Question

find an equation of the tangent line to the given curve at the specified point.
$y = \frac{x^2 - 1}{x^2 + x + 1}$, $\\ (1, 0)$
$y = \square$

Explanation:

Step1: Apply quotient rule for derivative

Let \( f(x) = x^2 - 1 \), \( g(x) = x^2 + x + 1 \). Quotient rule: \( y' = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \).
\( f'(x) = 2x \), \( g'(x) = 2x + 1 \).
So \( y' = \frac{2x(x^2 + x + 1) - (x^2 - 1)(2x + 1)}{(x^2 + x + 1)^2} \).

Step2: Substitute \( x = 1 \) into derivative

Numerator at \( x=1 \): \( 2(1)(1+1+1) - (1-1)(2+1) = 2(3) - 0 = 6 \).
Denominator at \( x=1 \): \( (1+1+1)^2 = 9 \).
Slope \( m = \frac{6}{9} = \frac{2}{3} \).

Step3: Use point-slope form

Point-slope: \( y - y_1 = m(x - x_1) \).
Substitute \( (1,0) \) and \( m=\frac{2}{3} \): \( y - 0 = \frac{2}{3}(x - 1) \).
Simplify: \( y = \frac{2}{3}x - \frac{2}{3} \).

Answer:

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