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

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

The inserted Socratic graph is tangent-inclusive.

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

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To find the equation of the tangent line to the curve y = x - √x at (1,0), we need to find the slope of the tangent line at that point.

First, we find the derivative of the curve by using the power rule and the chain rule. The derivative of y = x - √x is dy/dx = 1 - (1/2)√x.

Next, we substitute x = 1 into the derivative to find the slope at (1,0). dy/dx = 1 - (1/2)√1 = 1 - (1/2) = 1/2.

Therefore, the slope of the tangent line at (1,0) is 1/2.

Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we substitute the values: y - 0 = (1/2)(x - 1).

Simplifying the equation, we get y = (1/2)x - 1/2.

Thus, the equation of the tangent line to the curve y = x - √x at (1,0) is y = (1/2)x - 1/2.

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