Question

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

Answer 1

`y=0`

Explanation:

Find the derivative to calculate the slope of the line at the point `(1,0)`.

We will use the chain rule:

`f'(x)=6(x-1)*d/dx[x-1]=6x-6`

To find the slope of the line:

`f'(1)=6(1)-6=0`

Thus, the tangent line is horizontal, and its equation is `y=0`.