Function Transformations - IB Mathematics Applications and Interpretation

1. 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, $P_{new}(x)$?
   Answer: $P_{new}(x) = -1.5(x-35)^2 + 29$ The profit $P$ and advertising cost $x$ 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 $25 + 4 = 29$. 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 $(x-40)$ to $(x+5-40) = (x-35)$. Thus, the new function is $P_{new}(x) = -1.5(x-35)^2 + 29$.
2. Which transformation maps the graph of $P_1(t)$ to the graph of $P_2(t)$?
   Answer: A horizontal translation of 20 units to the right. To identify the horizontal translation, we need to write the exponent in the form $k(t-h)$. The exponent in $P_2(t)$ is $0.1t - 2$. We can factor out 0.1: $0.1t - 2 = 0.1(t - 2/0.1) = 0.1(t - 20)$. So, $P_2(t) = 500e^{0.1(t-20)}$. Comparing this to $P_1(t) = 500e^{0.1t}$, we see that $t$ has been replaced by $(t-20)$. This represents a horizontal translation of 20 units to the right.
3. Which of the following functions models the new temperature, $T_{new}(h)$?
   Answer: $T_{new}(h) = -8\cos(\frac{\pi}{12}(h-2)) + 21$ 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 $18 + 3 = 21$. The temperature cycle beginning 2 hours later corresponds to a horizontal translation of 2 units to the right. This means replacing $h$ with $(h-2)$. Therefore, the new function is $T_{new}(h) = -8\cos(\frac{\pi}{12}(h-2)) + 21$.
4. 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 $H(t)$?
   Answer: A vertical translation of 1.5 units downward. Permanently lowering the sea level by 1.5 m means that at any given time $t$, the height of the tide $H(t)$ 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 $H_{new}(t) = H(t) - 1.5$. A stretch or compression would change the amplitude (the difference between max and min tides), which is not stated to happen.
5. Which transformation on the graph of $V(d)$ represents the new value per share, $V_{new}(d)$?
   Answer: A vertical compression by a factor of 2. The value of the stock, $V$, is represented on the vertical axis. Halving the value of each share means that for any given day $d$, 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 $V_{new}(d) = \frac{1}{2}V(d)$.
6. The path of a remote-controlled boat is modelled by the graph of $y=f(x)$. The operator adjusts the controls. The new path, $y=g(x)$, 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 $g(x)$?
   Answer: $g(x) = f(2x) + 5$ Let's apply the transformations in order. 1. '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 $x$ with $2x$ in the function. So, the path becomes $y = f(2x)$. 2. 'Shifting the entire path 5 units upward' is a vertical translation. We add 5 to the function. The path becomes $y = f(2x) + 5$. Thus, $g(x) = f(2x) + 5$.

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