1. **State the problem:** We are given the function $$f(x) = 4 \cos\left(4\pi \left(x + \frac{1}{6}\right)\right) - 3$$ and want to understand its key features such as amplitude, period, phase shift, vertical shift, and verify given points. 2. **Recall the general form of a cosine function:** $$f(x) = A \cos(B(x - C)) + D$$ where: - $A$ is the amplitude (height from center line to peak), - $B$ affects the period $T = \frac{2\pi}{|B|}$, - $C$ is the horizontal shift (phase shift), - $D$ is the vertical shift. 3. **Identify parameters from the given function:** - Amplitude $A = 4$ - Inside cosine: $4\pi (x + \frac{1}{6}) = 4\pi x + \frac{4\pi}{6} = 4\pi x + \frac{2\pi}{3}$ - So $B = 4\pi$ - Phase shift $C = -\frac{1}{6}$ (since $x + \frac{1}{6} = x - (-\frac{1}{6})$) - Vertical shift $D = -3$ 4. **Calculate the period:** $$T = \frac{2\pi}{|B|} = \frac{2\pi}{4\pi} = \frac{1}{2}$$ 5. **Check the given points:** - At $x=0$: $$f(0) = 4 \cos\left(4\pi \left(0 + \frac{1}{6}\right)\right) - 3 = 4 \cos\left(\frac{4\pi}{6}\right) - 3 = 4 \cos\left(\frac{2\pi}{3}\right) - 3$$ Since $\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$, $$f(0) = 4 \times \left(-\frac{1}{2}\right) - 3 = -2 - 3 = -5$$ This matches the point $(0, -5)$. - At $x = -\frac{1}{6}$: $$f\left(-\frac{1}{6}\right) = 4 \cos\left(4\pi \left(-\frac{1}{6} + \frac{1}{6}\right)\right) - 3 = 4 \cos(0) - 3 = 4 \times 1 - 3 = 1$$ This matches the point $\left(-\frac{1}{6}, 1\right)$. - At $x=0$, the point $(0, -3)$ is not on the function since $f(0) = -5$, so $(0, -3)$ is not on the graph. - The point $(0,1)$ is also not on the graph since $f(0) = -5$. 6. **Summary:** - Amplitude: 4 - Period: $\frac{1}{2}$ - Phase shift: $-\frac{1}{6}$ - Vertical shift: $-3$ - Verified points: $(0, -5)$ and $\left(-\frac{1}{6}, 1\right)$ are on the graph. Final answer: The function is a cosine wave with amplitude 4, period $\frac{1}{2}$, phase shift $-\frac{1}{6}$, vertical shift $-3$, and the points $(0, -5)$ and $\left(-\frac{1}{6}, 1\right)$ lie on the graph.