Question

4 hot spot 1 point radian measure g period in radia select the graph with a period of π.

Explanation:

Step1: Recall the formula for the period of a sinusoidal function

The general form of a sinusoidal function is \(y = A\sin(Bx - C)+D\) or \(y = A\cos(Bx - C)+D\), and the period \(T=\frac{2\pi}{|B|}\). We want \(T = \pi\). So, if \(T=\frac{2\pi}{|B|}=\pi\), then solving for \(B\) gives \(|B| = 2\).

Step2: Analyze the graphs

• For the first graph (top - left): The period can be measured from one peak to the next. If we assume it is a sine - like function, by looking at the x - values of the peaks (e.g., from \(x=\frac{\pi}{2}\) to \(x=\frac{5\pi}{2}\)), the period \(T = 2\pi\) (since \(\frac{5\pi}{2}-\frac{\pi}{2}=2\pi\)).
• For the second graph (top - right): If we assume it is a sine - like function, by looking at the x - values of the peaks (e.g., from \(x = 2\pi\) to \(x=6\pi\)), the period \(T=4\pi\).
• For the third graph (bottom - left): If we assume it is a sine - like function, by looking at the x - values of the peaks (e.g., from \(x=\frac{\pi}{4}\) to \(x=\frac{3\pi}{4}\)), the period \(T=\frac{\pi}{2}\) (since \(\frac{3\pi}{4}-\frac{\pi}{4}=\frac{\pi}{2}\)).
• For the fourth graph (bottom - right): If we assume it is a sine - like function, by looking at the x - values of the peaks (e.g., from \(x=\frac{\pi}{2}\) to \(x=\frac{3\pi}{2}\)), the period \(T=\pi\) (since \(\frac{3\pi}{2}-\frac{\pi}{2}=\pi\)).

Answer:

The bottom - right graph.