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