Question

The curve $y=a{x}^3+bx+5$ touches the x-axis at $P(-2,0)$ and cuts y-axis at point $Q$ where its gradient is $3.$ Find $a$ and $b.$ Also find coordinates of point $Q.$

Answer 1

$y=a{x}^3+bx+5$

$P(-2,0)$

$0=a(-2{)}^3+b(-2)+5$

$-8a-2b+5=0$ ……..(i)

$\frac{dy}{dx}=3a{x}^2+b$ [In y-axis, $x=0$]

$3a{x}^2+b{|}_{x=0}=b$

$\Rightarrow b=3$

From equation (i)

$-8a-6+5=0$

$\Rightarrow a=-\frac{1}{8}$

$y=\frac{-x^3}{8}+3x+5$

$Q(k,0)$

$k=0+3\left(0\right)+5$

$\Rightarrow k=5$

$Q(5,0)\to$ coordinates of $Q.$