Answer
$y = 2x - 2 + 3 = 2x + 1$
Work Step by Step
first of all lets find the derivative of y:
$y = 2x^{3}-x^{2} + 2$
$y' = (2x^{3}-x^{2} + 2)'= (2x^{3})'-(x^{2})' + (2)'=6x^{2}-2x$
So the slope of the tangent line at $(1, 3)$ is (x = 1, and calculate the derivative)
$6\times1^{2}-2\times1=6-4=2$
We use the point-slope form to write an equation of the tangent line at $(1,3)$
$y - 3 = 2 (x-1)$
$y = 2x - 2 + 3 = 2x + 1$
