Question

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

Answer 1

`y=4x-4`. See process below

Explanation:

`f´(x_0)` gives the slope of tangent line to `f(x)` at `x=x_0`

General equation for a straigth line is `y=ax+b` where `a` is the slope and `b` is the y-intercept

By definition `f´(x)=2x` at `x_0=2` gives the value `f´(2)=4`

So, the slope `a=4`

By other hand `f(2)=2^2=4`, thus tangent line passes by `(2,4)`

Our Tangent line is `y=4x+b` passing thru `(2,4)`. Then

`4=4·2+b`, solving this simple equation `b=-4`

Finally tangent line at `x_0=2` is `y=4x-4`