Sketch the slope field for dy/dx = x – y

1.) g has a critical value when g'(x) = 0
(x^2 - 16)/(x - 2) = 0
(x - 4)(x + 4) = 0
x = 4 or x = -4
Therefore, g has critical values at x = 4 and x = -4

2.) g''(x) = [(x - 2)(2x) - (x^2 - 16)]/(x - 2)^2 = [2x^2 - 4x - x^2 + 16]/(x - 2)^2 = (x^2 - 4x + 16)/(x - 2)^2
g''(4) = 4^2 - 4(4) + 16 = 16 - 16 + 16 = 16 => local minimum
g''(-4) = (-4)^2 - 4(-4) + 16 = 16 + 16 + 16 = 48 => local minimum

3.) The graph of g is not concave down at any intervals.

4.) Required equation is y - 4 = g'(3)(x - 3)
y - 4 = [((3)^2 - 16)/(3 - 2)](x - 3)
y - 4 = -7(x - 3)
y = -7x + 25

5.) the tangent line lies below the graph.