The given equation is:

$y = x^2 + 2$

This is a quadratic function, which describes a parabola. The key features of this parabola are:

1. Vertex: The vertex of the parabola is at the point where the function reaches its minimum (or maximum, depending on the direction). Since the coefficient of $x^2$ is positive, the parabola opens upwards, and the vertex is the minimum point. In this case, the vertex is at $(0, 2)$.

2. Axis of symmetry: The axis of symmetry is a vertical line that passes through the vertex. For this equation, it is the line $x = 0$.

3. Y-intercept: The y-intercept occurs where $x = 0$. Substituting $x = 0$ into the equation gives $y = 2$, so the y-intercept is $(0, 2)$.

4. Direction of the parabola: Since the coefficient of $x^2$ is positive, the parabola opens upwards.

5. Shape of the graph: The parabola will be wider or narrower depending on the value of the coefficient in front of $x^2$. In this case, with a coefficient of 1, the graph has a standard shape.

Would you like me to plot this equation or dive into more details?

Here are 5 related questions to consider:

1. What is the general form of a quadratic equation, and how does it differ from this one?
2. How do you find the vertex of a quadratic equation in different forms?
3. What happens to the graph if we change the constant $+2$ to another value?
4. How can you solve for $x$ when $y$ is a specific value?
5. What is the significance of the axis of symmetry in a parabola?

Tip: The vertex form of a quadratic equation, $y = a(x-h)^2 + k$, is helpful for quickly identifying the vertex and the direction of the parabola.