identify the price range where attendance drops to zero and the optimal pricing for maximum attendance. What pricing analysis should management use?

  1. Zero attendance at $15 and $25 ticket prices, maximum attendance of $2000020000 $ people at $40 ticket price.
  2. Zero attendance at $15 and $25 ticket prices, maximum attendance of $50005000 $ people at $20 ticket price.
  3. Zero attendance at $10 and $30 ticket prices, maximum attendance of $2000020000 $ people at $20 ticket price.
  4. Zero attendance at $10 and $30 ticket prices, maximum attendance of $50005000 $ people at $20 ticket price. (correct answer)
Explanation: When analyzing quadratic functions that model real-world scenarios like attendance versus price, you need to find two key features: the zeros (where the function equals zero) and the vertex (maximum or minimum point). To find where attendance drops to zero, set A(p)=0A(p) = 0: −50p2+2000p−15000=0-50p^2 + 2000p - 15000 = 0 Dividing by -50: p2−40p+300=0p^2 - 40p + 300 = 0 Using the quadratic formula or factoring: (p−10)(p−30)=0(p - 10)(p - 30) = 0 This gives us p=10p = 10 and p=30p = 30, so attendance is zero at $10 and $30 ticket prices. For maximum attendance, find the vertex. Since the coefficient of $p2p^2 isnegative(−50),thisparabolaopensdownward,sothevertexrepresentsamaximum.Thevertexoccursatis negative (-50), this parabola opens downward, so the vertex represents a maximum. The vertex occurs at p=−b2a=−20002(−50)=20p = -\frac{b}{2a} = -\frac{2000}{2(-50)} = 20 $. At p = 20 : A(20) = -50(400) + 2000(20) - 15000 = -20000 + 40000 - 15000 = 5000 So maximum attendance is 5000 people at a $20 ticket price. Option A incorrectly identifies zeros at $15 and $25, and claims maximum attendance of 20,000 at $40. Option B has the wrong zeros but correct maximum values. Option C has correct zeros but wrong maximum attendance (20,000 instead of 5,000). When working with quadratic models, always double-check your arithmetic when finding zeros and vertices—small calculation errors can lead you to attractive but incorrect answer choices that test common computational mistakes.

Question 8

A company's monthly profit PP (in thousands of dollars) is modeled by P(t)=−2t2+16t−24P(t) = -2t^2 + 16t - 24, where tt is the number of months after January. The company wants to determine when they should expect to break even during the year. Which statement correctly interprets the relevant feature of this model?

  1. The company breaks even in February and October, since the zeros occur at t=2t = 2 and t=12t = 12.
  2. The company breaks even in March and September, since the zeros occur at t=2t = 2 and t=6t = 6. (correct answer)
  3. The company breaks even in April and August, since the vertex occurs at t=4t = 4 with maximum profit.
  4. The company breaks even only in April, since this is when maximum profit of 88 thousand dollars occurs.
Explanation: To find break-even points, set P(t) = 0: -2t² + 16t - 24 = 0. Dividing by -2: t² - 8t + 12 = 0. Factoring: (t-2)(t-6) = 0, so t = 2 and t = 6. Since t represents months after January, t = 2 is March and t = 6 is September. Choice A incorrectly calculates one zero as t = 12. Choice C confuses the vertex with break-even points. Choice D misunderstands that break-even means zero profit, not maximum profit.

Question 9

The height hh (in feet) of a projectile is given by h(t)=−16t2+64t+80h(t) = -16t^2 + 64t + 80, where tt is time in seconds. A safety engineer needs to determine the maximum height and when the projectile returns to ground level. What information should the engineer report?

  1. Maximum height of 144144 feet at t=2t = 2 seconds, hits ground at t=5t = 5 seconds only. (correct answer)
  2. Maximum height of 144144 feet at t=2t = 2 seconds, hits ground at t=−1t = -1 and t=5t = 5 seconds.
  3. Maximum height of 8080 feet at t=0t = 0 seconds, hits ground at t=4t = 4 seconds only.
  4. Maximum height of 208208 feet at t=4t = 4 seconds, hits ground at t=6t = 6 seconds only.
Explanation: The vertex occurs at t = -b/(2a) = -64/(2(-16)) = 2 seconds. Maximum height is h(2) = -16(4) + 64(2) + 80 = 144 feet. For ground level, solve -16t² + 64t + 80 = 0, which gives t = -1 and t = 5. Since negative time is not physically meaningful for this scenario, the projectile hits ground at t = 5 seconds. Choice B includes the mathematically correct but physically irrelevant negative time. Choice C uses the initial height as maximum. Choice D incorrectly calculates both vertex time and maximum height.

Question 10

A physics student launches a model rocket from ground level. The rocket's height h (in feet) at time t (in seconds) is given by the quadratic function h(t) = -16t² + 128t.

The student needs to determine both when the rocket will land and its maximum altitude for a safety report. However, the student is confused about which features of the quadratic model to analyze. What should the student report?

  1. The rocket reaches maximum height of 256256 feet at t=8t = 8 seconds and lands at t=4t = 4 seconds.
  2. The rocket reaches maximum height of 128128 feet at t=8t = 8 seconds and lands at t=4t = 4 seconds.
  3. The rocket reaches maximum height of 256256 feet at t=4t = 4 seconds and lands at t=8t = 8 seconds. (correct answer)
  4. The rocket reaches maximum height of 512512 feet at t=4t = 4 seconds and lands at t=8t = 8 seconds.
Explanation: When analyzing quadratic models of projectile motion, you need to find two key features: the vertex (maximum height and when it occurs) and the x-intercepts (when the object hits the ground). For the function h(t)=−16t2+128th(t) = -16t^2 + 128t, start by finding when the rocket lands. Set h(t)=0h(t) = 0: −16t2+128t=0-16t^2 + 128t = 0 t(−16t+128)=0t(-16t + 128) = 0 This gives t=0t = 0 (launch time) and t=8t = 8 seconds (landing time). To find maximum height, use the vertex formula. For h(t)=at2+bt+ch(t) = at^2 + bt + c, the vertex occurs at t=−b2at = -\frac{b}{2a}. Here, a=−16a = -16 and b=128b = 128, so: t=−1282(−16)=12832=4t = -\frac{128}{2(-16)} = \frac{128}{32} = 4 seconds The maximum height is h(4)=−16(4)2+128(4)=−256+512=256h(4) = -16(4)^2 + 128(4) = -256 + 512 = 256 feet. Therefore, the rocket reaches maximum height of 256 feet at t=4t = 4 seconds and lands at t=8t = 8 seconds, making C correct. Choice A incorrectly swaps the times for maximum height and landing. Choice B uses the correct timing but calculates the wrong maximum height (128 instead of 256). Choice D correctly identifies the timing but inflates the maximum height to 512 feet, likely from forgetting to subtract the −16t2-16t^2 term. Remember: for projectile motion problems, always find the vertex for maximum height and solve for zero to find landing time. The vertex formula t=−b2at = -\frac{b}{2a} is your most reliable tool.

Question 11

A projectile is launched from the top of a 64-foot building. Its height h(t)=−16t2+32t+64h(t) = -16t^2 + 32t + 64 feet after tt seconds. The projectile reaches its maximum height and then falls to the ground. What does the yy-intercept of this function represent in the context of this situation?

  1. The maximum height reached by the projectile during its flight
  2. The initial height of the projectile when it was launched (correct answer)
  3. The time when the projectile reaches its maximum height
  4. The horizontal distance traveled when the projectile hits the ground
Explanation: The y-intercept occurs when t = 0, which gives h(0) = -16(0)² + 32(0) + 64 = 64 feet. This represents the height of the projectile at time t = 0, which is the initial height when launched. Choice A is incorrect because the maximum height occurs at the vertex. Choice C is incorrect because the y-intercept represents a height value, not a time value. Choice D is incorrect because this function gives height, not horizontal distance.

Question 12

A ball is thrown upward from ground level. Its height is modeled by h(t)=−16t2+48th(t) = -16t^2 + 48t, where hh is height in feet and tt is time in seconds. The ball lands back on the ground after some time. What is the physical meaning of both zeros of this function?

  1. One zero represents when the ball is thrown, the other represents the maximum height reached
  2. Both zeros represent the same moment when the ball reaches its peak height
  3. One zero represents when the ball is thrown, the other represents when it lands back on the ground (correct answer)
  4. Both zeros represent different maximum heights the ball reaches during its flight
Explanation: Setting h(t) = 0: -16t² + 48t = 0, so t(-16t + 48) = 0, giving t = 0 and t = 3. The zero at t = 0 represents the initial moment when the ball is at ground level (being thrown), and t = 3 represents when the ball returns to ground level (landing). Choice A is wrong because zeros represent when h = 0, not maximum height. Choice B is wrong because there are two different times, not the same moment, and peak height isn't at ground level. Choice D is wrong because zeros indicate ground level (height = 0), not maximum heights.

Question 13

An object is launched from a height of 100 feet with an initial velocity. Its height function is h(t)=−16t2+64t+100h(t) = -16t^2 + 64t + 100 feet after tt seconds. The object eventually hits the ground. At what time does the object return to its initial launch height of 100 feet?

  1. At t=0t = 0 seconds only, since this is when it was launched
  2. At t=2t = 2 seconds only, since this is when it reaches maximum height
  3. At t=0t = 0 seconds and t=4t = 4 seconds (correct answer)
  4. At t=4t = 4 seconds only, since this is when it hits the ground
Explanation: To find when the object is at 100 feet, solve h(t) = 100: -16t² + 64t + 100 = 100, which simplifies to -16t² + 64t = 0, so t(-16t + 64) = 0. This gives t = 0 and t = 4. At t = 0, the object is at its launch height, and at t = 4, it returns to that same height on its way down. Choice A misses the second time. Choice B is incorrect because the maximum height occurs at t = 2 (the vertex) and equals h(2) = 164 feet, not 100 feet. Choice D misses the initial time and incorrectly assumes the object hits the ground at t = 4.

Question 14

A soccer ball is kicked from the ground. Its height follows h(t)=−5t2+20th(t) = -5t^2 + 20t meters, where tt is time in seconds. The ball reaches its maximum height at the vertex and then returns to the ground. How long is the ball above a height of 15 meters?

  1. 2 seconds total (correct answer)
  2. 1 second total
  3. 3 seconds total
  4. The ball never reaches 15 meters in height
Explanation: This quadratic function problem tests your ability to find when a projectile stays above a specific height. When you see questions about objects moving in parabolic paths, you need to set up an equation and solve for the time intervals. To find how long the ball stays above 15 meters, set the height function equal to 15: −5t2+20t=15-5t^2 + 20t = 15. Rearranging gives −5t2+20t−15=0-5t^2 + 20t - 15 = 0. Dividing by -5 simplifies this to t2−4t+3=0t^2 - 4t + 3 = 0. Factoring yields (t−1)(t−3)=0(t-1)(t-3) = 0, so t=1t = 1 and t=3t = 3. This means the ball reaches exactly 15 meters at t=1t = 1 second (going up) and again at t=3t = 3 seconds (coming down). Between these times, from t=1t = 1 to t=3t = 3, the ball stays above 15 meters. The duration is 3−1=23 - 1 = 2 seconds. Looking at the wrong answers: B) suggests 1 second, which might come from incorrectly thinking about just one of the crossing points rather than the interval between them. C) gives 3 seconds, which could result from misunderstanding that 3 is the endpoint time, not the duration. D) claims the ball never reaches 15 meters, but our factoring clearly shows two real solutions where it does. Remember: for "time above a certain height" problems, solve for when the object equals that height, then find the time interval between those crossing points. The ball follows a parabolic path, so it crosses any given height twice.

Question 15

A suspension bridge cable follows the path y=0.01x2−0.8x+20y = 0.01x^2 - 0.8x + 20 feet above the roadway, where xx is the horizontal distance in feet from one end of the bridge. The vertex of this parabola occurs at (40,4)(40, 4). What does this vertex represent in the context of the bridge?

  1. The lowest point of the cable, located 40 feet from one end at a height of 4 feet (correct answer)
  2. The point where the cable is attached to the bridge tower at maximum height
  3. The midpoint of the bridge span where the cable height equals the tower height
  4. The point where the cable crosses the roadway at ground level
Explanation: When you encounter a quadratic function modeling a real-world situation, the vertex represents the function's maximum or minimum point. For suspension bridge cables, which naturally sag under their own weight, the parabola opens upward, making the vertex the lowest point of the cable. The given vertex (40,4)(40, 4) tells you that at a horizontal distance of 40 feet from one end of the bridge, the cable reaches its minimum height of 4 feet above the roadway. This is where the cable hangs lowest between the supporting towers. Choice A correctly identifies this as the lowest point of the cable at the specified location and height. This makes physical sense—suspension bridge cables form parabolic curves with their lowest point typically near the center of the span. Choice B incorrectly suggests this is where the cable attaches to a tower at maximum height. Tower attachment points would be at the parabola's endpoints or edges, not at the vertex, and they'd be at maximum height, not minimum. Choice C wrongly claims this is where cable height equals tower height. The vertex represents the minimum cable height, which would be much lower than the towers supporting the bridge. Choice D incorrectly states the cable crosses the roadway at ground level. Since y=4y = 4 feet at the vertex, the cable is still 4 feet above the roadway, not at ground level. Remember: for parabolas opening upward in real-world contexts, the vertex typically represents a minimum value—whether it's the lowest point of a cable, the minimum cost in a business problem, or the closest approach in a physics scenario.

Question 16

A company's profit function is given by P(x)=−2x2+80x−600P(x) = -2x^2 + 80x - 600 dollars, where xx represents the number of units produced per day. The vertex of this parabola occurs at (20,200)(20, 200). If the company currently produces 15 units per day and wants to maximize profit, what should they conclude?

  1. They should decrease production to 10 units per day to maximize profit
  2. They should increase production to 20 units per day to maximize profit (correct answer)
  3. They should maintain current production since 15 units gives maximum profit
  4. They should increase production to 25 units per day to maximize profit
Explanation: Since the coefficient of x² is negative (-2), this parabola opens downward, making the vertex the maximum point. The vertex (20, 200) means maximum profit of $200 occurs when producing 20 units per day. Currently producing 15 units, they should increase to 20 units. Choice A is wrong because moving further from the vertex decreases profit. Choice C is wrong because 15 ≠ 20. Choice D is wrong because 25 units is past the maximum, resulting in lower profit than at the vertex.

Question 17

A farmer has 200 feet of fencing and wants to create a rectangular pen against an existing barn wall (so fencing is needed for only three sides). The area function is A(w)=w(200−2w)A(w) = w(200 - 2w) square feet, where ww is the width of the pen perpendicular to the barn. The vertex form of this function is A(w)=−2(w−50)2+5000A(w) = -2(w - 50)^2 + 5000. If the farmer chooses a width of 40 feet, how does the resulting area compare to the maximum possible area?

  1. The area will be 200 square feet less than the maximum possible area
  2. The area will be 1000 square feet less than the maximum possible area
  3. The area will be equal to the maximum possible area since 40 is close to 50
  4. The area will be 800 square feet less than the maximum possible area (correct answer)
Explanation: This is a classic optimization problem involving quadratic functions in vertex form. When you see a quadratic function already converted to vertex form f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k, the vertex is at (h,k)(h, k), which represents either the maximum or minimum value depending on the sign of aa. From the vertex form A(w)=−2(w−50)2+5000A(w) = -2(w - 50)^2 + 5000, you can immediately identify that the maximum area occurs at w=50w = 50 feet (the vertex), giving a maximum area of 5000 square feet. The negative coefficient −2-2 confirms this is a downward-opening parabola with a maximum. To find the area when w=40w = 40, substitute into the vertex form: A(40)=−2(40−50)2+5000=−2(−10)2+5000=−2(100)+5000=4200A(40) = -2(40 - 50)^2 + 5000 = -2(-10)^2 + 5000 = -2(100) + 5000 = 4200 square feet. The difference is 5000−4200=8005000 - 4200 = 800 square feet, making D correct. Let's examine why the other answers are wrong. A suggests 200 square feet less, which would give an area of 4800 — this doesn't match our calculation. B claims 1000 square feet less, yielding 4000 square feet, also incorrect. C incorrectly assumes that being "close" to the optimal width means the areas are equal, but quadratic functions can change significantly even with small input changes. Study tip: When working with vertex form quadratics, the vertex gives you the maximum or minimum immediately. Always substitute the given value directly into the vertex form rather than converting back to standard form — it's faster and less error-prone.

Question 18

A store's weekly revenue RR (in hundreds of dollars) depends on the price pp (in dollars) according to R(p)=−2p2+80p−600R(p) = -2p^2 + 80p - 600. The store manager observes that revenue becomes zero at certain price points and wants to find the optimal pricing strategy. What should the manager conclude about the relationship between price and revenue?

  1. Revenue is zero when price is $15 or $25, and maximum revenue occurs at $20 with $200 in revenue.
  2. Revenue is zero when price is $10 or $30, and maximum revenue occurs at $20 with $200 in revenue. (correct answer)
  3. Revenue is zero when price is $10 or $30, and maximum revenue occurs at $40 with $800 in revenue.
  4. Revenue is zero when price is $15 or $25, and maximum revenue occurs at $40 with $200 in revenue.
Explanation: For zero revenue, solve -2p² + 80p - 600 = 0. Dividing by -2: p² - 40p + 300 = 0. Factoring: (p-10)(p-30) = 0, so p = $10 or $30. The vertex occurs at p = 80/(2(2)) = $20. Maximum revenue is R(20) = -2(400) + 80(20) - 600 = -800 + 1600 - 600 = $200 (in hundreds). Choice A incorrectly factors the quadratic. Choice C has correct zeros but wrong vertex location and revenue calculation. Choice D has incorrect zeros and vertex location.