1. Problem statement: Analyze the function $f(x)=x^2$.\n2. Domain: The domain is all real numbers, written as $(-\infty,\infty)$, because squaring is defined for every real input.\n3. Zeros and factorization: Set $f(x)=0$ and solve $x^2=0$ which gives $x=0$. Factorization is $x^2=x\cdot x$.\n4. Symmetry: The function is even because $f(-x)=(-x)^2=x^2=f(x)$, so it is symmetric about the y-axis.\n5. Vertex and shape: Completing the square gives $f(x)=x^2$ which has vertex at $(0,0)$ and opens upward, so the vertex is a minimum.\n6. Derivative (slope): Using the power rule $\frac{d}{dx}x^n = n x^{n-1}$, we get $f'(x)=2x$. Critical point at $x=0$ gives $f'(0)=0$.\n7. Second derivative and concavity: $f''(x)=2>0$ for all x, so the graph is concave up everywhere and the critical point is a local and global minimum.\n8. Antiderivative: An antiderivative is $F(x)=\frac{x^3}{3}+C$, since $\frac{d}{dx} (\frac{x^3}{3})=x^2$.\n9. Intercepts: y-intercept is $f(0)=0$ and x-intercept is at $x=0$, so the graph passes through the origin.\n10. Graph description: This is a parabola with axis of symmetry the y-axis, minimum at $(0,0)$, and grows without bound as $x\to\pm\infty$.\nFinal answer: The function $f(x)=x^2$ has domain $(-\infty,\infty)$, range $[0,\infty)$, vertex $(0,0)$, derivative $f'(x)=2x$, and antiderivative $F(x)=\frac{x^3}{3}+C$.\n