Question

using the graph shown below, determine the function the graph depicts.

a. ( y = 3cos x )
b. ( y = 3sin x )
c. ( y = cos 3x )
d. ( y = sin 3x )

Explanation:

Step1: Recall the general form of sine and cosine functions

The general form of a sine function is \(y = A\sin(Bx)\) and for a cosine function is \(y = A\cos(Bx)\), where \(A\) is the amplitude (\(\vert A\vert=\frac{\text{max}-\text{min}}{2}\)) and the period \(T=\frac{2\pi}{\vert B\vert}\).
For the given graph, when \(x = 0\), \(y = 0\). The cosine function \(y = A\cos(Bx)\) has \(y=A\) when \(x = 0\) (since \(\cos(0)=1\)), and the sine function \(y=A\sin(Bx)\) has \(y = 0\) when \(x = 0\) (since \(\sin(0)=0\)). So we can eliminate options A and C.

Step2: Calculate the amplitude

The amplitude \(A=\frac{\text{max}-\text{min}}{2}\). From the graph, \(\text{max}=3\) and \(\text{min}=-3\). Then \(A=\frac{3 - (- 3)}{2}=\frac{6}{2}=3\).

Step3: Calculate the period

The period \(T\) of the function. The standard period of \(y=\sin(x)\) is \(2\pi\). For \(y = A\sin(Bx)\), \(T=\frac{2\pi}{\vert B\vert}\). From the graph, if we consider one - cycle. Let's find \(B\). We know \(A = 3\), and assume \(y=3\sin(Bx)\). The period of the given graph: if we look at the distance between two consecutive peaks or troughs. The standard \(y = \sin(x)\) has a period of \(2\pi\). For the given graph, when \(y = 3\sin(Bx)\), if we take \(x = 0\) (where \(y = 0\)) and the next \(x\) value where \(y = 0\) (after a full - cycle). The period of the given function is \(2\pi\) (from \(-2\pi\) to \(0\) is a half - cycle, \(0\) to \(2\pi\) is another half - cycle). Using the formula \(T=\frac{2\pi}{\vert B\vert}\), and \(T = 2\pi\), then \(\frac{2\pi}{\vert B\vert}=2\pi\), so \(B = 1\).

Answer:

B. \(y = 3\sin x\)