the following graph shows at least one complete cycle of the graph of an equation containing a trigonometric…

Explanation:

Step1: Determine the amplitude

The amplitude $A$ of a trig - function $y = A\sin(Bx - C)+D$ or $y = A\cos(Bx - C)+D$ is half of the vertical distance between the maximum and minimum values. The maximum value of the graph is $y = 1$ and the minimum value is $y=-6$. So, $A=\frac{1 - (-6)}{2}=\frac{7}{2}$.

Step2: Determine the vertical shift

The vertical shift $D$ is the mid - value between the maximum and minimum values. $D=\frac{1+( - 6)}{2}=-\frac{5}{2}$.

Step3: Determine the period

The period $P$ of the graph is the horizontal distance between two consecutive maxima or minima. Here, $P=\pi$. For a trig - function $y = A\sin(Bx - C)+D$ or $y = A\cos(Bx - C)+D$, the period formula is $P=\frac{2\pi}{|B|}$. Since $P = \pi$, then $\pi=\frac{2\pi}{|B|}$, so $|B| = 2$.

Step4: Determine the phase shift

The graph looks like a cosine function with no phase shift ($C = 0$) because it reaches its maximum at $x =-\frac{\pi}{2}, \frac{\pi}{2},\cdots$. The general form of a cosine function is $y=A\cos(Bx - C)+D$. Substituting $A=\frac{7}{2}$, $B = 2$, $C = 0$ and $D=-\frac{5}{2}$ into the formula, we get $y=\frac{7}{2}\cos(2x)-\frac{5}{2}$.

Answer:

$y=\frac{7}{2}\cos(2x)-\frac{5}{2}$