Given the curve:

`y=\frac{x^2-1}{x^2+x+1}`

We have to find an equation of the tangent line to the given curve at the specified point (1,0).

Let us find the slope of the tangent by taking the first derivative of the curve.

`y'=\frac{(x^2+x+1)\frac{d}{dx}(x^2-1)-(x^2-1)\frac{d}{dx}(x^2+x+1)}{(x^2+x+1)^2}`

`=\frac{2x(x^2+x+1)-(x^2-1)(2x+1)}{(x^2+x+1)^2}`

`=\frac{2x^3+2x^2+2x-2x^3-x^2+2x+1}{(x^2+x+1)^2}`

`=\frac{x^2+4x+1}{(x^2+x+1)^2}`

Now the slope of the tangent at point (1,0) is:

`y'=m=\frac{2}{3}`

The equation of the tangent line can now be written as,

`y-0=m(x-1)`

`y-0=\frac{2}{3}(x-1)`

`y=\frac{2}{3}x-\frac{2}{3}`

The general equation of a line passing through the point (a, b) and with slope m is `(y - b)/(x - a)` = m.

We need to determine the equation of a line that passes through the point=(1,0). The value of (a, b) here is (1,0).

Given a curve y = f(x), the slope of a tangent at the point (p, q) is f'(p).

For the curve y= (x^2-1)/(x^2+x+1)

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

(The entire section contains 4 answers and 563 words.)

## Unlock This Answer Now

Start your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.