Question

the movement of the progress bar may be uneven because questions can be worth more or less (including zero) depending on your answer. which of the following is the graph of $y = -3 \cdot \cos(3x)$?

Explanation:

Step1: Analyze the amplitude

The general form of a cosine function is \(y = A\cos(Bx)\). Here, \(A=-3\), so the amplitude \(|A| = 3\). This means the graph oscillates between \(y = 3\) and \(y=-3\).

Step2: Analyze the period

The period of the function \(y=\cos(Bx)\) is \(T=\frac{2\pi}{|B|}\). For \(y = - 3\cos(3x)\), \(B = 3\), so the period \(T=\frac{2\pi}{3}\).

Step3: Analyze the reflection

Since \(A=-3<0\), the graph of \(y = \cos(3x)\) is reflected about the \(x -\)axis. When \(x = 0\), \(y=-3\cos(0)=- 3\).

Answer:

The graph that has an amplitude of \(3\), a period of \(\frac{2\pi}{3}\), is reflected about the \(x -\)axis (starts at \((0,-3)\)) is the correct one. (Since the options are not labeled with letters, but based on the analysis of amplitude \(3\), period \(\frac{2\pi}{3}\) and reflection, the graph which has these characteristics is the answer. If we assume the first graph (left - most) has a very small period (maybe \(T=\frac{\pi}{3}\) if it's a standard multiple - choice with 4 graphs as in the problem statement's visual), the second graph has a period of \(6\pi\) (incorrect), the third graph has \(y(0)=3\) (incorrect as \(y(0)=-3\) for \(y = - 3\cos(3x)\)), the fourth graph (assuming it has \(y(0)=-3\), amplitude \(3\) and appropriate period) is the answer. But without proper labeling of options in the problem's image, the key is to check amplitude \(3\), period \(\frac{2\pi}{3}\) and \(y(0)=-3\))