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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.

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)’)
/((x^2+x+1)^2)

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

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

so
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
y'(1)=5

y=5(x-1)