1. The problem is to analyze and graph the function $f(x) = 2 \cos(5x - 1)$.\n\n2. The general form of a cosine function is $y = A \cos(\beta (x - c)) + D$, where:\n- $A$ is the amplitude (height of peaks),\n- $\beta$ affects the period,\n- $c$ is the phase shift,\n- $D$ is the vertical shift.\n\n3. From the given function, we identify:\n- Amplitude $A = 2$,\n- Inside the cosine, $5x - 1$ means $\beta = 5$ and phase shift $c = \frac{1}{5}$ (since $5(x - \frac{1}{5}) = 5x - 1$),\n- Vertical shift $D = 0$.\n\n4. The period $P$ of the cosine function is given by $P = \frac{2\pi}{\beta} = \frac{2\pi}{5}$.\n\n5. The phase shift is $c = \frac{1}{5}$ to the right.\n\n6. The amplitude is $2$, so the graph oscillates between $2$ and $-2$.\n\n7. The vertical shift is $0$, so the midline is $y=0$.\n\n8. The graph is a cosine wave starting at $x = \frac{1}{5}$ shifted right, with period $\frac{2\pi}{5}$, amplitude $2$, and no vertical shift.\n\n9. The horizontal line segment from $x = -3$ to $x = 1$ at $y = -1$ is a separate feature and not part of the cosine function graph.\n\nFinal answer: The function $f(x) = 2 \cos(5x - 1)$ has amplitude $2$, period $\frac{2\pi}{5}$, phase shift $\frac{1}{5}$ to the right, and vertical shift $0$.\n