Find the equation of the line tangent to the graph of f(x)=(2x-5)/(x+1) at the point where x=0

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The equation of the line to the curve represented by `y = (2x - 5)/(x+1)` has to be determined at the point where x = 0.

The slope of the tangent to the curve y at any point is the value of y' at that point.

y = `(2x - 5)/(x+1)`

=> `y' = 7/(x+1)^2`

at x = 0, y' = 7.

Also at x = 0, y = -5

The equation of the tangent is `(y + 5)/(x - 0) = 7`

=> y + 5 = 7x

=> 7x - y - 5 = 0

**The required equation of the tangent is 7x - y - 5 = 0**

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