Question

Find an equation of the tangent line to the curve at the given point. y = 7x² - x³, (1, 6)

Answer 1

The equation of the tangent line to the curve y = 7x² - x³ at the point (1, 6) is found by calculating the derivative, evaluating the slope at x = 1, and using the point-slope form. The final equation is y = 11x - 5.

Step-by-step explanation:

To find an equation of the tangent line to the curve y = 7x² - x³ at the given point (1, 6), we need to follow these steps:

1. Calculate the derivative of the function to find the slope of the tangent line.
2. Evaluate the derivative at the x-coordinate of the given point to find the slope at that point.
3. Use the point-slope form of a line equation with the given point and the calculated slope to write the equation of the tangent line.

The derivative of the function y = 7x² - x³ is y' = 14x - 3x². Evaluating at x = 1, we get y'(1) = 14(1) - 3(1)² = 14 - 3 = 11. This is the slope of the tangent line at the point (1, 6).

Using the point-slope form y - y1 = m(x - x1), where m is the slope and (x1, y1) is the given point, we substitute m = 11 and (x1, y1) = (1, 6) to get the equation of the tangent line:

y - 6 = 11(x - 1)

Therefore, the equation of the tangent line to the curve at the given point (1, 6) is y = 11x - 5.