Question

What is the equation of the line tangent to `f(x)=2x-x^2` at `x=0`?

Answer 1

y = 2x

Explanation:

To find the equation of the tangent : y - b = m(x-a)

Require to know m , the gradient and (a , b ) a point on the line.

Now, m = f'(x) and a is given in the question (x = 0 ). To find b ,

substitute x = 0 into f(x)

f'(x) = 2 - 2x → f'(0) = 2 = m.

and f(0) = 0 → (a , b ) = (0 , 0 )

This is a line passing thro' the origin with m =2 and hence

equation is y = 2x