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.

## Find equation of tangetn line to the curve

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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)’)

/((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

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

# Comments

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

^{2 }- 1/x^{2 }+ x + 1Rewrite 1/x

^{2}as x^{-2}so

y= x

^{2}- x^{-2}+ x + 1do derivative

y' = 2x+2x

^{-3}+ 1To get slope plug 1 into the x

y'(1)= 2(1)+2(1

^{-3) }+ 1y'(1)= 2+2+1

y'(1)=5

y=5(x-1)

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