Final answer:
The equation of the tangent line to the curve y=2x³ -x² +1 at the point (2,13) is y=20x-27, found by calculating the derivative for slope and applying the point-slope form of a line.
Step-by-step explanation:
To find an equation of the tangent line to the curve y=2x³ -x² +1 at the given point (2,13), we need to first calculate the derivative of the function to find the slope of the tangent line at x=2.
The derivative of the function is y' = 6x² - 2x. Plugging in x=2 gives us the slope m=6(2)² - 2(2) = 24 - 4 = 20. Now we have the slope of the tangent line at the point (2,13).
Using the point-slope form of a line y - y1 = m(x - x1), where (x1, y1) is the point on the curve and m is the slope, we get y - 13 = 20(x - 2). Simplifying, the equation of the tangent line is y = 20x - 27.