Mathematics - Application of Derivative Question with Solution | TestHub

As the x-axis touches the curve at $P(-2,0)$, we have ${\left(\frac{dy}{dx}\right)}_{(-2,0)}=0\dots (1)$ and $(y{)}_{x=-2}=0$ As the curve cuts y-axis at Q where the gradient of the curve is $3,{\left(\frac{dy}{dx}\right)}_{x=0}=3$ $\frac{dy}{dx}=3a{x}^2+2bx+c$ (1) $\Rightarrow 12a-4b+c=0$; (2) $\Rightarrow -8a+4b-2c+5=0$; (3) $\Rightarrow c=3$ Solving, $a=-\frac{1}{2},b=-\frac{3}{4}\therefore$ the curve is $y=-\frac{1}{2}{x}^3-\frac{3}{4}{x}^2+3x+5$