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.

Learn how to find the range of the quadratic function f(x) = x^2 + 12 with a detailed step-by-step explanation. Explore algebraic concepts like quadratic functions, vertex formula, and range determination. Suitable for students in Grades 8-10.

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