The curve $y=a x^{3}+b x^{2}+c x+5$ touches the $x$-axis at $(-2,0)$ and cuts the $y$ - axis at a po

Question

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

Professor PC Thomas Classes · Answer

Given curve is $y=a x^{3}+b x^{2}+c x+5$ The curve (1) touches $x$-axis at $(-2,0)$\ $\Rightarrow \quad-8 a+4 b-2 c+5=0$\ and $\left(\frac{d y}{d x}ight)_{(-2,0)}=0 \Rightarrow 12 a-4 b+c=0$ The curve (1) cuts the $y$-axis at $(0,5)$\ Gradient of the curve at $(0,5)$ is\ $3 \Rightarrow\left(\frac{d y}{d x}ight)_{(0,5)}=3$\ $\Rightarrow \quad\left[3 a x^{2}+2 b x+cight]_{(0,5)}=3$\ i.e. $c=3$ Solving (2.1), (2.2), (2.3), we get\ $a--\frac{1}{2}, b=-\frac{3}{4}, c=3$.\