1. The problem is to understand and analyze the function $y=\sin(x^2)$. 2. The function is $y=\sin(x^2)$, which means the sine of the square of $x$. 3. Important rule: The sine function, $\sin(\theta)$, oscillates between -1 and 1 for any real number $\theta$. Here, $\theta = x^2$, which is always non-negative. 4. To analyze the function, note that as $x$ increases or decreases, $x^2$ grows larger, causing the sine function to oscillate more rapidly. 5. The function has no simple algebraic simplification but can be graphed to observe its behavior. 6. The domain is all real numbers $x$, and the range is between -1 and 1. 7. The function is even because $\sin((-x)^2) = \sin(x^2)$. 8. The zeros of the function occur when $\sin(x^2) = 0$, which happens when $x^2 = n\pi$ for integer $n$, so $x = \pm \sqrt{n\pi}$. 9. The function oscillates faster as $|x|$ increases due to the $x^2$ inside the sine. 10. Final answer: The function is $y=\sin(x^2)$ with domain $(-\infty, \infty)$ and range $[-1,1]$.