Question

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

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, \( C \) is the phase shift, and \( D \) is the vertical shift.

Step2: Determine the amplitude

Looking at the graph, the maximum value is \( 1 \) and the minimum value is \( - 3 \). The amplitude \( A \) is half the distance between the maximum and minimum values. So, \( A=\frac{1 - (-3)}{2}=\frac{4}{2} = 2 \).

Step3: Determine the vertical shift \( D \)

The vertical shift \( D \) is the midline of the graph, which is the average of the maximum and minimum values. So, \( D=\frac{1+(-3)}{2}=\frac{-2}{2}=-1 \).

Step4: Determine the phase shift

We know the standard sine function \( y = \sin(x) \) has a phase shift of \( 0 \). Let's compare with our function. We can also check the phase shift by looking at the starting point. Let's consider the function \( y = 2\sin(x + C)-1 \). We know that when \( x =-\frac{\pi}{4} \), let's see the value. Alternatively, we can use the fact that for the function \( y = A\sin(x + C)+D \), the phase shift is \( - C \). We can check the options. Let's consider the options with \( D=-1 \) (from step 3, since \( D=-1 \), options A and B are candidates). Now, let's check the phase shift. The standard \( \sin(x) \) has a zero crossing at \( x = 0 \) going up. Our graph, let's see the zero crossing. Let's take the function \( y=2\sin(x+\frac{\pi}{4})-1 \). Let's find when \( y = 0 \):

\( 0=2\sin(x+\frac{\pi}{4})-1 \)

\( 2\sin(x+\frac{\pi}{4})=1 \)

\( \sin(x+\frac{\pi}{4})=\frac{1}{2} \)

One solution is \( x+\frac{\pi}{4}=\frac{\pi}{6} \), but maybe a better way is to check the phase shift. The standard \( \sin(x) \) shifted left by \( \frac{\pi}{4} \) would be \( \sin(x+\frac{\pi}{4}) \). Let's check the graph. The graph seems to have a phase shift to the left (since the cycle starts earlier). So the phase shift \( C=\frac{\pi}{4} \) (since the function is \( \sin(x+\frac{\pi}{4})=\sin(x - (-\frac{\pi}{4})) \), so phase shift is \( -\frac{\pi}{4} \), meaning left shift of \( \frac{\pi}{4} \)). And the vertical shift \( D=-1 \), amplitude \( A = 2 \). So the function should be \( y = 2\sin(x+\frac{\pi}{4})-1 \), which is option A.

Answer:

A. \( 2\sin(x+\frac{\pi}{4})-1 \)