Question

Suppose g(x) is a differentiable function such that g(–4) = 0 and g′(–4) = 3. Find an equation to the line tangent to y = g(x) at x = -4.

Answer 1

any idea?

Answer 2

Start with fact that tangent line has the slope of the derivative of the function at that point. slope = g'(-4) = -3.

Then put a line through the point of tangency (-4, 0) with that slope.