Let $$f(x)$$ be a quadratic polynomial in $$x$$ such that $$f(x) \geq 0$$ for all real numbers $$x$$. If $$f(2) = 0$$ and $$f(4) = 6$$, then $$f(-2)$$ is equal to:

Question

Accepted Answer

1. **Understanding the Problem:**  
   Since $$f(x) \geq 0$$ for all real numbers $$x$$, the graph of $$f(x)$$ is U-shaped and opens upwards. Given that $$f(2) = 0$$, the vertex of the parabola is at $$(2, 0)$$. This implies that the quadratic polynomial can be expressed in vertex form:  
   $$
   f(x) = a(x - 2)^2
   $$

---

2. **Finding the Coefficient $$a$$:**  
   Using the condition $$f(4) = 6$$:  
   $$
   f(4) = a(4 - 2)^2 = 6
   $$  
   Simplify:  
   $$
   a(2)^2 = 6
   $$  
   Solve for $$a$$:  
   $$
   a = \frac{6}{4} = \frac{3}{2}
   $$

Thus, the quadratic polynomial becomes:  
$$
f(x) = \frac{3}{2}(x - 2)^2
$$

---

3. **Finding $$f(-2)$$:**  
   Substitute $$x = -2$$ into the polynomial:  
   $$
   f(-2) = \frac{3}{2}((-2) - 2)^2
   $$  
   Simplify:  
   $$
   f(-2) = \frac{3}{2}(-4)^2
   $$  
   Calculate:  
   $$
   f(-2) = \frac{3}{2} 	imes 16 = 24
   $$

---

### Final Answer:  
The value of $$f(-2)$$ is **24**, and the correct answer is **C)**.

Answer

Answer

Answer