At what points on the curve

Solution:

Let $\left(x_{1}, y_{1}\right)$ be the required point.

The slope of line $y=3 x+4$ is 3 .

Since, the point lies on the curve.

Hence, $y_{1}=2 x_{1}^{2}-x_{1}+1$

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

$\frac{d y}{d x}=4 x-1$

Now,

Slope of the tangent at $\left(x_{1}, y_{1}\right)=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=4 x_{1}-1$

Slope of the tangent at $\left(x_{1}, y_{1}\right)=$ Slope of the given line [Given]

$\therefore 4 x_{1}-1=3$

$\Rightarrow 4 x_{1}=4$

$\Rightarrow x_{1}=1$

and

$y_{1}=2 x_{1}^{2}-x_{1}+1=2-1+1=2$

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