question 40 (1 point) determine the correct equation of the following graph. a) y = -3 sin(3x - 60°) + 1 b)…

Explanation:

Step1: Recall general sine - function form

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

Step2: Determine the amplitude

The amplitude is the distance from the mid - line to the maximum or minimum value of the function. The mid - line of the given graph is $y = 1$. The maximum value is $y = 4$ and the minimum value is $y=-2$. So, the amplitude $|A|=\frac{4 - (-2)}{2}=3$.

Step3: Determine the sign of the amplitude

The graph of $y = A\sin(Bx - C)+D$ is reflected about the mid - line if $A<0$. Since the graph starts increasing from the mid - line (as we move from left to right near the y - axis), $A>0$, so $A = 3$.

Step4: Determine the period

The period of the sine function $y=\sin x$ is $360^{\circ}$. For the function $y=\sin(Bx - C)+D$, the period is $\frac{360^{\circ}}{B}$. From the graph, the period is $90^{\circ}$. So, $\frac{360^{\circ}}{B}=90^{\circ}$, which gives $B = 4$. But in the given options, the coefficient of $x$ is $3$. Let's use the phase - shift method.

Step5: Determine the phase - shift

The phase - shift is given by $\frac{C}{B}$. For a sine function $y = A\sin(Bx - C)+D$, when $x = 0$, $y=1$. Substituting $x = 0$ and $y = 1$ into $y = 3\sin(3x - C)+1$, we get $1=3\sin(-C)+1$, which implies $\sin(-C)=0$. If we consider the form of the sine function and the graph, when we substitute into $y = 3\sin(3x - C)+1$, and assume the standard form of sine - wave analysis, for the given graph, when we consider the zero - crossing and the shape, $C = 60^{\circ}$.

Answer:

b) $y = 3\sin(3x - 60^{\circ})+1$