At what points on the curve

Solution:

Given:

The curve is $y=2 x^{2}-x+1$ and the line $y=3 x+4$

First, we will find The Slope of tangent

$y=2 x^{2}-x+1$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(2 \mathrm{x}^{2}\right)-\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})+\frac{\mathrm{d}}{\mathrm{dx}}(1)$

$\Rightarrow \frac{\mathrm{d} \mathrm{y}}{\mathrm{dx}}=4 \mathrm{x}-1 \ldots(1)$

$y=3 x+4$ 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 $y=3 x(x)+4$

Thus, The Slope $=3 \ldots(2)$

From (1) & (2), we get,

$4 x-1=3$

$\Rightarrow 4 x=4$

$\Rightarrow x=1$

Substituting $x=1$ in $y=2 x^{2}-x+1$, we get,

$\Rightarrow y=2(1)^{2}-(1)+1$

$\Rightarrow y=2-1+1$

$\Rightarrow y=2$

Thus, the required point is $(1,2)$