Question

Find the points on the curve $y=x^3-2x^2-x$, where the tangents are parallel to $3x-y+1=0$.

Answer 1

$\begin{array}{c}y=x3-2x2-x...\left(1\right)\\ letpointscurveos\left(x_1,y_1\right)\\ differentiatew.r.t.x\\ \frac{dy}{dx}=\frac{d\left(x3-2x2-x\right)}{dx}\\ \frac{dy}{dx}=3x^2-4x-1\\ slope=\frac{dy}{dx}\\ 3x-y+1=\mathrm{0....}\left(1\right)\\ y=3x+1\\ slop=3\\ whichisparallel\\ then\\ 3x^2-4x-1=3\\ 3x^2-4x-4=0\\ 3x^2-6x+2x-4=0\\ 3x\left(x-2\right)+2\left(x-2\right)=0\\ \left(3x+2\right)\left(x-2\right)=0\\ x=-\frac{2}{3}.orx=2\\ puttingx=-\frac{2}{3}orx=2inequatin\left(1\right)\\ y=-\frac{8}{27}-\frac{8}{9}+\frac{2}{3}\\ y=-\frac{14}{27}\\ and\\ y=-2\\ thereqpointsare\\ \left(-\frac{2}{3},-\frac{14}{27}\right)and\left(2,-2\right)\end{array}$