1. **State the problem:** We need to analyze the function $y = 2\sin(5x^2 + 4)$.\n\n2. **Formula and rules:** The sine function $\sin(\theta)$ oscillates between $-1$ and $1$. Multiplying by 2 scales the amplitude to oscillate between $-2$ and $2$. The argument inside the sine is $5x^2 + 4$, which affects the frequency and phase.\n\n3. **Amplitude:** The amplitude is the coefficient before sine, which is $2$. This means the graph oscillates between $-2$ and $2$.\n\n4. **Argument analysis:** The argument $5x^2 + 4$ is a quadratic expression in $x$. As $x$ changes, the input to sine changes non-linearly, causing the sine wave to compress or stretch differently than a simple linear argument.\n\n5. **Period:** Normally, sine has period $2\pi$. For $\sin(kx)$, period is $\frac{2\pi}{k}$. Here, the argument is $5x^2 + 4$, not linear in $x$, so the function is not periodic in the usual sense.\n\n6. **Intercepts:** To find $y$-intercept, set $x=0$: $$y=2\sin(5\cdot0^2 +4)=2\sin(4)\approx 2\times -0.7568 = -1.5136.$$\n\n7. **Summary:** The function oscillates between $-2$ and $2$ with a non-linear argument $5x^2 +4$, so it is not periodic in $x$. The $y$-intercept is approximately $-1.5136$.