Question

Find the points on the curve y = x³ − 2x² − 2x at which the tangent lines are parallel to the line y = 2x − 3.

Answer 1

Let (x₁, y₁) be the required point.
Given:

$$\begin{gathered} \mathrm{y}=2\mathrm{x}-3 \\ \therefore \mathrm{Slope}\mathrm{of}\mathrm{the}\mathrm{line}=\frac{dy}{dx}=2 \\ y=x^3-2x^2-2x \\ \mathrm{Since}\left(x_1{y}_1\right)\mathrm{lies}\mathrm{on}\mathrm{curve},y_1={x_1}^3-2{x_1}^2-2x_1...\left(1\right) \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=3{x_1}^2-4x_1-2 \\ \text{It is given that the tangent and the given line are parallel.} \\ \text{∴ Slope of the tangent = Slope of the given line} \\ 3{x_1}^2-4x_1-2=2 \\ \Rightarrow 3{x_1}^2-4x_1-4=0 \\ \Rightarrow 3{x_1}^2-6x_1+2x_1-4=0 \\ \Rightarrow 3x_1\left(x_1-2\right)+2\left(x_1-2\right)=0 \\ \Rightarrow \left(x_1-2\right)\left(3x_1+2\right)=0 \\ \Rightarrow x_1=2\text{or}{x}_1=\frac{-2}{3} \\ \text{Case}1 \\ \mathrm{When}{x}_1=2 \\ \mathrm{On}\mathrm{s}\text{ubstituting the value of}{\mathit{\text{x}}}_1\text{in eq. (1), we get} \\ {y}_1=8-8-4=-4 \\ \therefore \left(x_1,y_1\right)=\left(2,-4\right) \\ \text{Case}2 \\ \mathrm{When}{x}_1=\frac{-2}{3} \\ \mathrm{On}\mathrm{s}\text{ubstituting the value of}{x}_1\text{in eq. (1), we get} \\ {y}_1=\frac{-8}{27}-\frac{8}{9}+\frac{4}{3}=\frac{-8-24+36}{27}=\frac{4}{27} \\ \therefore \left(x_1,y_1\right)=\left(\frac{-2}{3},\frac{4}{27}\right) \end{gathered}$$