What is the equation of the line tangent to #f(x)=x^3-6x # at #x=2#?

1 Answer
May 3, 2016

The equation of the tangent at #f(2)# is #y=6x-16#

Explanation:

#f(x) = x^3-6x#
#f'(x) = 3x^2 - 6#

The equation of any tangent to #f(x)# in slope and intercept form is:
#y = mx+c#
Where: #m# = slope of #f(x)# and #c# = the intercept on the y axis

Slope of the tangent to #f(x)# at #x=2# is:
#f'(2)= 3*2^2-6#
#=6 -> m=6#

Since the tangent touches #f(x)# at #f(2)#
#f(2) = m*2 +c#

Replacing for #f(2)# and #m=6#:
#2^3-6*2 = 6*2 +c#

#c = 8-12-12#
#c=-16#

Hence the equation of the tangent at #x=2# is:

#y = 6x-16#