Question

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

Answer 1

`y=3x-4`

Explanation:

From the given, `f(x) = x^2 - x` at `x=2`

compute for the point `(x_1,y_1)`

`x_1=2`

`y_1= f(x_1)=f(2)=2^2-2=2`

therefore `(x_1,y_1) = (2, 2)`

find the first derivative then use `x=2` to find the slope `m`

`f' (x) = 2x-1`

slope `m=f(' 2)=2*2-1=3`

`m=3`

For the tangent line:

Use `y-y_1=m(x-x_1)`

`y-2=3*(x-2)`

`y=3x-6+2`

`y=3x-4` the equation of the tangent line.