Question

Find equations of the tangent lines to the curve

Answer 1

Procedure: derive the function y and equal the derivative to the slope of the line.

slope of x - 2y = 3.

y = x/2 - 3/2 => slope = 1/2

y' = [x-1] / [x+1] = [(x+1) - (x-1)] / (x+1)^2 = 2 / (x+1)^2

2 / (x+1)^2 = 1/2

(x+1)^2 = 4

x+1 = +/- 2

x = -1 + 2 and x =x = -1 -2

x = 1 and x = -3

Now find the y-coordinate of the function for both x-values

x = 1 => y = [1-1]/[1+1] = 0

x = -3 => y = [-3+1]/[-3-1] = -2 / -4 = -1/2

Now you find the equations of the tangent using the slope and the point.

(1,0) and slope 1/2 => y - 0 = [1/2] (x -1)

y = x/2 - 1/2

(-3, -1/2) and slope 1/2 => y - (-1/2) = [1/2] (x -(-3))

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

y = x/2 + 3/2 - 1/2 = x/2 +1

The tangent lines of the function have these equations:

   y = x/2 - 1/2

   y = x/2 +1