1. The problem is to analyze the function $f(x) = 1 + x^2$. 2. This is a quadratic function where the term $x^2$ dominates the shape. 3. The general form of a quadratic function is $f(x) = ax^2 + bx + c$. Here, $a=1$, $b=0$, and $c=1$. 4. Since $a=1 > 0$, the parabola opens upwards. 5. The vertex of the parabola is at $x = -\frac{b}{2a} = -\frac{0}{2 \times 1} = 0$. 6. Evaluating $f(0)$ gives the vertex value: $f(0) = 1 + 0^2 = 1$. 7. The vertex is therefore at $(0,1)$, which is the minimum point of the function. 8. The function has no real roots because $1 + x^2 = 0$ implies $x^2 = -1$, which is impossible for real $x$. 9. The y-intercept is at $f(0) = 1$. 10. The function is symmetric about the y-axis because it contains only even powers of $x$. Final answer: The function $f(x) = 1 + x^2$ is a parabola opening upwards with vertex at $(0,1)$, minimum value 1, no real roots, and y-intercept at 1.