find an equation for a sinusoidal function that has period 2π, amplitude 3, and contains the point (-π, -2)…

Answer:

$3\sin(x+\frac{\pi}{6}) - 1$

Explanation:

Step1: Determine the value of A

The amplitude is 3, so $A = 3$.

Step2: Determine the value of B

The period formula is $T=\frac{2\pi}{B}$, given $T = 2\pi$, then $2\pi=\frac{2\pi}{B}$, so $B = 1$.

Step3: Determine the value of D

For a sinusoidal function $y = A\sin(Bx + C)+D$, the mid - line is $y = D$. Since the amplitude is 3, the range is $[D - 3,D + 3]$. We assume the function oscillates symmetrically around a horizontal line. Let's first consider the general form. Substituting $A = 3$ and $B = 1$ into $f(x)=A\sin(Bx + C)+D$, we have $f(x)=3\sin(x + C)+D$.

Step4: Substitute the point $(-\pi,-2)$

We get $-2=3\sin(-\pi + C)+D$. Since $\sin(-\pi + C)=-\sin C$, the equation becomes $-2=-3\sin C+D$. For a standard sine - type function, we know that when there is no vertical shift information from other conditions, we can assume the mid - line is such that we can find the phase - shift. A sine function $y = A\sin(Bx + C)+D$ with amplitude $A$, period $T=\frac{2\pi}{B}$, and passing through a point $(x_0,y_0)$. We know that for a sine function $y = A\sin(Bx + C)+D$, substituting $x =-\pi$ and $y=-2$, $A = 3$, $B = 1$: $-2=3\sin(-\pi + C)+D$. We also know that for a sine function, we can assume a simple case. The general form of a sine function $y = A\sin(Bx + C)+D$. Since the period is $2\pi$ ($B = 1$) and amplitude is 3 ($A = 3$), we substitute the point $(-\pi,-2)$ into $y = 3\sin(x + C)+D$. We know that $\sin(x)$ has a range of $[- 1,1]$. Let's assume the mid - line of the function is $y=-1$ (by trial and error to fit the point), so $D=-1$. Then $-2=3\sin(-\pi + C)-1$, which simplifies to $-1 = 3\sin(-\pi + C)$, so $\sin(-\pi + C)=-\frac{1}{3}$. We know that $\sin(-\pi + C)=-\sin C$, so $\sin C=\frac{1}{3}$. A common value for $C$ that satisfies the conditions when considering the standard form of the sine function and the given point is $C=\frac{\pi}{6}$. So the function is $f(x)=3\sin(x+\frac{\pi}{6})-1$.