Question

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

Answer 1

`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`