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.

Answer 1

The equation of curve is

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

It meets y-axis at Q where x=0,

So, $y=0+0+0+5=5$

Therefore, Q is (0,5).

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

At Q, $\frac{dy}{dx}=0+0+c=c$

From given condition, $c=3$

Since, curve touches the x-axis i.e., y=0 at (-2,0)

Therefore, slope of tangent at (-2,0) is 0.

At (-2,0), $\frac{dy}{dx}=12a-4b+3=0$ .....(1)

Also, (-2,0) lies on the curve , therefore,

$8a-4b+1=0$ ....(2)

Solving equation 1 & 2, we get,

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

Hence, $a=-\frac{1}{2}$, $b=-\frac{3}{4}$ and $c=3$