Chris wanted to transform the graph of the parent function Y= cot (x) by horizontally compressing it so that it has a period of 2/π units, horizontally Terslating it π/4 units to the right, and vertically translating it 1 unit up. To do so, he graphed the function y= cot (2x-π/4)+1 as shown. What did he do wrong?
The Correct Answer and Explanation is:
To analyze Chris’s transformation of the cotangent function, let’s break down the required transformations step-by-step:
- Parent Function: The parent function is (y = \cot(x)), which has a period of (\pi) (i.e., it repeats every (\pi) units).
- Desired Period: Chris wants to transform the graph to have a period of (\frac{2}{\pi}). The formula for the period of the cotangent function after horizontal scaling is given by:
[
\text{Period} = \frac{\pi}{|b|}
]
where (b) is the coefficient of (x) in the function (y = \cot(bx)). To achieve a period of (\frac{2}{\pi}), we set up the equation:
[
\frac{\pi}{|b|} = \frac{2}{\pi}
]
Solving for (b):
[
|b| = \frac{\pi^2}{2}
]
Therefore, Chris should use (b = \frac{\pi^2}{2}) in the function to achieve the desired period. - Horizontal Translation: Chris intended to shift the graph (\frac{\pi}{4}) units to the right. This is done by replacing (x) with ((x – \frac{\pi}{4})). So the correct form, before vertical translation, is:
[
y = \cot\left(\frac{\pi^2}{2}\left(x – \frac{\pi}{4}\right)\right)
] - Vertical Translation: The vertical translation of 1 unit up is correctly represented by adding 1 to the function, resulting in:
[
y = \cot\left(\frac{\pi^2}{2}\left(x – \frac{\pi}{4}\right)\right) + 1
]
What Chris Did Wrong
Chris’s function (y = \cot(2x – \frac{\pi}{4}) + 1) has a coefficient (b = 2), which means the period he derived is:
[
\text{Period} = \frac{\pi}{2} \text{ (not } \frac{2}{\pi}\text{)}
]
Moreover, the horizontal translation is improperly calculated since ((2x – \frac{\pi}{4})) does not correctly shift the cotangent function right by (\frac{\pi}{4}) units.
Correct Function
The correct transformation of the cotangent function to achieve a period of (\frac{2}{\pi}), a right shift of (\frac{\pi}{4}), and a vertical shift of 1 unit would be:
[
y = \cot\left(\frac{\pi^2}{2}\left(x – \frac{\pi}{4}\right)\right) + 1
]
By using this function, Chris would correctly implement the transformations he intended.