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.

• A. $A=-1/2,B=-3/2,C=3$
  
• B. $A=-1/2,B=-3/2,C=8$
• C. $A=-1/2,B=-3/4,C=3$
• D. $A=1/2,B=-3/2,C=3$

Answer 1

The correct option is C $A=-1/2,B=-3/4,C=3$

A] Equation of curve $y=a{x}^3+b{x}^2+cx+5$ __ (1)

It meets y axis at o where x = 0

substitute x = 0 in (1)

$y=0+0+0+5\Rightarrow ,y=5\therefore p=(0,5)$

From (1),

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

at p, $y^{\prime }=0+0+c=c$

gradient is 3 $\therefore c=3$

(1) touches x axis at (-2,0)

$\therefore$ slope of tangent at (-2,0) is 0

at (-2,0)$y^{\prime }=12a-4b+3$ __ (4)

$\therefore 12a-4b+3=0$

(-2,0) lies on (1)

$-8a+4b-2c+5=0$ __ (3)

subtracting (3)-(4) $\Rightarrow 4a+2=0$ $\therefore a=-1/2$

$4b=-3\Rightarrow b=-3/2$ & $c=3$