If $ f(x)=x^3+ax^2+bx $ has a stationary point at (-2,0), then the values of $ a $ and $ b $ respectively are:

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We start by using the information that f(x) = x^3 + a x^2 + b x has a stationary point at x = -2. A stationary point occurs where the derivative is zero.

1) Compute the derivative: f'(x) = 3x^2 + 2a x + b.
   Evaluate at x = -2: f'(-2) = 3(-2)^2 + 2a(-2) + b = 12 - 4a + b.
   Setting f'(-2) = 0 gives the equation 12 - 4a + b = 0, hence b = 4a - 12.

2) Since the stationary point is at (-2, 0), the point lies on the graph as well. So f(-2) = 0.
   Compute f(-2): f(-2) = (-2)^3 + a(-2)^2 + b(-2) = -8 + 4a - 2b.
   Setting f(-2) = 0 yields -8 + 4a - 2b = 0, which simplifies to 4a - 2b = 8, or b = 2a - 4.

3) Now we have two expressions for b:
   - From the slope condition: b = 4a - 12.
   - From the point condition: b = 2a - 4.
   Equate them: 4a - 12 = 2a - 4.
   Solve for a: 2a = 8 => a = 4.

4) Substitute a = 4 into b = 2a - 4 (or b = 4a - 12):
   b = 2(4) - 4 = 8 - 4 = 4 (or b = 4(4) - 12 = 16 - 12 = 4).

Therefore, the pair is a = 4 and b = 4.

Option-by-option analysis:
- Option a: a = 4, b = 4. This matches the derived values and is consistent with both the derivative condition and the point condition.
- Option b: a = -4, b = 4. If a were -4, then b from b = 4a - 12 would be -16 - 12 = -28, which contradicts b = 4 in this option; also it would violate f(-2) = 0.
- Option c: a = -4, b = 2. Substituting a = -4 into b = 4a - 12 gives b = -28, not 2, so this is inconsistent with the derivative condition.
- Option d: a = 4, b = 2. With a = 4, the derivative condition requires b = 4a - 12 = 4, not 2, so this does not satisfy f'(-2) = 0.

In summary, only the pair a = 4 and b = 4 satisfies both the stationary condition f'(-2) = 0 and the point condition f(-2) = 0.

MUF0091 Mathematics Unit 1 - Semester 1, 2025 12.7 Study Area 3 revision quiz

If \( f(x)=x^3+ax^2+bx \) has a stationary point at (-2,0), then the values of \( a \) and \( b \) respectively are:

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