# `y = (x^2 - 1)/(x^2 + x + 1) , (1, 0)` Find an equation of the tangent line to the given curve at the specified point.

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Student Comments

balajia | Student

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

`(dy)/(dx)=(2x(x^2+x+1)-(2x+1)(x^2-1))/(x^2+x+1)^2`

By substituting (1,0) we get the derivative of the function which is equal to slope of the tangent at (1,0).

The slope is equal to 2/3.

The equation of the tangent is y-0=(2/3)(x-1)

that is 2x-3y-2=0.