If the curve $y = a x^{3} + b x^{2} + c x + 5$ touches the $x -$ axis at $A (- 2, 0)$ and cuts the $y -$ axis at point $B$ where its slope is $3,$ then the value of $(2 a - 4 b + c)$ is

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Solution

$y = a x^{3} + b x^{2} + c x + 5 \dots (1)$
$\Rightarrow \frac{d y}{d x} = 3 a x^{2} + 2 b x + c$
Since given curve touches the $x -$ axis at $A (- 2, 0),$
therefore $A$ lies on $(1)$ and ${(\frac{d y}{d x})}_{A} = 0$
$\Rightarrow - 8 a + 4 b - 2 c + 5 = 0 \dots (2)$
and $12 a - 4 b + c = 0 \dots (3)$

Curve $(1)$ cuts $y -$ axis at $B (0, 5) .$
It is given ${(\frac{d y}{d x})}_{B} = 3$
$\Rightarrow c = 3$
For $c = 3,$ from $(2)$ and $(3),$ we have
$a = - \frac{1}{2}, b = - \frac{3}{4}$
$∴ 2 a - 4 b + c = - 1 + 3 + 3 = 5$

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