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 a point $Q$, where its gradient is $3$. Find $a,b,c$.

• A. $a=-\frac{1}{5},b=1,c=3$
• B. $a=-\frac{1}{4},b=-1,c=4$
• C. $a=-\frac{1}{4},b=0,c=3$
• D. $a=-\frac{1}{3},b=1,c=-3$

Answer 1

The correct option is C $a=-\frac{1}{4},b=0,c=3$

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

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

Since the curve touches $x$-axis at $(-2,0)$ so

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

The curve cut the $y$-axis at $(0,8)$, so

$\frac{dy}{dx}{|}_{(0,8)}=3\Rightarrow c=3$

Also the curve passes through $(-2,0)$, so

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

Solving $\left(i\right)$ and $\left(ii\right)$ $a=-\frac{1}{4}$, $b=0$