Verified answer

To find lines that are parallel to the line whose equation is:

x - 2y = 3

You should write the equation in the slope-intercept y = mx + b:

-2y = -x + 3

y = (1/2)x - 3/2

The slope of this line is 1/2.

Now let's compute an equation that will give us the slope a tangent line to the given curve at any given point by computing the first derivative:

dy(x)/dx = d[(x -1)/(x + 1)]/dx

To compute this we will use the quotient rule f'(g(x)/h(x)) = {h(x)g'(x) - h'(x)g(x)}/{h(x)}²

let g(x) = x - 1, then g'(x) = 1

let h(x) = x + 1, then h'(x) = 1

dy(x)/dx = f'(g(x)/h(x)) = {(x + 1)(1) - (1)(x - 1)}/{x + 1}²

dy(x)/dx = {x + 1 - x + 1)}/{x + 1}²

dy(x)/dx = 2/{x + 1}²

Now we set the right-hand side equal to 1/2 and then solve for x:

2/{x + 1}² = 1/2

4 = {x + 1}²

{x + 1} = ±2

x = 1

x = -3

y(1) = (1 - 1)/(1 + 1) = 0

y(-3) = (-3 - 1)/(-3 + 1) = -4/-2 = 2

The points where the slope (m = 1/2) are (1, 0) and (-3, 2)

Using the point-slope for of a line, y - y1 = m(x - x1) we can write the equations for these two lines:

y - 0 = (1/2)(x - 1)

y - 2 = (1/2)(x - (-3))

y = (1/2)(x - 1)

y = (1/2)(x + 3)) + 2

y = (1/2)x - 1/2

y = (1/2)x + 7/2

The y -intercepts are -1/2 and + 7/2