Question

If the curve $y=a{x}^3+b{x}^2+cx+5$ 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 $(2a-4b+c)$ is

Answer 1

$y=a{x}^3+b{x}^2+cx+5\cdots \left(1\right)$
$\Rightarrow \frac{dy}{dx}=3a{x}^2+2bx+c$
Since given curve touches the $x-$axis at $A(-2,0),$
therefore $A$ lies on $\left(1\right)$ and ${\left(\frac{dy}{dx}\right)}_A=0$
$\Rightarrow -8a+4b-2c+5=0\cdots \left(2\right)$
and $12a-4b+c=0\cdots \left(3\right)$

Curve $\left(1\right)$ cuts $y-$axis at $B(0,5).$
It is given ${\left(\frac{dy}{dx}\right)}_B=3$
$\Rightarrow c=3$
For $c=3,$ from $\left(2\right)$ and $\left(3\right),$ we have
$a=-\frac{1}{2},b=-\frac{3}{4}$
$\therefore 2a-4b+c=-1+3+3=5$