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:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (D) only
$y=ax^2+bx$ has a minimum at $x=2$.
For a quadratic $ax^2+bx$, the vertex occurs at $x=-\frac{b}{2a}$.
$-\frac{b}{2a}=2$
$\Rightarrow b=-4a$
Minimum value at $x=2$ is given to be $-12$:
$y(2)=a(2)^2+b(2)=-12$
$4a+2b=-12$
Substitute $b=-4a$:
$4a+2(-4a)=-12$
$4a-8a=-12$
$-4a=-12$
$a=3$
$b=-4a=-12$
Thus, the correct options are A and D.