The slope of the tangent line to the graph of a function at the given point is the derivative of the graph at that point.

`f(x)=sqrt(x-1)`

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

Therefore Slope(m) of the tangent line at (5,2) =1/2(4)^(1/2)

m=1/4

The equation of the line can be written using point slope form of the equation.

y-y_1=m(x-x_1)

y-2=1/4(x-5)

4y-8=x-5

**4y-x=3**

` `

The given function is:-

f(x) = {(x-1)^(1/2)

differentiating both sides w.r.t 'x' we get

f'(x) = (1/2)*[1/{(x-1)^(1/2)}]

Now, slope of the tangent at the point (5,2) = f'(5) = 1/4

Thus, equation of the tangent at the point (5,2) and having slope = (1/4) is :-

y - 2 = (1/4)*(x - 5)

or, 4y - 8 = x - 5

or, 4y - x = 3 is the equation of the tangent to the given curve at (5,2)

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