The function f is defined by `f(x)=x^4-4x^4 -x^2+cx -12` , where c is a constant.In the xy-plane, the graph of f intersects the x-axis in the four points (-2,0), (1,0), (p,0), and (q,0). What is the value of c ?

Options

A. 16
B. `-16`
C. `p/q`
D. `q/p`

Answer

To find the value of the constant \( c \) in the function \( f(x) = x^4 - 4x^3 - x^2 + cx - 12 \), we know that the function intersects the x-axis at the points (-2, 0), (1, 0), (p, 0), and (q, 0). This means that \( f(x) = 0 \) at these points.

We can use one of the known points to find \( c \). Let's use the point (1, 0).

### Step-by-Step Solution:

1. **Substitute \( x = 1 \) into the function**:
\[ f(1) = 1^4 - 4(1^3) - 1^2 + c(1) - 12 \]

2. **Calculate each term**:
\[ f(1) = 1 - 4 - 1 + c - 12 \]

3. **Combine the constant terms**:
\[ f(1) = 1 - 4 - 1 - 12 + c = -16 + c \]

4. **Set \( f(1) \) equal to 0** (since the graph intersects the x-axis at this point):
\[ -16 + c = 0 \]

5. **Solve for \( c \)**:
\[ c = 16 \]

Thus, the value of \( c \) is \( \boxed{16} \).