Find equations of the tangent lines to the curve \[ y=\frac{x-1}{x+1} \] that are parallel to the line \( x-2 y=4 \). (Enter your answers as a comma-separated list of equations.) F

To find the tangent lines to the curve \( y=\frac{x-1}{x+1} \) that are parallel to the line \( x - 2y = 4 \), we first need to determine the slope of the given line. Rewriting it in slope-intercept form gives us: \[ 2y = x - 4 \] \[ y = \frac{1}{2}x - 2 \] Thus, the slope \( m \) is \( \frac{1}{2} \). Next, we need to find the derivative of the curve to determine the points where the slope of the tangent line equals \( \frac{1}{2} \). The derivative \( y' \) will give us the slope of the tangent line at any point on the curve. The derivative of \( y = \frac{x-1}{x+1} \) is obtained using the quotient rule: \[ y' = \frac{(x+1)(1) - (x-1)(1)}{(x+1)^2} = \frac{(x + 1) - (x - 1)}{(x + 1)^2} = \frac{2}{(x + 1)^2} \] Setting the derivative equal to the desired slope: \[ \frac{2}{(x + 1)^2} = \frac{1}{2} \] Cross-multiplying gives: \[ 2 \cdot 2 = (x + 1)^2 \] \[ 4 = (x + 1)^2 \] Taking the square root of both sides gives two solutions: \[ x + 1 = 2 \quad \text{or} \quad x + 1 = -2 \] Thus, 1. \( x = 1 \) leads to \( y = \frac{1-1}{1+1} = 0 \) 2. \( x = -3 \) leads to \( y = \frac{-3-1}{-3+1} = 2 \) Now we can write the equations of the tangent lines at these points, using point-slope form \( y - y_1 = m(x - x_1) \): 1. For the point \( (1, 0) \): \[ y - 0 = \frac{1}{2}(x - 1) \rightarrow y = \frac{1}{2}x - \frac{1}{2} \] 2. For the point \( (-3, 2) \): \[ y - 2 = \frac{1}{2}(x + 3) \rightarrow y = \frac{1}{2}x + \frac{1}{2} \] Thus, the equations of the tangent lines are: \[ y = \frac{1}{2}x - \frac{1}{2}, \quad y = \frac{1}{2}x + \frac{1}{2} \]