Question

The curve $y=a{x}^3+b{x}^2+cx+8$ touches $x$-axis at $P(-2,0)$ and cuts the $y$-axis at a point $Q(0,8)$ where its gradient is 3. The values of $a$, $b$, $c$ are respectively

• A. $-\frac{1}{2},-\frac{3}{4},3$
• B. $3,-\frac{1}{2},-4$
• C. $-\frac{1}{2},-\frac{7}{4},2$
• D. none of these

Answer 1

The correct option is D none of these

Given, $y=a{x}^3+b{x}^2+cx+8$

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

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 $\left(ii\right)$

$a=-1/4$, $b=0$