Math 2 Quiz: Building Quadratic Models
16 questions · exam conditions
0:00
Building Quadratic ModelsQuestion 1 of 16

The daily profit PP (in dollars) from selling xx units of a product is modeled by P=−2x2+60x−200P = -2x^2 + 60x - 200. On a particular day, the company wants to determine the minimum number of units they must sell to achieve a profit of at least $250. Which inequality correctly represents this situation?

−2x2+60x−200≥250-2x^2 + 60x - 200 \geq 250
−2x2+60x−200≤250-2x^2 + 60x - 200 \leq 250
−2x2+60x+50≥0-2x^2 + 60x + 50 \geq 0
−2x2+60x−450≥0-2x^2 + 60x - 450 \geq 0
← Back to quizzes

Math 2 Quiz

Math 2 Quiz: Building Quadratic Models

Practice Building Quadratic Models 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 Building Quadratic Models, 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

The daily profit PP (in dollars) from selling xx units of a product is modeled by P=−2x2+60x−200P = -2x^2 + 60x - 200. On a particular day, the company wants to determine the minimum number of units they must sell to achieve a profit of at least $250. Which inequality correctly represents this situation?

  1. −2x2+60x−200≥250-2x^2 + 60x - 200 \geq 250
  2. −2x2+60x−200≤250-2x^2 + 60x - 200 \leq 250
  3. −2x2+60x+50≥0-2x^2 + 60x + 50 \geq 0
  4. −2x2+60x−450≥0-2x^2 + 60x - 450 \geq 0 (correct answer)
Explanation: To achieve a profit of at least $250, we need $P≥250P \geq 250 ,so, so −2x2+60x−200≥250-2x^2 + 60x - 200 \geq 250 .Subtracting250frombothsides:. Subtracting 250 from both sides: −2x2+60x−200−250≥0-2x^2 + 60x - 200 - 250 \geq 0 ,whichsimplifiesto, which simplifies to −2x2+60x−450≥0-2x^2 + 60x - 450 \geq 0 .ChoiceAshowstheinequalitybeforealgebraicmanipulation.ChoiceBincorrectlyuses. Choice A shows the inequality before algebraic manipulation. Choice B incorrectly uses ≤\leq insteadofinstead of ≥\geq $. Choice C makes an error in the constant term when rearranging.

Question 2

A projectile is launched from ground level with an initial velocity of 80 feet per second at an angle. The horizontal distance xx and height yy are related by y=x−0.01x2y = x - 0.01x^2. What is the maximum horizontal distance the projectile can travel before hitting the ground?

  1. 80 feet
  2. 100 feet (correct answer)
  3. 120 feet
  4. 150 feet
Explanation: The projectile hits the ground when y=0y = 0. Setting 0=x−0.01x2=x(1−0.01x)0 = x - 0.01x^2 = x(1 - 0.01x), we get x=0x = 0 (launch point) or 1−0.01x=01 - 0.01x = 0, which gives x=100x = 100 feet. Choice A incorrectly uses the initial velocity as the distance. Choice C assumes the vertex x-coordinate is the answer. Choice D miscalculates by using 10.01×1.5\frac{1}{0.01} \times 1.5.

Question 3

A rectangular swimming pool is being designed with a surrounding walkway. The pool itself has length ll and width ww, while the walkway extends 3 feet on all sides. If the total area (pool plus walkway) is 600 square feet and the length of the pool is twice its width, which equation represents this constraint?

  1. 2w2+18w+36=6002w^2 + 18w + 36 = 600
  2. (2w+3)(w+3)=600(2w + 3)(w + 3) = 600
  3. (2w)(w)+6w+12=600(2w)(w) + 6w + 12 = 600
  4. (2w+6)(w+6)=600(2w + 6)(w + 6) = 600 (correct answer)
Explanation: This problem tests your ability to set up area equations for composite geometric figures. When you see a problem involving a shape with an extension (like a walkway around a pool), focus on finding the total dimensions of the enlarged figure. Start by identifying what you know: the pool has length l=2wl = 2w and width ww, and the walkway extends 3 feet on all sides. The key insight is that the walkway adds 3 feet to each side of the pool, meaning it adds 6 feet total to both the length and width (3 feet on the left + 3 feet on the right = 6 feet total). The total length becomes 2w+62w + 6 and the total width becomes w+6w + 6. Since the total area equals 600 square feet, the equation is (2w+6)(w+6)=600(2w + 6)(w + 6) = 600. Choice A (2w2+18w+36=6002w^2 + 18w + 36 = 600) appears to be the expanded form of some incorrect setup. Choice B ((2w+3)(w+3)=600(2w + 3)(w + 3) = 600) makes the common error of adding only 3 feet to each dimension instead of 6, forgetting that the walkway extends on both sides. Choice C ((2w)(w)+6w+12=600(2w)(w) + 6w + 12 = 600) incorrectly treats this as the pool area plus some additional terms, rather than recognizing that you need the area of the entire enlarged rectangle. When setting up area problems with borders or frames, always remember that an extension of xx units "on all sides" means adding 2x2x to each dimension, since the extension occurs on both ends of each side.

Question 4

A company manufactures rectangular storage containers where the length is 3 feet more than twice the width. If the total cost to produce each container is C=2A+50C = 2A + 50, where AA is the area of the base in square feet, which equation represents the total cost CC as a function of the width ww?

  1. C=4w2+6w+50C = 4w^2 + 6w + 50 (correct answer)
  2. C=2w2+6w+50C = 2w^2 + 6w + 50
  3. C=4w2+12w+50C = 4w^2 + 12w + 50
  4. C=2w2+3w+50C = 2w^2 + 3w + 50
Explanation: First, express the length in terms of width: l=2w+3l = 2w + 3. The area is A=w⋅l=w(2w+3)=2w2+3wA = w \cdot l = w(2w + 3) = 2w^2 + 3w. Substituting into the cost equation: C=2(2w2+3w)+50=4w2+6w+50C = 2(2w^2 + 3w) + 50 = 4w^2 + 6w + 50. Choice B incorrectly uses the area formula directly without multiplying by 2. Choice C incorrectly distributes 2(3w) as 12w instead of 6w. Choice D forgets to multiply the area by 2 in the cost equation.

Question 5

A rectangular garden has dimensions where the length exceeds the width by 4 meters. If gravel costing $3 per square meter is needed to cover the garden, and the total cost is represented by $C=3w2+12wC = 3w^2 + 12w ,whatdoesthevariable, what does the variable ww $ represent?

  1. The length of the garden in meters
  2. The area of the garden in square meters
  3. The perimeter of the garden in meters
  4. The width of the garden in meters (correct answer)
Explanation: When you encounter a problem involving cost functions and geometric relationships, you need to work backwards from the given equation to understand what each variable represents. The problem states that the length exceeds the width by 4 meters, so if ww is the width, then the length is w+4w + 4. The area of a rectangle equals length × width, giving us w(w+4)=w2+4ww(w + 4) = w^2 + 4w square meters. Since gravel costs $3 per square meter, the total cost would be $C=3(w2+4w)=3w2+12wC = 3(w^2 + 4w) = 3w^2 + 12w .Thismatchesthegivencostequationexactly,confirmingthat. This matches the given cost equation exactly, confirming that ww $ represents the width of the garden in meters. Looking at the wrong answers: Choice A is incorrect because the length would be w + 4 , not w . If w were the length, the width would be w - 4 , giving an area of w(w - 4) = w^2 - 4w , which doesn't match our equation. Choice B is wrong because w represents a linear dimension (width), not the area itself—the area is w^2 + 4w . Choice C is incorrect because the perimeter would be 2w + 2(w + 4) = 4w + 8 , and this linear measurement wouldn't produce the quadratic cost function we see. Remember: when analyzing cost functions involving area, identify the relationship between dimensions first, then express the area in terms of one variable. The coefficient structure in the resulting equation will confirm which dimension your variable represents.

Question 6

A farmer wants to create a rectangular pen against an existing barn wall, using the barn wall as one side. He has 240 feet of fencing for the other three sides. If xx represents the width of the pen (perpendicular to the barn), which expression gives the area AA of the pen?

  1. A=x(240−x)A = x(240 - x)
  2. A=x(240−2x)A = x(240 - 2x) (correct answer)
  3. A=x(120−x)A = x(120 - x)
  4. A=2x(240−2x)A = 2x(240 - 2x)
Explanation: The pen uses fencing for two widths and one length. So 2x+l=2402x + l = 240, which gives l=240−2xl = 240 - 2x. The area is A=x⋅l=x(240−2x)A = x \cdot l = x(240 - 2x). Choice A incorrectly assumes only one width needs fencing. Choice C incorrectly divides the total fencing by 2 before setting up the constraint. Choice D incorrectly includes an extra factor of 2 in front.

Question 7

A manufacturer produces custom picture frames. The cost CC (in dollars) to produce nn frames is given by C=0.5n2+10n+200C = 0.5n^2 + 10n + 200. If they sell each frame for $25, how many frames should they produce to maximize profit?

  1. 10 frames
  2. 20 frames
  3. 15 frames (correct answer)
  4. 25 frames
Explanation: When you encounter profit optimization problems, remember that profit equals revenue minus cost, and maximum profit occurs where the derivative of the profit function equals zero. First, let's set up the profit function. Revenue from selling nn frames at $25 each is $R=25nR = 25n .Giventhatcostis. Given that cost is C=0.5n2+10n+200C = 0.5n^2 + 10n + 200 ,theprofitfunctionbecomes:, the profit function becomes: P=R−C=25n−(0.5n2+10n+200)=−0.5n2+15n−200P = R - C = 25n - (0.5n^2 + 10n + 200) = -0.5n^2 + 15n - 200 $ To find the maximum, take the derivative and set it equal to zero: P' = -n + 15 = 0 n = 15 Since the second derivative P'' = -1 is negative, this confirms we have a maximum at n = 15 frames. Looking at the wrong answers: Choice A (10 frames) gives you a profit that's still increasing since you haven't reached the peak yet. Choice B (20 frames) represents overproduction—you've passed the optimal point and profit is declining due to the quadratic cost structure. Choice D (25 frames) is even further past the optimum, resulting in significantly lower profit. The key insight is that the quadratic cost function creates diminishing returns. Initially, fixed costs are spread over more units, but eventually the 0.5n^2 term dominates, making additional production unprofitable. Remember: for quadratic profit functions, always use calculus to find the exact maximum rather than guessing. The optimal production level balances increasing marginal costs against constant marginal revenue.

Question 8

A ball is dropped from a building and its height hh (in feet) after tt seconds is modeled by h=−16t2+h0h = -16t^2 + h_0, where h0h_0 is the initial height. If the ball takes 3 seconds to hit the ground, what quadratic equation represents the ball's height as a function of time?

  1. h=−16t2+144h = -16t^2 + 144 (correct answer)
  2. h=−16t2+48h = -16t^2 + 48
  3. h=−16t2+96h = -16t^2 + 96
  4. h=−16t2+72h = -16t^2 + 72
Explanation: When the ball hits the ground, h=0h = 0 and t=3t = 3. Substituting: 0=−16(3)2+h0=−16(9)+h0=−144+h00 = -16(3)^2 + h_0 = -16(9) + h_0 = -144 + h_0. Therefore h0=144h_0 = 144 feet. The equation is h=−16t2+144h = -16t^2 + 144. Choice B incorrectly calculates h0=16×3=48h_0 = 16 \times 3 = 48. Choice C uses h0=16×6=96h_0 = 16 \times 6 = 96. Choice D uses h0=24×3=72h_0 = 24 \times 3 = 72.

Question 9

A farmer is designing a rectangular field where the width must be at least 50 feet for equipment access. The length is constrained to be twice the width minus 20 feet. If the total area must be expressed as a function of width ww, which equation is correct?

  1. A=2w2−20wA = 2w^2 - 20w for w≥50w \geq 50 (correct answer)
  2. A=2w2+20wA = 2w^2 + 20w for w≥50w \geq 50
  3. A=2w2−20wA = 2w^2 - 20w for w≥10w \geq 10
  4. A=w2−10wA = w^2 - 10w for w≥50w \geq 50
Explanation: Length =2w−20= 2w - 20. Area =w×(2w−20)=2w2−20w= w \times (2w - 20) = 2w^2 - 20w. The constraint w≥50w \geq 50 comes from the equipment access requirement. Choice B incorrectly calculates length as 2w+202w + 20. Choice C uses the correct area formula but wrong domain (it uses w≥10w \geq 10 to ensure positive length, ignoring the equipment constraint). Choice D incorrectly simplifies the area calculation.

Question 10

A retail store finds that when they price an item at pp dollars, they sell (100−2p)(100 - 2p) items per day. The store pays $15 per item to their supplier. If they want to model their daily profit as a function of price, which equation should they use?

  1. Profit=p(100−2p)−15\text{Profit} = p(100 - 2p) - 15
  2. Profit=100p−2p2−15p\text{Profit} = 100p - 2p^2 - 15p
  3. Profit=(p−15)(100−2p)\text{Profit} = (p - 15)(100 - 2p) (correct answer)
  4. Profit=p(100−2p)−15(100−2p)\text{Profit} = p(100 - 2p) - 15(100 - 2p)
Explanation: When you encounter profit optimization problems, remember that profit equals revenue minus cost. You need to carefully identify what the business earns versus what it spends. Here, the store sells (100−2p)(100 - 2p) items at pp dollars each, so revenue is p(100−2p)p(100 - 2p). However, they pay \15\15 per item to their supplier. The key insight is that their profit per item is (p−15)(p - 15) - the selling price minus the cost per item. Since they sell (100−2p)(100 - 2p) items, total profit is (p−15)(100−2p)(p - 15)(100 - 2p), which is answer C. Let's examine why the other options are incorrect: A) Profit=p(100−2p)−15\text{Profit} = p(100 - 2p) - 15 treats the supplier cost as a fixed \15\15 regardless of how many items are sold. This ignores that cost should increase with quantity sold. B) Profit=100p−2p2−15p\text{Profit} = 100p - 2p^2 - 15p expands to 85p−2p285p - 2p^2, which subtracts only \15p\15p from revenue. This assumes they pay \15\15 per dollar of selling price, not per item. D) Profit=p(100−2p)−15(100−2p)\text{Profit} = p(100 - 2p) - 15(100 - 2p) correctly multiplies the \15\15 cost by the number of items sold, but it's structured as total revenue minus total cost rather than profit-per-item times quantity. Study tip: In profit problems, always identify the profit per unit first, then multiply by the quantity. This approach - (selling price - cost per unit) × quantity sold - will reliably lead you to the correct formula structure.

Question 11

A company manufactures boxes with square bases. The height of each box is 3 inches less than twice the side length of the base. If ss represents the side length of the base in inches, which expression represents the volume when the base dimensions are each increased by 2 inches and the height remains the same?

  1. (s+2)2(2s−3)(s + 2)^2(2s - 3) (correct answer)
  2. (s+2)2(2s−1)(s + 2)^2(2s - 1)
  3. (s+4)(2s−3)(s + 4)(2s - 3)
  4. (s+2)(2s+1)(s + 2)(2s + 1)
Explanation: The original box has a square base with side length ss and height 2s−32s - 3. When the base dimensions are increased by 2 inches each, the new base has side length s+2s + 2. The height remains 2s−32s - 3. The volume of the modified box is (s+2)2(2s−3)(s + 2)^2(2s - 3). Choice B incorrectly modifies the height. Choice C treats the base as having area s+4s + 4 instead of (s+2)2(s + 2)^2. Choice D incorrectly represents both the base area and height.

Question 12

A ball is thrown upward from ground level with an initial velocity of v0v_0 feet per second. Its height after tt seconds is given by h=−16t2+v0th = -16t^2 + v_0 t. If the ball reaches a maximum height of 64 feet, what was the initial velocity v0v_0?

  1. 32 feet per second
  2. 48 feet per second
  3. 64 feet per second (correct answer)
  4. 96 feet per second
Explanation: The maximum height occurs at the vertex of the parabola. For h=−16t2+v0th = -16t^2 + v_0 t, the vertex occurs at t=−v02(−16)=v032t = -\frac{v_0}{2(-16)} = \frac{v_0}{32}. The maximum height is hmax=−16(v032)2+v0(v032)=−16⋅v021024+v0232=−v0264+v0232=−v02+2v0264=v0264h_{max} = -16\left(\frac{v_0}{32}\right)^2 + v_0\left(\frac{v_0}{32}\right) = -16 \cdot \frac{v_0^2}{1024} + \frac{v_0^2}{32} = -\frac{v_0^2}{64} + \frac{v_0^2}{32} = \frac{-v_0^2 + 2v_0^2}{64} = \frac{v_0^2}{64}. Setting this equal to 64: v0264=64\frac{v_0^2}{64} = 64, so v02=4096v_0^2 = 4096, giving v0=64v_0 = 64 feet per second. Choice A would give a maximum height of 16 feet. Choice B would give a maximum height of 36 feet. Choice D would give a maximum height of 144 feet.

Question 13

A rectangular garden has a length that is 6 feet longer than twice its width. If the width is ww feet, which expression represents the area when the garden is expanded by adding a uniform border of xx feet around the entire perimeter?

  1. (w+2x)(2w+6+2x)(w + 2x)(2w + 6 + 2x) (correct answer)
  2. (w+x)(2w+6+x)(w + x)(2w + 6 + x)
  3. w(2w+6)+4x(w+x)w(2w + 6) + 4x(w + x)
  4. (w+2x)(2w+6)+4x(w + 2x)(2w + 6) + 4x
Explanation: The original garden has width ww and length 2w+62w + 6. When a uniform border of xx feet is added around the entire perimeter, the border adds xx feet to each side. So the new width becomes w+x+x=w+2xw + x + x = w + 2x and the new length becomes (2w+6)+x+x=2w+6+2x(2w + 6) + x + x = 2w + 6 + 2x. The area of the expanded garden is (w+2x)(2w+6+2x)(w + 2x)(2w + 6 + 2x). Choice B only adds xx to each dimension instead of 2x2x. Choice C attempts to add the border area separately but doesn't account for the corner overlaps correctly. Choice D incorrectly adds the border area as a simple linear term.

Question 14

A projectile is launched from the top of a 48-foot building with an initial upward velocity of 32 feet per second. The height hh (in feet) of the projectile tt seconds after launch is given by h=−16t2+32t+48h = -16t^2 + 32t + 48. After how many seconds will the projectile return to the same height as the building from which it was launched?

  1. 1 second
  2. 2 seconds (correct answer)
  3. 3 seconds
  4. 4 seconds
Explanation: The projectile was launched from a building that is 48 feet tall. To find when it returns to this height, we set h=48h = 48: 48=−16t2+32t+4848 = -16t^2 + 32t + 48. Subtracting 48 from both sides: 0=−16t2+32t0 = -16t^2 + 32t. Factoring: 0=−16t(t−2)=16t(2−t)0 = -16t(t - 2) = 16t(2 - t). This gives t=0t = 0 or t=2t = 2. Since t=0t = 0 corresponds to the initial launch time, t=2t = 2 seconds is when the projectile returns to the building height. Choice A (1 second) would be the time to reach maximum height. Choice C (3 seconds) and Choice D (4 seconds) are incorrect; the projectile would be below the building height or have already hit the ground by then.

Question 15

A landscaping company is designing a rectangular garden bed that will be surrounded by a decorative stone border. The garden bed itself will have dimensions of length LL feet and width WW feet. The stone border will extend bb feet outward from each edge of the garden bed on all sides.

If the total cost of materials is $8 per square foot for the garden bed and $12 per square foot for the stone border, which expression represents the total cost $CC $ of the project?

  1. C=8LW+12(L+2b)(W+2b)C = 8LW + 12(L + 2b)(W + 2b)
  2. C=8LW+12[(L+2b)(W+2b)−LW]C = 8LW + 12[(L + 2b)(W + 2b) - LW] (correct answer)
  3. C=8LW+12⋅2b(L+W)C = 8LW + 12 \cdot 2b(L + W)
  4. C=8LW+12b(2L+2W+4b)C = 8LW + 12b(2L + 2W + 4b)
Explanation: The total cost consists of the cost for the garden bed plus the cost for the stone border. The garden bed area is LWLW at $8 per square foot, so that cost is $8LW8LW .Thetotalareaincludingborderis. The total area including border is (L+2b)(W+2b)(L + 2b)(W + 2b) .Thestoneborderareaaloneis. The stone border area alone is (L+2b)(W+2b)−LW(L + 2b)(W + 2b) - LW .At$12persquarefootfortheborder,thetotalcostis$. At $12 per square foot for the border, the total cost is $C = 8LW + 12[(L + 2b)(W + 2b) - LW]$$. Choice A incorrectly applies the border cost to the entire area including the garden. Choice C underestimates the border area by not accounting for corners. Choice D attempts to calculate border area as a perimeter-based formula but makes errors in the calculation.

Question 16

A swimming pool has a rectangular surface with length ll and width ww. A concrete walkway of uniform width xx surrounds the pool. If the total area of the pool and walkway combined is 525 square feet, and the pool itself has an area of lw=180lw = 180 square feet, which equation correctly models this situation?

  1. (l+x)(w+x)=525(l + x)(w + x) = 525
  2. (l+2x)(w+2x)=525(l + 2x)(w + 2x) = 525 (correct answer)
  3. lw+2x(l+w)=525lw + 2x(l + w) = 525
  4. lw+4x(l+w+x)=525lw + 4x(l + w + x) = 525
Explanation: When a walkway of uniform width xx surrounds the rectangular pool, it adds xx feet to each side of the pool. The total length becomes l+x+x=l+2xl + x + x = l + 2x and the total width becomes w+x+x=w+2xw + x + x = w + 2x. The total area of pool plus walkway is (l+2x)(w+2x)=525(l + 2x)(w + 2x) = 525. Choice A only adds xx to each dimension instead of 2x2x. Choice C attempts to calculate the walkway area separately but misses the corner areas. Choice D incorrectly calculates the walkway area and includes extra terms.