1. **State the problem:** We are given a sinusoidal graph of $f(x)$ with amplitude 9 and need to find the amplitude, period, and a formula for $f(x)$ in the form $A\cos(Bx)$, $A\sin(Bx)$, or $\pm \tan(Bx)$. 2. **Find the amplitude:** The amplitude is the maximum absolute value of the function. Given the peak is 9 and trough is -9, the amplitude is $$\text{Amplitude} = 9$$ 3. **Find the period:** The period is the length of one full cycle along the x-axis. The graph peaks at $x = -\frac{\pi}{4}$ and again at $x = \frac{\pi}{4}$. Calculate the period: $$\text{Period} = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$$ 4. **Find $B$ in the formula:** The period $T$ relates to $B$ by $$T = \frac{2\pi}{B} \implies B = \frac{2\pi}{T} = \frac{2\pi}{\frac{\pi}{2}} = 4$$ 5. **Determine the function form:** Since the graph has a peak at $x = -\frac{\pi}{4}$ (not at zero), and the wave looks like a cosine shifted, but we want a formula of the form $A\cos(Bx)$ or $A\sin(Bx)$ without phase shift. Check if $f(x) = 9\cos(4x)$ fits: - At $x=0$, $f(0) = 9\cos(0) = 9$ (maximum), but the graph peaks at $x = -\frac{\pi}{4}$, so this is shifted. Check $f(x) = 9\sin(4x)$: - At $x=0$, $f(0) = 0$, which is a zero crossing, consistent with sine. - The peak of sine occurs at $x = \frac{\pi}{8}$ since $\sin(4x)$ peaks when $4x = \frac{\pi}{2} \Rightarrow x = \frac{\pi}{8}$. Given the graph peaks at $x = -\frac{\pi}{4}$, which is $-\frac{\pi}{4} = -2 \times \frac{\pi}{8}$, the function is likely a cosine shifted or negative sine. Try $f(x) = 9\cos(4x)$ shifted by $\frac{\pi}{4}$: $$f(x) = 9\cos\left(4\left(x + \frac{\pi}{4}\right)\right) = 9\cos(4x + \pi) = -9\cos(4x)$$ So $f(x) = -9\cos(4x)$ matches the peak at $x = -\frac{\pi}{4}$. 6. **Final formula:** $$f(x) = -9\cos(4x)$$ **Summary:** - Amplitude: $9$ - Period: $\frac{\pi}{2}$ - Formula: $f(x) = -9\cos(4x)$