Calculus problem

Question

Use the limit definition of the derivative to find the standard form equation of the tangent line to f(x) = 2x^2-5x-9 at x=3

Answer 1

Limit definition:

The derivative is the limit (as h→0) of [f(x+h) - f(x)]/h

f(x) = 2x² - 5x - 9

f(x+h) = 2(x+h)² - 5(x+h) - 9 = 2(x² + 2xh + h²) - 5x - 5h - 9 = 2x² + 4xh + 2h² - 5x - 5h - 9

So f(x+h) - f(x) = 2x² + 4xh + 2h² - 5x - 5h - 9 - (2x² - 5x - 9) = 2x² + 4xh + 2h² - 5x - 5h - 9 - 2x² + 5x + 9 = 4xh + 2h² - 5h

And [f(x+h) - f(x)]/h = (4xh + 2h² - 5h)/h = h(4x +2h - 5)/h = 4x + 2h - 5

And Lim (h→0) of [f(x+h) - f(x)]/h = Lim (h→0) of 4x + 2h - 5 = 4x - 5

So f '(x) = 4x - 5 and f' (3) = 4(3) - 5 = 12 - 5 = 7

That means the slope is 7.

At x = 3, f(3) = 2(3)² - 5(3) - 9 = 18 - 15 - 9 = -6 so a point on the line is (3, -6) so the point slope form of the equation of the tangent line is y + 6 = 7(x - 3). This converts to y + 6 = 7x -21 and then to 7x - y = 27