Question

How do you find the equation of the tangent line to the curve `y=x-sqrtx` at (1,0)?

Answer 1

Just to welcome a newcomer, with an illustrative graph, for the nice answer,

Explanation:

To make y real, `x >=0`.

The inserted Socratic graph is tangent-inclusive.

graph{(2y-x+1)(y-x+sqrtx)=0x^2 [-2.5, 2.5, -1.25, 1.25]}