Let the point be (x,y) such that the tangent line is parallel to the secant that cuts f(x) in x= -1 and x= 2

First we will find the endpoints of the secant.

==> `x=-1 ==> f(-1)= -1^2 -4(-1)+1= 6==> (-1,6)`

`==> x= 2 ==> f(2)= 2^2 -4(2) +1 = -3 ==> (2,-3)`

`` Now we will find the slope of the line that passes through (-1,6) and (2,-3)

==> `m= (6+3)/(-1-2)= 9/-3 = -3`

`==> f'(x)= 2x-4 = -3`

`==> 2x= 1`

`==> x= 1/2`

`==> f(1/2)= (1/2)^2 - 4(1/2)+1 = 1/4 -2+1 = 1/4 -1= -3/4`

`==> (1/2, -3/4) `

``

Given the functions f(x) such that:

`f(x)= x^2 -4x+1`

The slope of the tangent line is 0.

`f'(x)= 2x-4 = 0`

`==> 2x-4 = 0`

`==> 2x = 4`

`==> x = 2`

`==> f(2)= 2^2 -4(2) +1 = 4-8+1 = -3`

**Then, the point such that the tangent is horizontal is `(2,-3)`**

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now