1. The problem is to analyze the function $y = x^4$ and understand its properties. 2. This function is a polynomial of degree 4. It is even because $y(-x) = (-x)^4 = x^4 = y(x)$. 3. To find intercepts, set $y=0$: $$x^4 = 0 \\ x = 0$$ The graph passes through the origin (0,0). 4. To find critical points, compute the derivative: $$y' = 4x^3$$ Setting $y'=0$ gives $4x^3=0 \\ x=0$. 5. To determine the nature of critical point at $x=0$, compute the second derivative: $$y'' = 12x^2$$ At $x=0$, $y''=0$ so the second derivative test is inconclusive. However, since $y = x^4$ is positive for all $x eq 0$ and zero only at 0, the point at (0,0) is a minimum. 6. The graph shape is a parabola-like curve opening upward, flatter near the origin compared to $x^2$, symmetric about the y-axis. Final answer: The function $y = x^4$ has a minimum at (0,0), is symmetric, with intercept at origin, and derivative $y' = 4x^3$.