If $y = ax^2 + bx$ has minima at $x = 2$ and the minimum value is -12, then which of the following are correct?

(A) $a = 3$

(B) $a = -3$

(C) $b = 12$

(D) $b = -12$

Choose the correct answer from the options given below:

The function $y = ax^2 + bx$ has a minimum at $x = 2$ with minimum value $-12$.

When a parabola has a minimum (lowest point), it opens upward.

Therefore, $a > 0$.

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At a minimum point, the derivative equals zero.

$\frac{dy}{dx} = 2ax + b$

At $x = 2$:

$2a(2) + b = 0$

$4a + b = 0$

$b = -4a$ ...(Equation 1)

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At $x = 2$, the value $y = -12$:

$y = ax^2 + bx$

$-12 = a(2)^2 + b(2)$

$-12 = 4a + 2b$ ...(Equation 2)

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From Equation 1: $b = -4a$

Substituting into Equation 2:

$-12 = 4a + 2(-4a)$

$-12 = 4a - 8a$

$-12 = -4a$

$a = 3$

Now finding $b$:

$b = -4a$

$b = -4(3)$

$b = -12$

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Checking the options:

(A) $a = 3$ ✓ Correct

(B) $a = -3$ ✗ Incorrect (would give maximum, not minimum)

(C) $b = 12$ ✗ Incorrect

(D) $b = -12$ ✓ Correct

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The correct options are (A) and (D).

Therefore, $a = 3$ and $b = -12$.