If a,b,c ` in R,` then for which of the following graphs of the quadratic polynomial `y = ax ^(2) -2bx+ c (a ne 0),` the product (abc) is negative ?

Options

A.
B.
C.
D.

Answer

To solve the problem, we need to analyze the quadratic polynomial given by \( y = ax^2 - 2bx + c \) where \( a, b, c \in \mathbb{R} \) and \( a \neq 0 \). We want to determine for which of the given graphs the product \( abc \) is negative.

### Step-by-Step Solution:

1. **Identify the Sign of \( a \)**:
- Since the problem states that the graphs are downward-opening parabolas, we know that \( a < 0 \).

**Hint**: Remember that the direction of the parabola (upward or downward) is determined by the sign of \( a \).

2. **Determine Conditions for \( abc < 0 \)**:
- For the product \( abc \) to be negative, given that \( a < 0 \), it follows that \( bc \) must be positive (since a negative times a positive is negative).

**Hint**: If \( a \) is negative, then \( bc \) must be positive for the overall product \( abc \) to be negative.

3. **Analyze Each Option**:
- **Option A**:
- \( C \) (the value of the polynomial at \( x = 0 \)) is negative. Thus, \( c < 0 \).
- The x-coordinate of the vertex \( -\frac{b}{2a} > 0 \), which implies \( b < 0 \) (since \( a < 0 \)).
- Therefore, \( a < 0 \), \( b < 0 \), \( c < 0 \) implies \( abc < 0 \) is not satisfied (as \( bc > 0 \)).
- **Conclusion**: \( abc \) is positive, so this option is not correct.

**Hint**: Check the signs of \( b \) and \( c \) carefully based on the vertex condition.

- **Option B**:
- \( C \) is negative, so \( c < 0 \).
- The x-coordinate of the vertex \( -\frac{b}{2a} < 0 \), which implies \( b > 0 \) (since \( a < 0 \)).
- Thus, \( a < 0 \), \( b > 0 \), \( c < 0 \) implies \( abc < 0 \) is satisfied.
- **Conclusion**: This option is correct.

**Hint**: Ensure that the signs of \( b \) and \( c \) are consistent with the conditions given.

- **Option C**:
- \( C \) is positive, so \( c > 0 \).
- The x-coordinate of the vertex \( -\frac{b}{2a} < 0 \), which implies \( b > 0 \).
- Thus, \( a < 0 \), \( b > 0 \), \( c > 0 \) implies \( abc > 0 \).
- **Conclusion**: This option is not correct.

**Hint**: Remember that if \( c \) is positive, the product \( abc \) cannot be negative.

- **Option D**:
- \( C \) is negative, so \( c < 0 \).
- The x-coordinate of the vertex \( -\frac{b}{2a} > 0 \), which implies \( b < 0 \).
- Thus, \( a < 0 \), \( b < 0 \), \( c < 0 \) implies \( abc < 0 \) is not satisfied (as \( bc > 0 \)).
- **Conclusion**: This option is not correct.

**Hint**: Again, check the signs of \( b \) and \( c \) carefully.

### Final Conclusion:
The only option for which the product \( abc \) is negative is **Option B**.