Question

which of the following is an equation for the graph? a. 2 sin(x + \frac{\pi}{4}) - 1 b. 2 sin(x - \frac{\pi}{4}) - 1 c. 2 sin(x + \frac{\pi}{4}) + 1 d. 2 sin(x - \frac{\pi}{4}) + 1

Explanation:

Step1: Recall the general form of a sine - function

The general form of a sine - function is $y = A\sin(B(x - C))+D$, where $A$ is the amplitude, $B$ affects the period ($T=\frac{2\pi}{|B|}$), $C$ is the phase - shift, and $D$ is the vertical shift.

Step2: Determine the amplitude

The amplitude $A$ is half of the vertical distance between the maximum and minimum values of the function. The maximum value of the given graph is $y = 3$ and the minimum value is $y=-1$. So, $A=\frac{3 - (-1)}{2}=\frac{4}{2}=2$.

Step3: Determine the vertical shift

The mid - line of the sine function is the average of the maximum and minimum values. The mid - line is $y=\frac{3+( - 1)}{2}=1$, so $D = 1$.

Step4: Determine the phase - shift

The standard sine function $y=\sin(x)$ has a zero - crossing at $x = 0$. For the given function, when $x=-\frac{\pi}{4}$, $y = 1$ (a point on the mid - line). The general form of the sine function with phase - shift $C$ is $y = A\sin(B(x - C))+D$. For $A = 2$, $B = 1$, $D = 1$, when $x=-\frac{\pi}{4}$ and $y = 1$, substituting into $y=2\sin(x - C)+1$ gives $1=2\sin(-\frac{\pi}{4}-C)+1$. Then $\sin(-\frac{\pi}{4}-C)=0$. A solution is $C=-\frac{\pi}{4}$, which means the function is $y = 2\sin(x+\frac{\pi}{4})+1$.

Answer:

C. $2\sin(x+\frac{\pi}{4})+1$