IB Mathematics: Applications and Interpretation Quiz: Function Transformations
12 questions · exam conditions
0:00
Function TransformationsQuestion 1 of 12

A company's monthly profit, PP, in thousands of dollars, is modeled by P(x)=−1.5(x−40)2+25P(x) = -1.5(x-40)^2 + 25, where xx is the amount spent on advertising, in thousands of dollars.

The company implements a new efficiency strategy that increases the maximum possible profit by $4000. It also finds that the same profit levels can be achieved by spending $5000 less on advertising. What is the equation of the new profit function, Pnew(x)P_{new}(x)?

Pnew(x)=−1.5(x−35)2+29P_{new}(x) = -1.5(x-35)^2 + 29
Pnew(x)=−1.5(x−45)2+29P_{new}(x) = -1.5(x-45)^2 + 29
Pnew(x)=−1.5(x−35)2+21P_{new}(x) = -1.5(x-35)^2 + 21
Pnew(x)=−1.5(x+5)2+29P_{new}(x) = -1.5(x+5)^2 + 29
← Back to quizzes

IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Function Transformations

Practice Function Transformations 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 Function Transformations, 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.

All questions

Question 1

A company's monthly profit, PP, in thousands of dollars, is modeled by P(x)=−1.5(x−40)2+25P(x) = -1.5(x-40)^2 + 25, where xx is the amount spent on advertising, in thousands of dollars.

The company implements a new efficiency strategy that increases the maximum possible profit by $4000. It also finds that the same profit levels can be achieved by spending $5000 less on advertising. What is the equation of the new profit function, Pnew(x)P_{new}(x)?

  1. Pnew(x)=−1.5(x−35)2+29P_{new}(x) = -1.5(x-35)^2 + 29 (correct answer)
  2. Pnew(x)=−1.5(x−45)2+29P_{new}(x) = -1.5(x-45)^2 + 29
  3. Pnew(x)=−1.5(x−35)2+21P_{new}(x) = -1.5(x-35)^2 + 21
  4. Pnew(x)=−1.5(x+5)2+29P_{new}(x) = -1.5(x+5)^2 + 29
Explanation: The profit PP and advertising cost xx are in thousands of dollars. An increase in maximum profit by $4000 means a vertical shift of 4 units up (since P is in thousands). The new maximum profit term will be 25+4=2925 + 4 = 29. Spending $5000 less on advertising to achieve the same results means the graph shifts 5 units to the left (since x is in thousands). A shift to the left by 5 units changes (x−40)(x-40) to (x+5−40)=(x−35)(x+5-40) = (x-35). Thus, the new function is Pnew(x)=−1.5(x−35)2+29P_{new}(x) = -1.5(x-35)^2 + 29.

Question 2

The daily temperature, TT, in degrees Celsius, in a greenhouse is modelled by the function T(h)=−8cos⁡(π12h)+18T(h) = -8\cos(\frac{\pi}{12}h) + 18, where hh is the number of hours after midnight, 0≤h≤240 \le h \le 24. The owner upgrades the heating system, which keeps the temperature consistently 3°C warmer at all times. They also adjust the lighting timer so that the temperature cycle begins 2 hours later.

Which of the following functions models the new temperature, Tnew(h)T_{new}(h)?

  1. Tnew(h)=−8cos⁡(π12(h+2))+15T_{new}(h) = -8\cos(\frac{\pi}{12}(h+2)) + 15
  2. Tnew(h)=−8cos⁡(π12(h−2))+21T_{new}(h) = -8\cos(\frac{\pi}{12}(h-2)) + 21 (correct answer)
  3. Tnew(h)=−8cos⁡(π12h−2)+21T_{new}(h) = -8\cos(\frac{\pi}{12}h - 2) + 21
  4. Tnew(h)=−5cos⁡(π12(h−2))+18T_{new}(h) = -5\cos(\frac{\pi}{12}(h-2)) + 18
Explanation: A consistently warmer temperature of 3°C corresponds to a vertical translation upwards by 3 units. The original vertical shift was +18, so the new one is 18+3=2118 + 3 = 21. The temperature cycle beginning 2 hours later corresponds to a horizontal translation of 2 units to the right. This means replacing hh with (h−2)(h-2). Therefore, the new function is Tnew(h)=−8cos⁡(π12(h−2))+21T_{new}(h) = -8\cos(\frac{\pi}{12}(h-2)) + 21.

Question 3

The value of a particular stock, VV, in dollars, is modelled by a function V(d)V(d), where dd is the number of days after it was first traded. Due to a stock split, each share is divided into two, halving the value of each individual share.

Which transformation on the graph of V(d)V(d) represents the new value per share, Vnew(d)V_{new}(d)?

  1. A horizontal compression by a factor of 2.
  2. A horizontal stretch by a factor of 2.
  3. A vertical compression by a factor of 2. (correct answer)
  4. A vertical stretch by a factor of 2.
Explanation: The value of the stock, VV, is represented on the vertical axis. Halving the value of each share means that for any given day dd, the new value will be half of the original value. This corresponds to a vertical compression of the graph by a factor of 2 (or a vertical stretch by a factor of 1/2). The new function is Vnew(d)=12V(d)V_{new}(d) = \frac{1}{2}V(d).

Question 4

The population of insects in a colony is modelled by P1(t)=500e0.1tP_1(t) = 500e^{0.1t}. After a change in environmental conditions, the population is modelled by P2(t)=500e0.1t−2P_2(t) = 500e^{0.1t-2}.

Which transformation maps the graph of P1(t)P_1(t) to the graph of P2(t)P_2(t)?

  1. A horizontal translation of 2 units to the right.
  2. A horizontal translation of 20 units to the right. (correct answer)
  3. A horizontal translation of 2 units to the left.
  4. A vertical translation of 2 units downward.
Explanation: To identify the horizontal translation, we need to write the exponent in the form k(t−h)k(t-h). The exponent in P2(t)P_2(t) is 0.1t−20.1t - 2. We can factor out 0.1: 0.1t−2=0.1(t−2/0.1)=0.1(t−20)0.1t - 2 = 0.1(t - 2/0.1) = 0.1(t - 20). So, P2(t)=500e0.1(t−20)P_2(t) = 500e^{0.1(t-20)}. Comparing this to P1(t)=500e0.1tP_1(t) = 500e^{0.1t}, we see that tt has been replaced by (t−20)(t-20). This represents a horizontal translation of 20 units to the right.

Question 5

The path of a remote-controlled boat is modelled by the graph of y=f(x)y=f(x). The operator adjusts the controls. The new path, y=g(x)y=g(x), is found by first halving the boat's horizontal distance from the y-axis at every point, and then shifting the entire path 5 units upward. Which expression represents g(x)g(x)?

  1. g(x)=f(2x)+5g(x) = f(2x) + 5 (correct answer)
  2. g(x)=f(12x)+5g(x) = f(\frac{1}{2}x) + 5
  3. g(x)=12f(x)+5g(x) = \frac{1}{2}f(x) + 5
  4. g(x)=2f(x)+5g(x) = 2f(x) + 5
Explanation: Let's apply the transformations in order.\n1. 'Halving the boat's horizontal distance from the y-axis' means the x-coordinates are halved. This is a horizontal compression by a factor of 2. To achieve this, we replace xx with 2x2x in the function. So, the path becomes y=f(2x)y = f(2x).\n2. 'Shifting the entire path 5 units upward' is a vertical translation. We add 5 to the function. The path becomes y=f(2x)+5y = f(2x) + 5.\nThus, g(x)=f(2x)+5g(x) = f(2x) + 5.

Question 6

The height of a tide, HH, in metres, is modelled by a sinusoidal function H(t)H(t), where tt is the number of hours after midnight. On a certain day, the maximum height is 10 m and the minimum height is 4 m.

A coastal engineering project permanently lowers the sea level in the harbour by 1.5 m. How does this transform the graph of the original function H(t)H(t)?

  1. A vertical compression by a factor of 1.5.
  2. A vertical stretch by a factor of 1.5.
  3. A vertical translation of 1.5 units upward.
  4. A vertical translation of 1.5 units downward. (correct answer)
Explanation: Permanently lowering the sea level by 1.5 m means that at any given time tt, the height of the tide H(t)H(t) will be 1.5 m less than it was originally. This corresponds to a vertical translation of the entire graph by 1.5 units downward. The new function would be Hnew(t)=H(t)−1.5H_{new}(t) = H(t) - 1.5. A stretch or compression would change the amplitude (the difference between max and min tides), which is not stated to happen.

Question 7

The population of insects in a colony is modelled by P1(t)=500e0.1tP_1(t) = 500e^{0.1t}. After a change in environmental conditions, the population is modelled by P2(t)=500e0.1t−2P_2(t) = 500e^{0.1t-2}.

Which transformation maps the graph of P1(t)P_1(t) to the graph of P2(t)P_2(t)?

  1. A horizontal translation of 2 units to the right.
  2. A horizontal translation of 20 units to the right. (correct answer)
  3. A horizontal translation of 2 units to the left.
  4. A vertical translation of 2 units downward.
Explanation: To identify the horizontal translation, we need to write the exponent in the form k(t−h)k(t-h). The exponent in P2(t)P_2(t) is 0.1t−20.1t - 2. We can factor out 0.1: 0.1t−2=0.1(t−2/0.1)=0.1(t−20)0.1t - 2 = 0.1(t - 2/0.1) = 0.1(t - 20). So, P2(t)=500e0.1(t−20)P_2(t) = 500e^{0.1(t-20)}. Comparing this to P1(t)=500e0.1tP_1(t) = 500e^{0.1t}, we see that tt has been replaced by (t−20)(t-20). This represents a horizontal translation of 20 units to the right.

Question 8

The path of a remote-controlled boat is modelled by the graph of y=f(x)y=f(x). The operator adjusts the controls. The new path, y=g(x)y=g(x), is found by first halving the boat's horizontal distance from the y-axis at every point, and then shifting the entire path 5 units upward. Which expression represents g(x)g(x)?

  1. g(x)=f(2x)+5g(x) = f(2x) + 5 (correct answer)
  2. g(x)=f(12x)+5g(x) = f(\frac{1}{2}x) + 5
  3. g(x)=12f(x)+5g(x) = \frac{1}{2}f(x) + 5
  4. g(x)=2f(x)+5g(x) = 2f(x) + 5
Explanation: Let's apply the transformations in order.\n1. 'Halving the boat's horizontal distance from the y-axis' means the x-coordinates are halved. This is a horizontal compression by a factor of 2. To achieve this, we replace xx with 2x2x in the function. So, the path becomes y=f(2x)y = f(2x).\n2. 'Shifting the entire path 5 units upward' is a vertical translation. We add 5 to the function. The path becomes y=f(2x)+5y = f(2x) + 5.\nThus, g(x)=f(2x)+5g(x) = f(2x) + 5.

Question 9

A company's monthly profit, PP, in thousands of dollars, is modeled by P(x)=−1.5(x−40)2+25P(x) = -1.5(x-40)^2 + 25, where xx is the amount spent on advertising, in thousands of dollars.

The company implements a new efficiency strategy that increases the maximum possible profit by $4000. It also finds that the same profit levels can be achieved by spending $5000 less on advertising. What is the equation of the new profit function, Pnew(x)P_{new}(x)?

  1. Pnew(x)=−1.5(x−35)2+29P_{new}(x) = -1.5(x-35)^2 + 29 (correct answer)
  2. Pnew(x)=−1.5(x−45)2+29P_{new}(x) = -1.5(x-45)^2 + 29
  3. Pnew(x)=−1.5(x−35)2+21P_{new}(x) = -1.5(x-35)^2 + 21
  4. Pnew(x)=−1.5(x+5)2+29P_{new}(x) = -1.5(x+5)^2 + 29
Explanation: The profit PP and advertising cost xx are in thousands of dollars. An increase in maximum profit by $4000 means a vertical shift of 4 units up (since P is in thousands). The new maximum profit term will be 25+4=2925 + 4 = 29. Spending $5000 less on advertising to achieve the same results means the graph shifts 5 units to the left (since x is in thousands). A shift to the left by 5 units changes (x−40)(x-40) to (x+5−40)=(x−35)(x+5-40) = (x-35). Thus, the new function is Pnew(x)=−1.5(x−35)2+29P_{new}(x) = -1.5(x-35)^2 + 29.

Question 10

The value of a particular stock, VV, in dollars, is modelled by a function V(d)V(d), where dd is the number of days after it was first traded. Due to a stock split, each share is divided into two, halving the value of each individual share.

Which transformation on the graph of V(d)V(d) represents the new value per share, Vnew(d)V_{new}(d)?

  1. A horizontal compression by a factor of 2.
  2. A horizontal stretch by a factor of 2.
  3. A vertical compression by a factor of 2. (correct answer)
  4. A vertical stretch by a factor of 2.
Explanation: The value of the stock, VV, is represented on the vertical axis. Halving the value of each share means that for any given day dd, the new value will be half of the original value. This corresponds to a vertical compression of the graph by a factor of 2 (or a vertical stretch by a factor of 1/2). The new function is Vnew(d)=12V(d)V_{new}(d) = \frac{1}{2}V(d).

Question 11

The daily temperature, TT, in degrees Celsius, in a greenhouse is modelled by the function T(h)=−8cos⁡(π12h)+18T(h) = -8\cos(\frac{\pi}{12}h) + 18, where hh is the number of hours after midnight, 0≤h≤240 \le h \le 24. The owner upgrades the heating system, which keeps the temperature consistently 3°C warmer at all times. They also adjust the lighting timer so that the temperature cycle begins 2 hours later.

Which of the following functions models the new temperature, Tnew(h)T_{new}(h)?

  1. Tnew(h)=−8cos⁡(π12(h+2))+15T_{new}(h) = -8\cos(\frac{\pi}{12}(h+2)) + 15
  2. Tnew(h)=−8cos⁡(π12(h−2))+21T_{new}(h) = -8\cos(\frac{\pi}{12}(h-2)) + 21 (correct answer)
  3. Tnew(h)=−8cos⁡(π12h−2)+21T_{new}(h) = -8\cos(\frac{\pi}{12}h - 2) + 21
  4. Tnew(h)=−5cos⁡(π12(h−2))+18T_{new}(h) = -5\cos(\frac{\pi}{12}(h-2)) + 18
Explanation: A consistently warmer temperature of 3°C corresponds to a vertical translation upwards by 3 units. The original vertical shift was +18, so the new one is 18+3=2118 + 3 = 21. The temperature cycle beginning 2 hours later corresponds to a horizontal translation of 2 units to the right. This means replacing hh with (h−2)(h-2). Therefore, the new function is Tnew(h)=−8cos⁡(π12(h−2))+21T_{new}(h) = -8\cos(\frac{\pi}{12}(h-2)) + 21.

Question 12

The height of a tide, HH, in metres, is modelled by a sinusoidal function H(t)H(t), where tt is the number of hours after midnight. On a certain day, the maximum height is 10 m and the minimum height is 4 m.

A coastal engineering project permanently lowers the sea level in the harbour by 1.5 m. How does this transform the graph of the original function H(t)H(t)?

  1. A vertical compression by a factor of 1.5.
  2. A vertical stretch by a factor of 1.5.
  3. A vertical translation of 1.5 units upward.
  4. A vertical translation of 1.5 units downward. (correct answer)
Explanation: Permanently lowering the sea level by 1.5 m means that at any given time tt, the height of the tide H(t)H(t) will be 1.5 m less than it was originally. This corresponds to a vertical translation of the entire graph by 1.5 units downward. The new function would be Hnew(t)=H(t)−1.5H_{new}(t) = H(t) - 1.5. A stretch or compression would change the amplitude (the difference between max and min tides), which is not stated to happen.