find the points on f(x)=(x^2-2)(x-4) where the tangent line is parallel to the line y=x+3?

Question

Geronimo · Accepted Answer

f(x) = (x² – 2) • (x – 4)

 f(x) = x³ – 4x² – 2x + 8

   f  '  (x) = 3x² – 8x – 2  ... since: y = x + 3  has a slope = 1 , then:

     1 = 3x² – 8x – 2

     0 = 3x² – 8x – 3

     0 = (3x + 1) • (x – 3)

 x = 3   and  x = -⅓

 f{3} = -7  and  f{-⅓} = 221 ⁄ 27 

 ... and the points are: (x, f(x)) = (3, -7) and (-⅓, 221 ⁄ 27)

find the points on f(x)=(x^2-2)(x-4) where the tangent line is parallel to the line y=x+3?

Trending Questions