the figure shows the graph of a trigonometric function f. which of the following could be an expression for…

Explanation:

Step1: Determine the amplitude

The amplitude $A$ of a trig - function $y = A\sin(B(x - C))+D$ or $y = A\cos(B(x - C))+D$ is given by the distance from the mid - line to the maximum or minimum. The mid - line of the given graph is $y=-1$, the maximum is $y = 2$ and the minimum is $y=-4$. So, $A=\frac{2-(-4)}{2}=3$.

Step2: Determine the period

The period $T$ of the function is the distance between two consecutive maxima or minima. From the graph, $T=\pi$. Since the formula for the period of $y = A\sin(B(x - C))+D$ or $y = A\cos(B(x - C))+D$ is $T=\frac{2\pi}{B}$, and $T = \pi$, we have $\pi=\frac{2\pi}{B}$, so $B = 2$.

Step3: Determine the phase - shift and the function type

The graph looks like a cosine function shifted to the right. The general form of a cosine function is $y=A\cos(B(x - C))+D$. We know $A = 3$, $B = 2$ and $D=-1$. When $x=\frac{\pi}{4}$, the function reaches a maximum. For the cosine function $y = A\cos(B(x - C))+D$, when $B(x - C)=2k\pi,k\in\mathbb{Z}$, the function reaches a maximum. Substituting $x = \frac{\pi}{4}$, $A = 3$, $B = 2$ and $D=-1$ into $y=A\cos(B(x - C))+D$, we get $2(\frac{\pi}{4}-C)=2k\pi$. Let $k = 0$, then $\frac{\pi}{2}-2C = 0$, so $C=\frac{\pi}{4}$. The function is $y = 3\cos\left(2\left(x-\frac{\pi}{4}\right)\right)-1$.

Answer:

A. $3\cos\left(2\left(x-\frac{\pi}{4}\right)\right)-1$