1. **Problem Statement:** We are given the function $y = \sec \theta \tan \theta$ and a graph showing this function between $\theta = -\frac{\pi}{4}$ and $\theta = \frac{\pi}{4}$. We want to understand the behavior of this function and the shaded area under the curve from $\theta = 0$ to $\theta = \frac{\pi}{4}$.\n\n2. **Recall the definitions:** \n- $\sec \theta = \frac{1}{\cos \theta}$\n- $\tan \theta = \frac{\sin \theta}{\cos \theta}$\n\n3. **Rewrite the function:**\n$$y = \sec \theta \tan \theta = \frac{1}{\cos \theta} \times \frac{\sin \theta}{\cos \theta} = \frac{\sin \theta}{\cos^2 \theta}$$\nThis means the function depends on sine and cosine values of $\theta$.\n\n4. **Domain considerations:**\nThe function is defined where $\cos \theta \neq 0$. Between $-\frac{\pi}{4}$ and $\frac{\pi}{4}$, $\cos \theta$ is positive and nonzero, so the function is continuous there.\n\n5. **Behavior at key points:**\n- At $\theta = 0$, $\sin 0 = 0$, so $y = 0$.\n- At $\theta = \frac{\pi}{4}$, $\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$ and $\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$, so\n$$y = \frac{\frac{\sqrt{2}}{2}}{\left(\frac{\sqrt{2}}{2}\right)^2} = \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}} = \sqrt{2}$$\nThis matches the graph's top boundary.\n\n6. **Shaded area:**\nThe shaded region is bounded by the curve from $\theta = 0$ to $\theta = \frac{\pi}{4}$, the vertical line at $\theta = 0$, and the horizontal line at $y = \sqrt{2}$. This area can be found by integrating the function over $[0, \frac{\pi}{4}]$.\n\n7. **Summary:**\n- The function $y = \sec \theta \tan \theta$ can be rewritten as $\frac{\sin \theta}{\cos^2 \theta}$.\n- It is continuous and defined on the interval $[-\frac{\pi}{4}, \frac{\pi}{4}]$.\n- The function value at $\frac{\pi}{4}$ is $\sqrt{2}$, matching the graph.\n- The shaded area under the curve from $0$ to $\frac{\pi}{4}$ represents the integral of the function over that interval.\n\nThis explanation helps understand the graph and the function's behavior step by step.