## Calculus 8th Edition

$y = 2x - 2 + 3 = 2x + 1$
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 $(1,3)$ $y - 3 = 2 (x-1)$ $y = 2x - 2 + 3 = 2x + 1$