Let f(x) = x2 − 2x − 3. The secant line through (2, f(2)) and (2 + h, f(2 + h)) for f(x) has slope h + 2. Use this formula to compute the following.

(a) The slope of the secant line through (2, f(2)) and (3, f(3))
(b) The slope of the tangent line at x = 2 (by taking a limit)

a) slope of secant line = 3

b) slope of tangent line = 2

Step-by-step explanation:

Given:

- The function:

                           f(x) = x^2 -2*x - 3

- The slope for f(x) @ x = 2 is:

                           slope = h + 2

Find:

a) The slope of the secant line through (2, f(2)) and (3, f(3))

b) The slope of the tangent line at x = 2

Solution:

- Since we are given the slope of the line computed via secant method. All we need to do is evaluate the slope given for respective question.

- The slope of secant line between points ( 2 , f(2) ) and ( 3 , f(3) ) is:

                             slope = h + 2

Where,  h is the step size between two points. h = 3 - 2 = 1

                             slope = 1 + 2 = 3

Hence, the slope of the secant is 3.

- The slope of tangent line @ points ( 2 , f(2) ) is:

                             slope = Lim _ h-->0 (h + 2)

Where,  h step size is reduced to infinitesimal small number. Hence, h = 0

                             slope = 0 + 2 = 2

Hence, the slope of the tangent is 2.

idk i would have to be there to under stand it

step-by-step explanation:

idk i would have to be there to under stand it