Math 2 Quiz: Completing The Square For Vertex Form
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Completing The Square For Vertex FormQuestion 1 of 20

A ball is thrown upward from a platform. Its height hh (in feet) above the ground after tt seconds is given by h(t)=−16t2+32t+48h(t) = -16t^2 + 32t + 48. To find when the ball reaches its maximum height, a student begins completing the square. What is the correct vertex form, and what does the vertex represent in this context?

h(t)=−16(t−1)2+64h(t) = -16(t - 1)^2 + 64; the ball reaches a maximum height of 64 feet after 1 second
h(t)=−16(t+1)2+64h(t) = -16(t + 1)^2 + 64; the ball reaches a maximum height of 64 feet after 1 second
h(t)=−16(t−1)2+32h(t) = -16(t - 1)^2 + 32; the ball reaches a maximum height of 32 feet after 1 second
h(t)=−16(t−2)2+48h(t) = -16(t - 2)^2 + 48; the ball reaches a maximum height of 48 feet after 2 seconds
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Math 2 Quiz: Completing The Square For Vertex Form

Practice Completing The Square For Vertex Form in Math 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Completing The Square For Vertex Form, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

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Question 1

A ball is thrown upward from a platform. Its height hh (in feet) above the ground after tt seconds is given by h(t)=−16t2+32t+48h(t) = -16t^2 + 32t + 48. To find when the ball reaches its maximum height, a student begins completing the square. What is the correct vertex form, and what does the vertex represent in this context?

  1. h(t)=−16(t−1)2+64h(t) = -16(t - 1)^2 + 64; the ball reaches a maximum height of 64 feet after 1 second (correct answer)
  2. h(t)=−16(t+1)2+64h(t) = -16(t + 1)^2 + 64; the ball reaches a maximum height of 64 feet after 1 second
  3. h(t)=−16(t−1)2+32h(t) = -16(t - 1)^2 + 32; the ball reaches a maximum height of 32 feet after 1 second
  4. h(t)=−16(t−2)2+48h(t) = -16(t - 2)^2 + 48; the ball reaches a maximum height of 48 feet after 2 seconds
Explanation: To complete the square for h(t)=−16t2+32t+48h(t) = -16t^2 + 32t + 48, factor out -16 from the first two terms: h(t)=−16(t2−2t)+48h(t) = -16(t^2 - 2t) + 48. To complete the square inside the parentheses, take half of the coefficient of tt and square it: (−2/2)2=1(-2/2)^2 = 1. Add and subtract this inside: h(t)=−16(t2−2t+1−1)+48=−16((t−1)2−1)+48=−16(t−1)2+16+48=−16(t−1)2+64h(t) = -16(t^2 - 2t + 1 - 1) + 48 = -16((t-1)^2 - 1) + 48 = -16(t-1)^2 + 16 + 48 = -16(t-1)^2 + 64. The vertex (1,64)(1, 64) means the maximum height of 64 feet occurs at t=1t = 1 second. Choice B has the wrong sign in the vertex form. Choice C incorrectly calculates the maximum height. Choice D uses the wrong time value.

Question 2

An engineer designs a suspension bridge cable following y=0.1x2−2x+15y = 0.1x^2 - 2x + 15, where xx is horizontal distance (in meters) from the left tower and yy is height (in meters) above the roadway. To find the cable's lowest point for roadway clearance, what is the vertex form and its practical meaning?

  1. y=0.1(x+10)2+5y = 0.1(x + 10)^2 + 5; lowest point is 5 meters above roadway, 10 meters before left tower
  2. y=0.1(x−10)2+5y = 0.1(x - 10)^2 + 5; lowest point is 5 meters above roadway, 10 meters from left tower (correct answer)
  3. y=0.1(x−10)2+15y = 0.1(x - 10)^2 + 15; lowest point is 15 meters above roadway, 10 meters from left tower
  4. y=0.1(x−20)2+5y = 0.1(x - 20)^2 + 5; lowest point is 5 meters above roadway, 20 meters from left tower
Explanation: When you encounter a quadratic function in standard form like y=ax2+bx+cy = ax^2 + bx + c, converting to vertex form y=a(x−h)2+ky = a(x - h)^2 + k reveals the parabola's vertex at (h,k)(h, k). For real-world applications like bridge design, the vertex represents the maximum or minimum point—crucial for engineering specifications. To convert y=0.1x2−2x+15y = 0.1x^2 - 2x + 15 to vertex form, complete the square. Factor out the coefficient of x2x^2 from the first two terms: y=0.1(x2−20x)+15y = 0.1(x^2 - 20x) + 15. To complete the square inside the parentheses, take half of the xx-coefficient (−20÷2=−10-20 ÷ 2 = -10), square it ((−10)2=100(-10)^2 = 100), then add and subtract: y=0.1(x2−20x+100−100)+15=0.1((x−10)2−100)+15y = 0.1(x^2 - 20x + 100 - 100) + 15 = 0.1((x - 10)^2 - 100) + 15. Distribute: y=0.1(x−10)2−10+15=0.1(x−10)2+5y = 0.1(x - 10)^2 - 10 + 15 = 0.1(x - 10)^2 + 5. The vertex is at (10,5)(10, 5), meaning the cable's lowest point is 5 meters above the roadway, located 10 meters from the left tower. This confirms answer B. Answer A incorrectly shows (x+10)2(x + 10)^2, placing the vertex at x=−10x = -10 (10 meters before the left tower). Answer C uses the original constant term 15 instead of the correct vertex yy-coordinate 5. Answer D miscalculates the xx-coordinate as 20 instead of 10. Remember: when completing the square, the vertex form directly gives you (h,k)(h, k) from y=a(x−h)2+ky = a(x - h)^2 + k, making it invaluable for optimization problems in engineering contexts.

Question 3

A water fountain shoots water in a parabolic arc given by h(t)=−16t2+24t+4h(t) = -16t^2 + 24t + 4, where hh is height in feet and tt is time in seconds. To design a decorative element at the water's peak, what is the vertex form and when does the peak occur?

  1. h(t)=−16(t+0.75)2+13h(t) = -16(t + 0.75)^2 + 13; peak height of 13 feet occurs 0.75 seconds before launch
  2. h(t)=−16(t−0.75)2+13h(t) = -16(t - 0.75)^2 + 13; peak height of 13 feet at 0.75 seconds (correct answer)
  3. h(t)=−16(t−0.75)2+4h(t) = -16(t - 0.75)^2 + 4; peak height of 4 feet at 0.75 seconds
  4. h(t)=−16(t−1.5)2+13h(t) = -16(t - 1.5)^2 + 13; peak height of 13 feet at 1.5 seconds
Explanation: When you encounter a parabolic function in standard form, converting to vertex form reveals the maximum or minimum point directly. The vertex form is f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k, where (h,k)(h, k) is the vertex coordinates. To convert h(t)=−16t2+24t+4h(t) = -16t^2 + 24t + 4 to vertex form, you'll complete the square. First, factor out the coefficient of t2t^2 from the first two terms: h(t)=−16(t2−1.5t)+4h(t) = -16(t^2 - 1.5t) + 4. To complete the square inside the parentheses, take half of the coefficient of tt (which is −1.5-1.5), square it: (−1.5÷2)2=(−0.75)2=0.5625(-1.5 ÷ 2)^2 = (-0.75)^2 = 0.5625. Add and subtract this inside the parentheses: h(t)=−16(t2−1.5t+0.5625−0.5625)+4=−16((t−0.75)2−0.5625)+4h(t) = -16(t^2 - 1.5t + 0.5625 - 0.5625) + 4 = -16((t - 0.75)^2 - 0.5625) + 4. Distributing: h(t)=−16(t−0.75)2+16(0.5625)+4=−16(t−0.75)2+9+4=−16(t−0.75)2+13h(t) = -16(t - 0.75)^2 + 16(0.5625) + 4 = -16(t - 0.75)^2 + 9 + 4 = -16(t - 0.75)^2 + 13. This gives us vertex (0.75,13)(0.75, 13), meaning the peak occurs at t=0.75t = 0.75 seconds with height 13 feet. Answer B is correct. Answer A incorrectly uses (t+0.75)(t + 0.75), which would place the vertex at t=−0.75t = -0.75 (before launch). Answer C has the right time but wrong peak height (4 instead of 13). Answer D doubles the time to 1.5 seconds, likely from confusing −1.5t-1.5t with the vertex location. Remember: in vertex form a(x−h)2+ka(x - h)^2 + k, the vertex is at (h,k)(h, k), and the sign inside the parentheses is opposite to the hh-coordinate.

Question 4

A satellite dish has a parabolic cross-section described by y=0.25x2−3x+12y = 0.25x^2 - 3x + 12, where xx and yy are in feet. To position the receiver at the focus, engineers first need the vertex form to locate the vertex of the parabola. What is the correct vertex form and vertex location?

  1. y=0.25(x−6)2+12y = 0.25(x - 6)^2 + 12; vertex at (6,12)(6, 12)
  2. y=0.25(x+6)2+3y = 0.25(x + 6)^2 + 3; vertex at (−6,3)(-6, 3)
  3. y=0.25(x−6)2+3y = 0.25(x - 6)^2 + 3; vertex at (6,3)(6, 3) (correct answer)
  4. y=0.25(x−3)2+10.75y = 0.25(x - 3)^2 + 10.75; vertex at (3,10.75)(3, 10.75)
Explanation: When you encounter a parabola in standard form and need the vertex form, you're working with completing the square—a crucial skill for finding the vertex coordinates and understanding parabolic behavior. To convert y=0.25x2−3x+12y = 0.25x^2 - 3x + 12 to vertex form, start by factoring out the coefficient of x2x^2 from the first two terms: y=0.25(x2−12x)+12y = 0.25(x^2 - 12x) + 12. Now complete the square inside the parentheses. Take half of the xx-coefficient: (−12)÷2=−6(-12) ÷ 2 = -6, then square it: (−6)2=36(-6)^2 = 36. Add and subtract this inside the parentheses: y=0.25(x2−12x+36−36)+12=0.25((x−6)2−36)+12y = 0.25(x^2 - 12x + 36 - 36) + 12 = 0.25((x - 6)^2 - 36) + 12. Distribute the 0.25: y=0.25(x−6)2−9+12=0.25(x−6)2+3y = 0.25(x - 6)^2 - 9 + 12 = 0.25(x - 6)^2 + 3. The vertex is at (6,3)(6, 3). Choice A incorrectly keeps the original constant term (12) instead of completing the square properly. Choice B has the wrong sign in the squared term—it should be (x−6)(x - 6), not (x+6)(x + 6)—and places the vertex in the wrong quadrant. Choice D makes an error in the completing the square process, using 3 instead of 6 as the hh-value and miscalculating the kk-value. Remember: in vertex form y=a(x−h)2+ky = a(x - h)^2 + k, the vertex is at (h,k)(h, k). When completing the square, always double-check your arithmetic, especially when factoring out coefficients and distributing them back through.

Question 5

A quadratic function has the form f(x)=3x2−18x+35f(x) = 3x^2 - 18x + 35. A student attempts to complete the square but makes an error in the process. Which of the following correctly shows the vertex form and identifies the vertex?

  1. f(x)=3(x−3)2+8f(x) = 3(x - 3)^2 + 8; vertex at (3,8)(3, 8) (correct answer)
  2. f(x)=3(x+3)2+8f(x) = 3(x + 3)^2 + 8; vertex at (−3,8)(-3, 8)
  3. f(x)=3(x−3)2+35f(x) = 3(x - 3)^2 + 35; vertex at (3,35)(3, 35)
  4. f(x)=3(x−6)2+35f(x) = 3(x - 6)^2 + 35; vertex at (6,35)(6, 35)
Explanation: For f(x)=3x2−18x+35f(x) = 3x^2 - 18x + 35, factor out 3: f(x)=3(x2−6x)+35f(x) = 3(x^2 - 6x) + 35. Complete the square: (−6/2)2=9(-6/2)^2 = 9. So f(x)=3(x2−6x+9−9)+35=3((x−3)2−9)+35=3(x−3)2−27+35=3(x−3)2+8f(x) = 3(x^2 - 6x + 9 - 9) + 35 = 3((x-3)^2 - 9) + 35 = 3(x-3)^2 - 27 + 35 = 3(x-3)^2 + 8. The vertex is (3,8)(3, 8). Choice B has wrong sign for the xx-coordinate. Choice C forgets to distribute the 3 through the subtracted 9. Choice D incorrectly uses 6 instead of 3 as the xx-coordinate of the vertex.

Question 6

The path of a diving board's end can be modeled by y=x2−8x+19y = x^2 - 8x + 19, where xx is the horizontal distance (in feet) from the pool's edge and yy is the height (in feet) above water. To find the lowest point of the dive, what is the vertex form and its interpretation?

  1. y=(x−8)2+19y = (x - 8)^2 + 19; lowest point is 19 feet above water, 8 feet from pool edge
  2. y=(x+4)2+3y = (x + 4)^2 + 3; lowest point is 3 feet above water, 4 feet before pool edge
  3. y=(x−4)2+19y = (x - 4)^2 + 19; lowest point is 19 feet above water, 4 feet from pool edge
  4. y=(x−4)2+3y = (x - 4)^2 + 3; lowest point is 3 feet above water, 4 feet from pool edge (correct answer)
Explanation: When you encounter a quadratic function and need to find its vertex (maximum or minimum point), you need to convert from standard form to vertex form by completing the square. Starting with y=x2−8x+19y = x^2 - 8x + 19, you complete the square on the xx terms. Take half of the coefficient of xx (which is −8-8), so half is −4-4. Square this to get 1616. Now rewrite: y=x2−8x+16+19−16=(x−4)2+3y = x^2 - 8x + 16 + 19 - 16 = (x - 4)^2 + 3. The vertex form y=(x−h)2+ky = (x - h)^2 + k immediately tells you the vertex is at point (h,k)(h, k). Since the coefficient of (x−4)2(x - 4)^2 is positive (it's 1), the parabola opens upward, making this vertex the minimum point. So the lowest point is at (4,3)(4, 3), meaning 4 feet from the pool edge and 3 feet above water. Choice A incorrectly completes the square—you can verify that (x−8)2+19(x - 8)^2 + 19 doesn't equal the original expression when expanded. Choice B has the wrong sign in the squared term; (x+4)2(x + 4)^2 would mean the vertex is at x=−4x = -4, which doesn't make sense for this problem setup. Choice C correctly identifies the xx-coordinate but gives the wrong yy-coordinate—it uses 19 instead of 3, suggesting the student didn't properly subtract the 16 when completing the square. Remember: when completing the square, whatever you add inside the parentheses must be subtracted outside to keep the equation balanced. The vertex form immediately reveals the turning point without additional calculation.

Question 7

A quadratic function modeling population growth is given by P(t)=−t2+14t+40P(t) = -t^2 + 14t + 40, where PP is population (in hundreds) and tt is years after 2020. A demographer needs to determine when maximum population occurs and what that maximum will be. What does completing the square reveal?

  1. P(t)=−(t+7)2+89P(t) = -(t + 7)^2 + 89; maximum population of 8,900 in year 2013
  2. P(t)=−(t−7)2+89P(t) = -(t - 7)^2 + 89; maximum population of 8,900 in year 2027 (correct answer)
  3. P(t)=−(t−7)2+40P(t) = -(t - 7)^2 + 40; maximum population of 4,000 in year 2027
  4. P(t)=−(t−14)2+89P(t) = -(t - 14)^2 + 89; maximum population of 8,900 in year 2034
Explanation: When you encounter a quadratic function in vertex form problems, you're looking to rewrite ax2+bx+cax^2 + bx + c as a(x−h)2+ka(x - h)^2 + k, where the vertex is at (h,k)(h, k). For population models with a negative leading coefficient, the vertex represents the maximum point. To complete the square for P(t)=−t2+14t+40P(t) = -t^2 + 14t + 40, first factor out the negative from the first two terms: P(t)=−(t2−14t)+40P(t) = -(t^2 - 14t) + 40. Take half of the coefficient of tt (which is −14-14), square it: (−14/2)2=49(-14/2)^2 = 49. Add and subtract this inside the parentheses: P(t)=−(t2−14t+49−49)+40=−(t2−14t+49)+49+40=−(t−7)2+89P(t) = -(t^2 - 14t + 49 - 49) + 40 = -(t^2 - 14t + 49) + 49 + 40 = -(t - 7)^2 + 89. This vertex form shows the maximum occurs at t=7t = 7 with a population value of 8989. Since tt represents years after 2020, this means year 2020+7=20272020 + 7 = 2027. The population is in hundreds, so 89×100=8,90089 × 100 = 8,900 people. Choice A incorrectly has (t+7)2(t + 7)^2, which would place the vertex at t=−7t = -7 (year 2013). Choice C correctly identifies the year but uses the wrong kk-value of 40 instead of 89, missing the constant term arithmetic. Choice D places the vertex at t=14t = 14 (year 2034), confusing the coefficient −14-14 with the vertex location. Remember: in vertex form a(x−h)2+ka(x - h)^2 + k, the vertex is at (h,k)(h, k), and when a<0a < 0, you have a maximum at that point.

Question 8

The temperature TT (in degrees Fahrenheit) in a greenhouse hh hours after midnight follows T(h)=2h2−16h+98T(h) = 2h^2 - 16h + 98. The greenhouse manager needs to know when the minimum temperature occurs and what that temperature is. What does completing the square reveal?

  1. T(h)=2(h−4)2+98T(h) = 2(h - 4)^2 + 98; minimum temperature of 98°F occurs at 4:00 AM
  2. T(h)=2(h+4)2+66T(h) = 2(h + 4)^2 + 66; minimum temperature of 66°F occurs at 4:00 PM
  3. T(h)=2(h−4)2+66T(h) = 2(h - 4)^2 + 66; minimum temperature of 66°F occurs at 4:00 AM (correct answer)
  4. T(h)=2(h−8)2+66T(h) = 2(h - 8)^2 + 66; minimum temperature of 66°F occurs at 8:00 AM
Explanation: When you encounter a quadratic function in vertex form questions, you're looking to rewrite the equation as a(x−h)2+ka(x-h)^2 + k where the vertex is at point (h,k)(h,k). Since the coefficient of h2h^2 is positive (2), this parabola opens upward, making the vertex the minimum point. To complete the square for T(h)=2h2−16h+98T(h) = 2h^2 - 16h + 98, first factor out the coefficient from the first two terms: T(h)=2(h2−8h)+98T(h) = 2(h^2 - 8h) + 98. To complete the square inside the parentheses, take half of the coefficient of hh (which is -8), square it: (−8/2)2=16(-8/2)^2 = 16. Add and subtract this inside: T(h)=2(h2−8h+16−16)+98=2((h−4)2−16)+98=2(h−4)2−32+98=2(h−4)2+66T(h) = 2(h^2 - 8h + 16 - 16) + 98 = 2((h-4)^2 - 16) + 98 = 2(h-4)^2 - 32 + 98 = 2(h-4)^2 + 66. This reveals the vertex at (4,66)(4, 66), meaning the minimum temperature of 66°F occurs at h=4h = 4 (4:00 AM). Answer A incorrectly keeps the constant as 98 instead of properly distributing the factored 2. Answer B has the wrong sign in the vertex form (h+4h + 4 instead of h−4h - 4), which would place the minimum at a negative time. Answer D incorrectly identifies the vertex's x-coordinate as 8 instead of 4, likely from confusing the coefficient -8 with the vertex location. Remember: when completing the square, carefully track your arithmetic, especially when factoring out coefficients and distributing them back through the expression.

Question 9

The height hh (in meters) of a rocket tt seconds after launch is given by h(t)=−5t2+30t+80h(t) = -5t^2 + 30t + 80. Mission control needs to determine the rocket's maximum altitude and the time it occurs. After completing the square, what information can be extracted?

  1. h(t)=−5(t−3)2+125h(t) = -5(t - 3)^2 + 125; maximum altitude of 125 meters at 3 seconds after launch (correct answer)
  2. h(t)=−5(t+3)2+125h(t) = -5(t + 3)^2 + 125; maximum altitude of 125 meters at 3 seconds before launch
  3. h(t)=−5(t−3)2+80h(t) = -5(t - 3)^2 + 80; maximum altitude of 80 meters at 3 seconds after launch
  4. h(t)=−5(t−6)2+125h(t) = -5(t - 6)^2 + 125; maximum altitude of 125 meters at 6 seconds after launch
Explanation: For h(t)=−5t2+30t+80h(t) = -5t^2 + 30t + 80, factor out -5: h(t)=−5(t2−6t)+80h(t) = -5(t^2 - 6t) + 80. Complete the square: (−6/2)2=9(-6/2)^2 = 9. So h(t)=−5(t2−6t+9−9)+80=−5((t−3)2−9)+80=−5(t−3)2+45+80=−5(t−3)2+125h(t) = -5(t^2 - 6t + 9 - 9) + 80 = -5((t-3)^2 - 9) + 80 = -5(t-3)^2 + 45 + 80 = -5(t-3)^2 + 125. The vertex (3,125)(3, 125) means the rocket reaches maximum altitude of 125 meters at t=3t = 3 seconds after launch. Choice B has wrong sign giving negative time. Choice C doesn't properly account for the distributed constant. Choice D has wrong vertex xx-coordinate.

Question 10

A parabolic arch for a bridge has the equation y=−0.5x2+4x+6y = -0.5x^2 + 4x + 6, where xx and yy are measured in meters. An engineer needs the vertex form to determine the arch's highest point and optimal support placement. What is the vertex form, and what is the practical meaning of the vertex coordinates?

  1. y=−0.5(x−4)2+14y = -0.5(x - 4)^2 + 14; highest point is 14 meters at horizontal position 4 meters (correct answer)
  2. y=−0.5(x+4)2+14y = -0.5(x + 4)^2 + 14; highest point is 14 meters at horizontal position -4 meters
  3. y=−0.5(x−4)2+6y = -0.5(x - 4)^2 + 6; highest point is 6 meters at horizontal position 4 meters
  4. y=−0.5(x−2)2+8y = -0.5(x - 2)^2 + 8; highest point is 8 meters at horizontal position 2 meters
Explanation: For y=−0.5x2+4x+6y = -0.5x^2 + 4x + 6, factor out -0.5: y=−0.5(x2−8x)+6y = -0.5(x^2 - 8x) + 6. Complete the square: (−8/2)2=16(-8/2)^2 = 16. So y=−0.5(x2−8x+16−16)+6=−0.5((x−4)2−16)+6=−0.5(x−4)2+8+6=−0.5(x−4)2+14y = -0.5(x^2 - 8x + 16 - 16) + 6 = -0.5((x-4)^2 - 16) + 6 = -0.5(x-4)^2 + 8 + 6 = -0.5(x-4)^2 + 14. The vertex (4,14)(4, 14) means the highest point of the arch is 14 meters high, located 4 meters from the reference point horizontally. Choice B has incorrect sign. Choice C doesn't properly add the constant term. Choice D has calculation errors in both coordinates.

Question 11

A company's monthly revenue RR (in thousands) is modeled by R(p)=−4p2+32p−60R(p) = -4p^2 + 32p - 60, where pp is the price per unit in dollars. To maximize revenue, the company needs the vertex form. What price should they charge, and what will be the maximum monthly revenue?

  1. R(p)=−4(p−8)2+4R(p) = -4(p - 8)^2 + 4; charge $8 per unit for maximum revenue of $4,000
  2. R(p)=−4(p+4)2+4R(p) = -4(p + 4)^2 + 4; charge $4 per unit for maximum revenue of $4,000
  3. R(p)=−4(p−4)2−60R(p) = -4(p - 4)^2 - 60; charge $4 per unit for maximum revenue of $60,000
  4. R(p)=−4(p−4)2+4R(p) = -4(p - 4)^2 + 4; charge $4 per unit for maximum revenue of $4,000 (correct answer)
Explanation: When you encounter a quadratic revenue function, you need to convert it to vertex form to find the maximum revenue and optimal price. The vertex form f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k immediately shows the vertex at (h,k)(h, k), where hh is the price that maximizes revenue and kk is the maximum revenue. To convert R(p)=−4p2+32p−60R(p) = -4p^2 + 32p - 60 to vertex form, complete the square. First, factor out the coefficient of p2p^2 from the first two terms: R(p)=−4(p2−8p)−60R(p) = -4(p^2 - 8p) - 60. To complete the square inside the parentheses, take half of the coefficient of pp (which is −8-8), square it: (−8/2)2=16(-8/2)^2 = 16. Add and subtract this inside: R(p)=−4(p2−8p+16−16)−60=−4((p−4)2−16)−60R(p) = -4(p^2 - 8p + 16 - 16) - 60 = -4((p - 4)^2 - 16) - 60. Distribute the −4-4: R(p)=−4(p−4)2+64−60=−4(p−4)2+4R(p) = -4(p - 4)^2 + 64 - 60 = -4(p - 4)^2 + 4. The vertex is at (4,4)(4, 4), meaning charge $4 per unit for maximum revenue of $4,000 (remember the revenue is in thousands). Choice A incorrectly identifies the optimal price as $8 instead of $4. Choice B has the wrong sign in the vertex form, showing $(p+4)2(p + 4)^2 insteadofinstead of (p−4)2(p - 4)^2 .ChoiceCfailstocompletethesquareproperly,showing. Choice C fails to complete the square properly, showing −60-60 astheas the kk −valueinsteadof-value instead of +4+4 $. Study tip: Always double-check your completed square by expanding it back to standard form. This catches sign errors and ensures your vertex coordinates are correct.

Question 12

The profit PP (in thousands of dollars) for a small business is modeled by P(x)=−2x2+16x−30P(x) = -2x^2 + 16x - 30, where xx is the number of years after 2020. A consultant needs to determine when the business will achieve maximum profit. After completing the square, what form should be used, and what business recommendation follows?

  1. P(x)=−2(x−4)2+2P(x) = -2(x - 4)^2 + 2; maximum profit of $2,000 occurs in 2024 (correct answer)
  2. P(x)=−2(x+4)2+2P(x) = -2(x + 4)^2 + 2; maximum profit of $2,000 occurs in 2016
  3. P(x)=−2(x−4)2−30P(x) = -2(x - 4)^2 - 30; maximum profit of $30,000 occurs in 2024
  4. P(x)=−2(x−8)2+2P(x) = -2(x - 8)^2 + 2; maximum profit of $2,000 occurs in 2028
Explanation: Starting with P(x)=−2x2+16x−30P(x) = -2x^2 + 16x - 30, factor out -2: P(x)=−2(x2−8x)−30P(x) = -2(x^2 - 8x) - 30. Complete the square: take half of -8, square it: (−8/2)2=16(-8/2)^2 = 16. Add and subtract: P(x)=−2(x2−8x+16−16)−30=−2((x−4)2−16)−30=−2(x−4)2+32−30=−2(x−4)2+2P(x) = -2(x^2 - 8x + 16 - 16) - 30 = -2((x-4)^2 - 16) - 30 = -2(x-4)^2 + 32 - 30 = -2(x-4)^2 + 2. The vertex (4,2)(4, 2) means maximum profit of $2,000 occurs 4 years after 2020, which is 2024. Choice B has wrong sign. Choice C fails to distribute the -2 correctly. Choice D uses wrong vertex $xx $-coordinate.

Question 13

A farmer wants to fence a rectangular field using 200 feet of fencing, with one side against an existing barn (requiring no fence). If ww is the width perpendicular to the barn, the area is A(w)=w(200−2w)=−2w2+200wA(w) = w(200 - 2w) = -2w^2 + 200w. What is the vertex form, and what dimensions maximize the area?

  1. A(w)=−2(w−100)2+5000A(w) = -2(w - 100)^2 + 5000; width = 100 feet, length = 0 feet, maximum area = 5000 sq ft
  2. A(w)=−2(w+50)2+5000A(w) = -2(w + 50)^2 + 5000; width = 50 feet, length = 100 feet, maximum area = 5000 sq ft
  3. A(w)=−2(w−50)2+200A(w) = -2(w - 50)^2 + 200; width = 50 feet, length = 100 feet, maximum area = 200 sq ft
  4. A(w)=−2(w−50)2+5000A(w) = -2(w - 50)^2 + 5000; width = 50 feet, length = 100 feet, maximum area = 5000 sq ft (correct answer)
Explanation: When you encounter optimization problems involving quadratic functions, you're looking for the maximum or minimum value by converting to vertex form: f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k, where the vertex is at (h,k)(h, k). To convert A(w)=−2w2+200wA(w) = -2w^2 + 200w to vertex form, complete the square. First, factor out the coefficient of w2w^2: A(w)=−2(w2−100w)A(w) = -2(w^2 - 100w). To complete the square inside the parentheses, take half of the coefficient of ww and square it: (−100÷2)2=502=2500(-100 ÷ 2)^2 = 50^2 = 2500. Add and subtract this inside: A(w)=−2(w2−100w+2500−2500)=−2((w−50)2−2500)=−2(w−50)2+5000A(w) = -2(w^2 - 100w + 2500 - 2500) = -2((w - 50)^2 - 2500) = -2(w - 50)^2 + 5000. The vertex is at (50,5000)(50, 5000), meaning maximum area occurs when w=50w = 50 feet. Since the total fencing is 200 feet and covers width + width + length (one width is against the barn), we have 2w+length=2002w + \text{length} = 200, so length =200−2(50)=100= 200 - 2(50) = 100 feet. The maximum area is 5000 square feet. Choice A incorrectly places the vertex at w=100w = 100, which would give length = 0 (impossible). Choice B has the wrong sign in the vertex form, placing the vertex at w=−50w = -50 (meaningless for width). Choice C has the correct vertex location but wrong maximum value, likely from calculation errors in completing the square. Remember: when completing the square, carefully track the algebra, especially when factoring out coefficients. The vertex of a parabola opening downward gives you the maximum value.

Question 14

The cost CC (in dollars) to produce nn items per day at a small factory is C(n)=0.5n2−20n+250C(n) = 0.5n^2 - 20n + 250. The manager wants to find the production level that minimizes cost per day. What does the vertex form tell us about optimal production?

  1. C(n)=0.5(n−20)2+250C(n) = 0.5(n - 20)^2 + 250; produce 20 items for minimum cost of $250 per day
  2. C(n)=0.5(n+20)2+50C(n) = 0.5(n + 20)^2 + 50; produce 20 items for minimum cost of $50 per day
  3. C(n)=0.5(n−20)2+50C(n) = 0.5(n - 20)^2 + 50; produce 20 items for minimum cost of $50 per day (correct answer)
  4. C(n)=0.5(n−10)2+200C(n) = 0.5(n - 10)^2 + 200; produce 10 items for minimum cost of $200 per day
Explanation: When you encounter a quadratic cost function, you're looking at an optimization problem where the vertex form reveals the minimum or maximum point. Since the coefficient of n2n^2 is positive (0.5), this parabola opens upward, meaning the vertex represents the minimum cost. To convert C(n)=0.5n2−20n+250C(n) = 0.5n^2 - 20n + 250 to vertex form, you need to complete the square. First, factor out the coefficient of n2n^2 from the first two terms: C(n)=0.5(n2−40n)+250C(n) = 0.5(n^2 - 40n) + 250. To complete the square inside the parentheses, take half of the coefficient of nn (which is -40), square it: (−20)2=400(-20)^2 = 400. Add and subtract this inside: C(n)=0.5(n2−40n+400−400)+250=0.5((n−20)2−400)+250=0.5(n−20)2−200+250=0.5(n−20)2+50C(n) = 0.5(n^2 - 40n + 400 - 400) + 250 = 0.5((n - 20)^2 - 400) + 250 = 0.5(n - 20)^2 - 200 + 250 = 0.5(n - 20)^2 + 50. This confirms answer C is correct: produce 20 items for a minimum cost of $50. Answer A uses the wrong constant term (250 instead of 50), likely from forgetting to distribute the 0.5 through the -400. Answer B has the wrong sign in the vertex form ($n+20n + 20 insteadofinstead of n−20n - 20 ),whichwouldincorrectlyplacetheminimumat), which would incorrectly place the minimum at n=−20n = -20 $. Answer D shows a calculation error in finding the vertex's x-coordinate, using 10 instead of 20. Always remember: in vertex form a(x - h)^2 + k , the vertex is at (h, k) , and when a > 0 , this point represents the minimum value of the function.

Question 15

The area AA of a rectangular garden with a fixed perimeter of 40 feet can be expressed as A(w)=w(20−w)=−w2+20wA(w) = w(20 - w) = -w^2 + 20w, where ww is the width in feet. A gardener wants to find the dimensions that maximize the area. What is the completed square form, and what dimensions should be used?

  1. A(w)=−(w−10)2+100A(w) = -(w - 10)^2 + 100; width = 10 feet, length = 10 feet for maximum area of 100 sq ft (correct answer)
  2. A(w)=−(w+10)2+100A(w) = -(w + 10)^2 + 100; width = 10 feet, length = 10 feet for maximum area of 100 sq ft
  3. A(w)=−(w−10)2+20A(w) = -(w - 10)^2 + 20; width = 10 feet, length = 10 feet for maximum area of 20 sq ft
  4. A(w)=−(w−20)2+100A(w) = -(w - 20)^2 + 100; width = 20 feet, length = 0 feet for maximum area of 100 sq ft
Explanation: Starting with A(w)=−w2+20wA(w) = -w^2 + 20w, factor out -1: A(w)=−(w2−20w)A(w) = -(w^2 - 20w). Complete the square: (20/2)2=100(20/2)^2 = 100. So A(w)=−(w2−20w+100−100)=−((w−10)2−100)=−(w−10)2+100A(w) = -(w^2 - 20w + 100 - 100) = -((w-10)^2 - 100) = -(w-10)^2 + 100. The vertex (10,100)(10, 100) means maximum area of 100 sq ft when width = 10 feet. Since perimeter is 40 feet, length = 20−10=1020 - 10 = 10 feet. The optimal garden is square. Choice B has wrong sign. Choice C has wrong maximum value. Choice D gives impossible dimensions (length can't be 0).

Question 16

A quadratic function f(x)=x2−8x+19f(x) = x^2 - 8x + 19 models the height of a projectile, where xx represents time in seconds and f(x)f(x) represents height in feet. To find when the projectile reaches its minimum height, Maria completes the square. What is the vertex form of this function, and what does the vertex represent in this context?

  1. f(x)=(x−4)2+3f(x) = (x - 4)^2 + 3; the projectile reaches its minimum height of 3 feet at 4 seconds (correct answer)
  2. f(x)=(x−4)2+19f(x) = (x - 4)^2 + 19; the projectile reaches its minimum height of 19 feet at 4 seconds
  3. f(x)=(x+4)2+3f(x) = (x + 4)^2 + 3; the projectile reaches its minimum height of 3 feet at -4 seconds
  4. f(x)=(x−8)2+3f(x) = (x - 8)^2 + 3; the projectile reaches its minimum height of 3 feet at 8 seconds
Explanation: To complete the square for f(x)=x2−8x+19f(x) = x^2 - 8x + 19: Take half of the coefficient of xx (which is −8/2=−4-8/2 = -4), square it to get 16, then add and subtract 16: f(x)=x2−8x+16+19−16=(x−4)2+3f(x) = x^2 - 8x + 16 + 19 - 16 = (x - 4)^2 + 3. The vertex is at (4,3)(4, 3), meaning the minimum height of 3 feet occurs at 4 seconds. Choice B incorrectly keeps 19 as the constant term. Choice C has the wrong sign in the binomial. Choice D uses the wrong value from completing the square process.

Question 17

The profit function for a small business is given by P(x)=−2x2+16x−26P(x) = -2x^2 + 16x - 26, where xx represents the number of items produced (in hundreds) and P(x)P(x) represents profit in thousands of dollars. If the business owner completes the square to find the vertex form, which statement correctly describes the maximum profit scenario?

  1. Maximum profit of $6,000 occurs when producing 400 items (correct answer)
  2. Maximum profit of $6,000 occurs when producing 800 items
  3. Maximum profit of $32,000 occurs when producing 400 items
  4. Maximum profit of $6,000 occurs when producing 4,000 items
Explanation: Completing the square for P(x)=−2x2+16x−26P(x) = -2x^2 + 16x - 26: Factor out -2 from the first two terms: P(x)=−2(x2−8x)−26P(x) = -2(x^2 - 8x) - 26. Complete the square inside: x2−8x=(x−4)2−16x^2 - 8x = (x - 4)^2 - 16. So P(x)=−2[(x−4)2−16]−26=−2(x−4)2+32−26=−2(x−4)2+6P(x) = -2[(x - 4)^2 - 16] - 26 = -2(x - 4)^2 + 32 - 26 = -2(x - 4)^2 + 6. The vertex is at (4,6)(4, 6). Since xx is in hundreds of items, this means 400 items, and since P(x)P(x) is in thousands of dollars, this means $6,000 profit. Choice B uses 800 items (double). Choice C misinterprets the scale. Choice D uses 4,000 items (ten times too large).

Question 18

The function h(t)=−16t2+64t+80h(t) = -16t^2 + 64t + 80 represents the height of a ball thrown upward, where tt is time in seconds and h(t)h(t) is height in feet. After completing the square to find vertex form, at what time does the ball reach its maximum height, and what is that maximum height?

  1. Maximum height of 144 feet at t=2t = 2 seconds (correct answer)
  2. Maximum height of 144 feet at t=4t = 4 seconds
  3. Maximum height of 80 feet at t=2t = 2 seconds
  4. Maximum height of 64 feet at t=2t = 2 seconds
Explanation: To complete the square for h(t)=−16t2+64t+80h(t) = -16t^2 + 64t + 80: Factor out -16 from the first two terms: h(t)=−16(t2−4t)+80h(t) = -16(t^2 - 4t) + 80. Complete the square inside: add and subtract (4/2)2=4(4/2)^2 = 4: h(t)=−16(t2−4t+4−4)+80=−16((t−2)2−4)+80=−16(t−2)2+64+80=−16(t−2)2+144h(t) = -16(t^2 - 4t + 4 - 4) + 80 = -16((t - 2)^2 - 4) + 80 = -16(t - 2)^2 + 64 + 80 = -16(t - 2)^2 + 144. The vertex is at (2,144)(2, 144), so maximum height is 144 feet at t=2t = 2 seconds. Choice B has the wrong time. Choice C gives the initial height instead of maximum height. Choice D uses the coefficient value as the height.

Question 19

The area of a rectangular garden is modeled by A(w)=w2+8w−20A(w) = w^2 + 8w - 20, where ww represents the width in feet and the length is (w+10)(w + 10) feet. By completing the square, determine the width that minimizes the area and interpret why this result might seem unusual in a real-world context.

  1. Width of 8 feet minimizes area to -36 square feet; this is unusual because the area should be maximized, not minimized
  2. Width of 4 feet minimizes area to -36 square feet; this is unusual because negative area is not physically meaningful
  3. Width of -4 feet minimizes area to 20 square feet; this is unusual because negative width is not physically meaningful
  4. Width of -4 feet minimizes area to -36 square feet; this is unusual because negative dimensions and area are not physically meaningful (correct answer)
Explanation: When you encounter a quadratic function representing area in a real-world context, you need to find the vertex using completing the square, then carefully interpret whether the mathematical result makes physical sense. To complete the square for A(w)=w2+8w−20A(w) = w^2 + 8w - 20, you take half the coefficient of ww (which is 8), square it to get 16, then rewrite: A(w)=(w2+8w+16)−16−20=(w+4)2−36A(w) = (w^2 + 8w + 16) - 16 - 20 = (w + 4)^2 - 36. The vertex form shows the minimum occurs at w=−4w = -4 with an area of −36-36 square feet. Choice A incorrectly identifies the width as 8 feet, likely confusing the coefficient 8 with the vertex location. The interpretation about maximizing versus minimizing area misses the real issue. Choice B gets the correct minimum area of -36 but incorrectly calculates the width as 4 feet, probably forgetting the negative sign when solving (w+4)2=0(w + 4)^2 = 0. The interpretation only identifies negative area as problematic. Choice C makes an error in calculating the minimum area as 20 instead of -36, possibly substituting incorrectly into the original function. Choice D correctly identifies that the width of -4 feet produces a minimum area of -36 square feet, and properly recognizes that both negative width and negative area are physically meaningless in this context. Study tip: When completing the square in real-world problems, always check if your mathematical answer makes physical sense. Negative dimensions, areas, or volumes should trigger you to reconsider the domain or recognize limitations of the mathematical model.

Question 20

A company's daily revenue function is R(x)=−3x2+30x+48R(x) = -3x^2 + 30x + 48, where xx represents the number of workers and R(x)R(x) represents revenue in hundreds of dollars. If the company completes the square to find the optimal number of workers, what is the vertex form, and how should management interpret this result for staffing decisions?

  1. R(x)=−3(x−5)2+123R(x) = -3(x - 5)^2 + 123; hire 5 workers for maximum daily revenue of $123
  2. R(x)=−3(x−10)2+123R(x) = -3(x - 10)^2 + 123; hire 10 workers for maximum daily revenue of $12,300
  3. R(x)=−3(x−5)2+123R(x) = -3(x - 5)^2 + 123; hire 5 workers for maximum daily revenue of $12,300 (correct answer)
  4. R(x)=−3(x−5)2+75R(x) = -3(x - 5)^2 + 75; hire 5 workers for maximum daily revenue of $7,500
Explanation: When you encounter a quadratic revenue function, you're looking at an optimization problem where completing the square reveals the maximum or minimum point. The vertex form f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k immediately shows the vertex at (h,k)(h, k), which represents the optimal input and output values. To complete the square for R(x)=−3x2+30x+48R(x) = -3x^2 + 30x + 48, first factor out the coefficient of x2x^2 from the first two terms: R(x)=−3(x2−10x)+48R(x) = -3(x^2 - 10x) + 48. To complete the square inside the parentheses, take half of the coefficient of xx (which is −10/2=−5-10/2 = -5) and square it: (−5)2=25(-5)^2 = 25. Add and subtract this inside: R(x)=−3(x2−10x+25−25)+48=−3((x−5)2−25)+48=−3(x−5)2+75+48=−3(x−5)2+123R(x) = -3(x^2 - 10x + 25 - 25) + 48 = -3((x - 5)^2 - 25) + 48 = -3(x - 5)^2 + 75 + 48 = -3(x - 5)^2 + 123. The vertex is at (5,123)(5, 123), meaning 5 workers generate maximum revenue of 123 hundreds of dollars, or $12,300. Choice A has the correct vertex form but misinterprets the units—forgetting that revenue is in hundreds of dollars. Choice B incorrectly identifies 10 workers as optimal, likely confusing the coefficient manipulation during completing the square. Choice D makes an arithmetic error in the final calculation, getting 75 instead of 123. Always pay attention to units in word problems—they're often the difference between answer choices and reflect real-world interpretation skills that math exams prioritize.