1. The problem is to analyze the function $y = 2\sin(5x^2 + 4)$.\n\n2. The general form of a sine function is $y = A\sin(Bx + C)$, where $A$ is the amplitude, $B$ affects the period, and $C$ is the phase shift. Here, the argument of sine is $5x^2 + 4$, which is a quadratic expression, making this a composite function.\n\n3. The amplitude $A$ is the coefficient before sine, which is 2. This means the sine wave oscillates between $-2$ and $2$.\n\n4. Since the argument is $5x^2 + 4$, the function is not periodic in the usual sense because $x^2$ is not linear. The sine function will oscillate faster as $|x|$ increases because $5x^2$ grows quadratically.\n\n5. To find intercepts, set $y=0$: \n$$0 = 2\sin(5x^2 + 4) \implies \sin(5x^2 + 4) = 0.$$\n\n6. The sine function is zero at integer multiples of $\pi$: \n$$5x^2 + 4 = k\pi, \quad k \in \mathbb{Z}.$$\n\n7. Solve for $x^2$: \n$$x^2 = \frac{k\pi - 4}{5}.$$\n\n8. Real solutions exist only if $\frac{k\pi - 4}{5} \geq 0$, so \n$$k\pi \geq 4.$$\n\n9. For such $k$, \n$$x = \pm \sqrt{\frac{k\pi - 4}{5}}.$$\n\n10. Extrema occur where the derivative is zero. The derivative is \n$$y' = 2\cos(5x^2 + 4) \cdot 10x = 20x \cos(5x^2 + 4).$$\n\n11. Set $y' = 0$: \n$$20x \cos(5x^2 + 4) = 0 \implies x=0 \text{ or } \cos(5x^2 + 4) = 0.$$\n\n12. For $\cos(5x^2 + 4) = 0$, \n$$5x^2 + 4 = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}.$$\n\n13. Solve for $x^2$: \n$$x^2 = \frac{\frac{\pi}{2} + k\pi - 4}{5}.$$\n\n14. Real solutions exist if the numerator is non-negative.\n\n15. The function has amplitude 2, oscillates with argument $5x^2 + 4$, and has intercepts and extrema as described.\n\nFinal answer: The function $y=2\sin(5x^2 + 4)$ has amplitude 2, zeros at $x=\pm \sqrt{\frac{k\pi - 4}{5}}$ for integers $k$ with $k\pi \geq 4$, and critical points at $x=0$ and $x=\pm \sqrt{\frac{\frac{\pi}{2} + k\pi - 4}{5}}$ where $\cos(5x^2 + 4)=0$.