Question

What is the equation of the tangent line of `f(x) =x^2-2x+8` at `x=2`?

Answer 1

Equation of the tangent line is `2x-y+4=0`

Explanation:

As `y=x^2-2x+8` and slope of tangent s given by function's derivative, let us first find `(dy)/(dx)`, which is given by `2x-2` and at `x=2` slope is `(2xx2-2)=2`

Further at `x=2`, `f(2)=2^2-2xx2+8=4-4+8=8`, hence tangent passes through `(2,8)`.

Hence the tangent will be `(y-8)=2(x-2)` or

`(y-8)=2x-4` or `2x-y+4=0`