The function you provided is:

$f(x) = x^2 + 12$

Step-by-step explanation:

1. Form of the function: This is a quadratic function, $f(x) = x^2 + 12$, which is a parabola that opens upwards because the coefficient of $x^2$ is positive.

2. Minimum value of $f(x)$: The vertex of a quadratic function $f(x) = ax^2 + bx + c$ occurs at $x = -\frac{b}{2a}$. In this case, $a = 1$, $b = 0$, and $c = 12$. Therefore, the vertex is at $x = 0$.

   Substituting $x = 0$ into the function: $f(0) = 0^2 + 12 = 12$ So, the minimum value of $f(x)$ is 12.

3. Behavior as $x \to \infty$: As $x$ increases or decreases, $x^2$ grows larger, and since $f(x) = x^2 + 12$, $f(x) \to \infty$.

4. Range of the function: The smallest value of $f(x)$ is 12, and it increases without bound. Therefore, the range of the function is:

   $\boxed{[12, \infty)}$

Would you like more details or have any other questions?

Here are five related questions:

1. What is the vertex form of a quadratic function?
2. How do you find the domain of a quadratic function?
3. Can the range of any quadratic function ever be all real numbers?
4. What is the difference between the domain and the range of a function?
5. How does shifting a parabola affect its range?

Tip: The range of a quadratic function can be found by determining whether it opens upwards or downwards and locating its vertex.