select the correct answer from each drop - down menu.\nlaura and becky are each graphing a transformation of…
select the correct answer from each drop - down menu.\nlaura and becky are each graphing a transformation of the parent cosine function.\nlauras function is a transformation where the parent function is horizontally compressed by a factor of \\(\\frac{1}{3}\\) and is reflected over the x - axis.\nbeckys function is defined by the equation \\(f(x)=3\\cos(x - \\pi)\\).\ndetermine which graph belongs to each student.\ngraph student answer
Answer
Explanation:
Step1: Analyze Laura's function
The parent - cosine function is (y = \cos(x)). A horizontal compression by a factor of (\frac{1}{3}) gives (y=\cos(3x)), and a reflection over the (x) - axis gives (y =-\cos(3x)). The period of (y = \cos(x)) is (2\pi), and the period of (y=-\cos(3x)) is (\frac{2\pi}{3}) (using the formula (T=\frac{2\pi}{|b|}) for (y = A\cos(bx - c)+d), here (b = 3)).
Step2: Analyze Becky's function
The function (f(x)=3\cos(x-\pi)). Using the identity (\cos(A - B)=\cos A\cos B+\sin A\sin B), (\cos(x-\pi)=-\cos x), so (f(x)=- 3\cos(x)). The period of (y=-3\cos(x)) is (2\pi).
Step3: Match the graphs
The graph with a period of (2\pi) and an amplitude of (3) (the peaks and troughs at (y = 3) and (y=-3)) belongs to Becky. The graph with a period of (\frac{2\pi}{3}) (more frequent cycles) belongs to Laura.
Answer:
| Graph | Student | Answer |
|---|---|---|
| First graph (period (2\pi), amplitude (3)) | Becky | Becky's function (f(x)=3\cos(x - \pi)) |
| Second graph (period (\frac{2\pi}{3})) | Laura | Laura's function (y =-\cos(3x)) |
| Third graph (not relevant based on above analysis) | None |