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Find equation of tangetn line to the curve

I need to see the steps for solving the following equation:  Find the equation of the tangent line to the curve at the given points for y=x^2-1/x^2+x+1  (1,0).  I know the answer is 2/3x-2/3 but I'm just not getting that when I try to derive the equation and then plug that answer into the line formula.


You should place parenthesis over the numerator and denominator.
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2 Answers

Find the equation of the tangent line to
f(x)=(x^2-1)/(x^2+x+1) at the point (1,0).

m = f’(1)

Use Quotient Rule:

f’(x) = ((x^2-1)’(x^2+x+1)–(x^2-1)(x^2+x+1)’)

f’(x) = (2x(x^2+x+1)–(x^2-1)(2x+1))

m = f’(1) = (2(3)–0)/(3^2) = 6/9 = 2/3

Tangent Line:

y–0 = 2/3 (x–1)

y = 2/3 x - 2/3


thankyou for responding timely, I have a calc exam coming up this Monday. It took me a bit to realize to plug in "1" for x to get the slope. It all worked out thanks again!!!!!
So to get the slope at (1,0) you need to find derivative of equation
y= x- 1/x+ x + 1
Rewrite 1/x2 as x-2
y= x2 - x-2 + x + 1
do derivative
y' = 2x+2x-3 + 1
To get slope plug 1 into the x
y'(1)= 2(1)+2(1-3) + 1
y'(1)= 2+2+1