1. The problem is to analyze the function $y = x^2 + 1$. 2. This is a quadratic function in the form $y = ax^2 + bx + c$ where $a=1$, $b=0$, and $c=1$. 3. The graph of this function is a parabola opening upwards because $a > 0$. 4. The vertex of the parabola is at the point where $x = -\frac{b}{2a} = -\frac{0}{2 \times 1} = 0$. 5. Substitute $x=0$ into the function to find the vertex's $y$-coordinate: $y = 0^2 + 1 = 1$. 6. So, the vertex is at $(0,1)$, which is the minimum point of the parabola. 7. The $y$-intercept is the value of $y$ when $x=0$, which is $1$. 8. The $x$-intercepts are the solutions to $x^2 + 1 = 0$. Since $x^2 = -1$ has no real solutions, there are no real $x$-intercepts. 9. Therefore, the parabola lies entirely above the $x$-axis, touching the $y$-axis at $(0,1)$. Final answer: The function $y = x^2 + 1$ has a vertex at $(0,1)$, no real $x$-intercepts, and a $y$-intercept at $(0,1)$.