What is the equation of the line tangent to #f(x)=(x-4)^2-x # at #x=-1#?

1 Answer
May 18, 2017

#y = f(x) = -11 x + 15#

Explanation:

#f(x) = (x - 4)^2 - x = x^2 -9x + 16#
at #x = -1, y = f(-1) =(-5)^2 - (-1) = 26#

gradient function, #f'(x) = 2x - 9#, therefore it gradient at #x = -1#,
#f'(-1) = 2(-1) - 9 = -11#

with using standard form of line equation, #f(x)= y = mx +c#, plug in #x = -1, y = 26 and m = -11#, into this equation to find #c#

#26 = -11(-1) + c#
#15 = c#

therefore it line tangent,
#y = f(x) = -11 x + 15#