Find a point on the curve


Find a point on the curve $y=x^{3}-2 x^{2}-2 x$ at which the tangent lines are parallel to the line $y=2 x-3$.



The curve $y=x^{3}-2 x^{2}-2 x$ and a line $y=2 x-3$

First, we will find The Slope of tangent

$y=x^{3}-2 x^{2}-2 x$

$\frac{d y}{d x}=\frac{d}{d x}\left(x^{3}\right)-\frac{d}{d x}\left(2 x^{2}\right)-\frac{d}{d x}(2 x)$

$\Rightarrow \frac{d y}{d x}=3 x^{3}-1-2 \times 2\left(x^{2}-1\right)-2 \times x^{1-1}$

$\Rightarrow \frac{d y}{d x}=3 x^{2}-4 x-2 \ldots(1)$

$y=2 x-3$ is the form of equation of a straight line $y=m x+c$,

where $m$ is the The Slope of the line. so the The Slope of the line is $y=2 x(x)-3$

Thus, The Slope $=2 \ldots(2)$

From (1) \& (2)

$\Rightarrow 3 x^{2}-4 x-2=2$

$\Rightarrow 3 x^{2}-4 x=4$

$\Rightarrow 3 x^{2}-4 x-4=0$

We will use factorization method to solve the above Quadratic equation.

$\Rightarrow 3 x^{2}-6 x+2 x-4=0$

$\Rightarrow 3 x(x-2)+2(x-2)=0$

$\Rightarrow(x-2)(3 x+2)=0$

$\Rightarrow(x-2)=0 \&(3 x+2)=0$

$\Rightarrow x=2$ or


Substitute $x=2 \& x=\frac{-2}{3}$ in $y=x^{3}-2 x^{2}-2 x$

when $x=2$

$\Rightarrow y=(2)^{3}-2 \times(2)^{2}-2 \times(2)$

$\Rightarrow y=8-(2 \times 4)-4$

$\Rightarrow y=8-8-4$

$\Rightarrow y=-4$

when $x=\frac{-2}{3}$

$\Rightarrow y=\left(\frac{-2}{3}\right)^{3}-2 \times\left(\frac{-2}{3}\right)^{2}-2 \times\left(\frac{-2}{3}\right)$

$\Rightarrow y=\left(\frac{-8}{27}\right)-2 \times\left(\frac{4}{9}\right)+\left(\frac{4}{3}\right)$

$\Rightarrow y=\left(\frac{-8}{27}\right)-\left(\frac{8}{9}\right)+\left(\frac{4}{3}\right)$

taking $\mathrm{lcm}$

$\Rightarrow y=\frac{(-8 \times 1)-(8 \times 3)+(4 \times 9)}{27}$

$\Rightarrow y=\frac{-8-24+36}{27}$

$\Rightarrow y=\frac{4}{27}$

Thus, the points are $(2,-4) \&\left(\frac{2}{3}, \frac{4}{27}\right)$

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