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
