Question

At what points on the curve y = 2x² − x + 1 is the tangent parallel to the line y = 3x + 4?

Answer 1

Let (x₁, y₁) be the required point.
The slope of line y = 3x + 4 is 3.

$$\begin{gathered} \mathrm{Since},\mathrm{the}\mathrm{point}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}. \\ \mathrm{Hence},y_1=2{x_1}^2-x_1+1 \\ \mathrm{Now},y=2x^2-x+1 \\ \frac{dy}{dx}=4x-1 \\ \mathrm{Now}, \\ \text{Slope of the tangent at}\left(x_1,y_1\right)\text{=}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\text{=4}{\mathit{\text{x}}}_{\mathit{1}}\text{-1} \\ \text{Slope of the tangent at}\left(x_1,y_1\right)\text{= Slope of the given line [Given]} \\ \therefore 4x_1-1=3 \\ \Rightarrow 4x_1=4 \\ \Rightarrow x_1=1 \\ \mathrm{and} \\ {y}_1=2{x_1}^2-x_1+1=2-1+1=2 \\ \text{Thus, the required point is}\left(1,2\right)\text{.} \end{gathered}$$