Lost at limits and derivatives

Question

Assume that f(x) is everywhere continuous and it is given to you that

lim f(x)+6

--------- = -10

x->3 x-3

It follows that y = _______ is the equation of the tangent line to y = f(x) at the point (___,___)

Answer 1

f'(c) = lim_x→c [(f(x) - f(c)) / (x-c)]

So, if we let c = 3 and f(c) = f(3) = -6, then the given limit is equal to f'(3).

Tangent line to the graph of y = f(x) at the point (3,-6) has slope -10.

Equation of tangent line: y - (-6) = -10(x - 3)

y + 10 = -10x + 30

y = -10x + 20