Question

Can you help me find the equation of the tangent line?

If y =`(x^2 -1)/(x^2+x+1)` given the point (1,0), what is y ' ?

Answer 1

`y = 2/3x - 2/3`

Explanation:

Finding `y'` requires the use of the Quotient Rule:

If `f(x) = (g(x))/(h(x))`, then `f'(x)=((h(x)*g'(x) - g(x)*h'(x)))/([h(x)]^2)`

Thus:

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

`y' = (2x^3 +2x^2 + 2x -2x^3-x^2+2x+1)/(x^2+x+1)^2`

`y' = (x^2 + 4x +1)/(x^2+x+1)^2`

`y'(1) = (1^2 + 4(1) + 1)/(1^2 + 1 + 1)^2 = 6/9=2/3`

Since we have the point at `(1,0)` and the slope is `2/3`, we can determine the tangent line:

`y-0 = 2/3(x-1)`

`y = 2/3x - 2/3`

graph{((x^2-1)/(x^2+x+1)-y)(2/3x-2/3-y)=0 [-2.772, 4.157, -1.885, 1.58]}