find an equation for the sine graph, $f(x)$:\nwrite your answer in the form $f(x) = a\\sin(bx + c) + d$…

Explanation:

Step1: Find vertical shift (D)

The midline is the average of max/min values. Max = 0, Min = -2. $$D = \frac{0 + (-2)}{2} = -1$$

Step2: Find amplitude (A)

Amplitude is half the range. Range = $0 - (-2) = 2$. $$A = \frac{2}{2} = -1$$ (negative because the graph is inverted from standard sine at x=0)

Step3: Find period and B

The period is the distance between two peaks: $\pi - (-\pi) = 2\pi$. Period formula: $\frac{2\pi}{|B|}=2\pi$. $$B = \frac{2\pi}{2\pi} = 1$$

Step4: Find phase shift (C)

Use the point $(0, -1)$: substitute into $f(x)=A\sin(Bx+C)+D$. $$-1 = -1\sin(0 + C) - 1$$ $$0 = -\sin(C) \implies \sin(C)=0$$ Use the behavior at $x=0$ (graph increasing), so $C = -\frac{\pi}{2}$ (to shift sine left/right to match the curve). $$-1\sin\left(1\cdot0 -\frac{\pi}{2}\right) -1 = -1(-1)-1=0? \text{ No, correct: use } f(0)=-1$$ $$-1 = -1\sin(C) -1 \implies \sin(C)=0, \text{ and at } x=\frac{\pi}{2}, f(x)=0:$$ $$0 = -1\sin\left(\frac{\pi}{2}+C\right)-1 \implies \sin\left(\frac{\pi}{2}+C\right)=-1$$ $$\frac{\pi}{2}+C = \frac{3\pi}{2} \implies C = \pi$$ Verify: $f(0)=-1\sin(\pi)-1=0-1=-1$, correct.

Answer:

$$f(x) = -\sin(x + \pi) - 1$$ Or simplified (using $\sin(x+\pi)=-\sin x$): $f(x)=\sin(x)-1$, which also matches the graph.