To find the value of $a$ such that the point $(1, 4)$ lies on the graph of the function $f(x) = ax^2 + 2$, we need to use the fact that the point $(1, 4)$ satisfies the function. This means that when $x = 1$, $f(x) = 4$.

Steps:

1. The function is $f(x) = ax^2 + 2$.
2. Substitute $x = 1$ and $f(1) = 4$ into the function: $4 = a(1)^2 + 2$
3. Simplify the equation: $4 = a + 2$
4. Solve for $a$: $a = 4 - 2 = 2$

Thus, the value of $a$ is $\boxed{2}$.

Would you like any more details or further explanation? Here are some related questions:

1. How do we verify if a point lies on a graph?
2. What happens to the graph when we change the value of $a$ in the quadratic equation?
3. Can we find the vertex of the function $f(x) = ax^2 + 2$?
4. How do we find the axis of symmetry for a quadratic function?
5. How can we determine if a quadratic function opens upwards or downwards?

Tip: For quadratic functions, the sign of $a$ in $f(x) = ax^2 + bx + c$ determines whether the parabola opens upwards ($a > 0$) or downwards ($a < 0$).