Question

The curve $y=a{x}^3+b{x}^2+cx+5$ touches the x -axis at P(-2,0) and cuts the y-axis at the point Q, where its gradient is 3. Find the equation of the curve completely.

Answer 1

$y=a{x}^3+b{x}^2+cx+5$

$(-z,0)$ lies on curve $y=a{x}^3+b{x}^2+cx+5d$

$\Rightarrow 0=-8a+4b-2c+5.....\left(1\right)$

Also at $(-2,0)$ curve touches x-axis ie

$\frac{dy}{dx}$ at $(-2,0)$ is 0

$\therefore \frac{dy}{dx}=3a{x}^2+2bx+c$

$\frac{dy}{dx}(-2,0)=0=12a=-4b+c....\left(2\right)$

Since curve crosses y-axis at Q, x-

coordinate of point Q must be zero

$\Rightarrow y=0+0+0+5\Rightarrow y=5$

$\therefore$ coordinate of $Q=(0,5)$

According to Question $\frac{dy}{dx}=3$ at $Q(0,5)$

$\Rightarrow 3=0+0+c$

$\Rightarrow c=3$

putting the value of c in $e{q}^n\left(1\right)$ & $e{q}^n\left(2\right)$

and gurthare solving we get

$a=\frac{-1}{2}$ & $b=\frac{-3}{4}$

$\therefore E{q}^n$ of cover is

$y=\frac{-x^3}{2}-\frac{3}{4}{x}^2+3x+5$