Answer
The equation of the tangent line to the curve at $(1,3)$ is $$(l):y=4x-1$$
Work Step by Step
$$y=2x^3-x^2+2$$
1) First, find the derivative of $y$ $$y'=\frac{d}{dx}(2x^3)-\frac{d}{dx}(x^2)+\frac{d}{dx}(2)$$
$$y'=2\times3x^2-2x+0$$
$$y'=6x^2-2x$$
2) Find the slope of the tangent line to the curve at the point $(1,3)$, or in fact, $y'(1)$: $$y'(1)=6\times1^2-2\times1=4$$
3) The tangent line to the curve at the point $(1,3)$ would have the following equation: $$(l):y=y'(1)x+b$$
$$(l):y=4x+b$$
Since the point $(1,3)$ also lies in the tangent line $(l)$, we can apply that point to the equation of $(l)$ to find $b$. In detail, $$4\times1+b=3$$$$4+b=3$$$$b=-1$$
Therefore, the equation of the tangent line to the curve at $(1,3)$ is $$(l):y=4x-1$$
