Question

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

Answer 1

As the point where its meet the y-axis its x-coordinates = 0

$y=0+0+0+5$

$y=5$

The point is $(0,5)$

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

$3=c$

As point $(-2,0)$ lie on the curve.

$0=-8a+4b-6+5$

$4b-8a=1$ $\dots \left(1\right)$

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

At the point $(-2,0)$ its gradient is $0.$

$0=12a-4b+3$ $\dots \left(2\right)$

Add $\left(1\right)$ and $\left(1\right)$,we get

$12a-4b+3+4b-8a-1=0\Rightarrow 4a+2=0\Rightarrow a=-\frac{1}{2}$

substitute in $\left(2\right)$ we get $b=-\frac{3}{4}$

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