which of the following is the graph of y = cos(2x)?

Answer:

The general form of a cosine - function is $y = A\cos(Bx - C)+D$. For the function $y=\cos(2x)$, we have $A = 1$, $B = 2$, $C = 0$, and $D = 0$.

The period of the cosine function $y=\cos(x)$ is $T = 2\pi$. For the function $y=\cos(Bx)$, the period is given by the formula $T=\frac{2\pi}{|B|}$.

Step1: Calculate the period

For $y=\cos(2x)$, since $B = 2$, the period $T=\frac{2\pi}{2}=\pi$.

The standard cosine function $y = \cos(x)$ has a period of $2\pi$, and the function $y=\cos(2x)$ oscillates twice as fast as $y=\cos(x)$ (because of the coefficient 2 in front of $x$), so its period is $\pi$.

The graph of $y = \cos(2x)$ will complete one full cycle in the interval $[0,\pi]$. Among the possible graphs (assuming the first graph has a period of $\pi$ and the second graph has a non - standard shape for $y=\cos(2x)$), the graph with a period of $\pi$ is the correct one for $y=\cos(2x)$. Without seeing all the options clearly, if one of the graphs has a period of $\pi$ (i.e., the distance between two consecutive peaks or troughs is $\pi$), that is the graph of $y=\cos(2x)$.

If we assume the first graph has a period of $\pi$ and the second graph has a different shape (not a cosine - like shape with the correct period), the answer is the graph with a period of $\pi$.

If we consider the key points of $y=\cos(2x)$: When $x = 0$, $y=\cos(0)=1$; when $x=\frac{\pi}{4}$, $y=\cos(\frac{\pi}{2}) = 0$; when $x=\frac{\pi}{2}$, $y=\cos(\pi)=-1$; when $x=\frac{3\pi}{4}$, $y=\cos(\frac{3\pi}{2}) = 0$; when $x=\pi$, $y=\cos(2\pi)=1$.

The graph of $y = \cos(2x)$ is a cosine - wave that oscillates with a period of $\pi$ and has an amplitude of 1 (since $A = 1$).

If the first graph shows a cosine - wave with a period of $\pi$ (peaks at $x = 0,\pi,2\pi$ etc. and troughs at $x=\frac{\pi}{2},\frac{3\pi}{2}$ etc.), then the answer is the first graph.