Question

What is the equation of the line tangent to `f(x)=(x-4)^2-3x` at `x=1`?

Answer 1

Tangent line: `y = -9x + 15`

Explanation:

Given: `f(x) = (x-4)^2 - 3x`

The tangent line is: `y = mx + b`, where `m = f'(1)`

Use the point `(1, f(1))` to find `b`.

Find the first derivative:
`f'(x) = 2(x-4)^1 - 3`

`f'(x) = 2(x-4) - 3 = 2x - 8 - 3 = 2x - 11`

`f'(1) = 2(1) - 11 = -9`

Find the `y` intercept:
`y = -9x + b`

`f(1) = (1-4)^2 - 3(1) = 9 - 3 = 6`

Substitute the point `(1, 6)` into the equation of the line:

`6 = -9(1) + b`

`b = 15`

Tangent line: `y = -9x + 15`

In General form:

`9x + y -15 = 0`