The equation of this tangent line can be written in the form

Question

f (x) = 4x²-4

(A) Find the slope of the line tangent to f(x) at x = - 2: _______

(B) Find the instantaneous rate of change of f (x) at x - 2: ________

(C) Find the equation of the line tangent to f (x) at x - 2: y = ________

Answer 1

1. Slope = rate of change = (change in y) / (change in x) = dy / dx
2. Rate of change at a single point = derivative at that point because the derivative is the equation of the slope
3. Drawing a tangent line that touches the graph at one point must have the same slope at that point (and a line has a constant slope because it's linear)

A) slope @ -2 = f'(x) @ -2 = 8*x @ 2 = 8(-2) = -16

B) Same as A (slope of tangent line is equal to the derivative (instant rate of change = slope @ that point)

C) Slope = m = (y2-y1) / (x2 - x1) right? That's the change in y over change in x. multiply over the (x2-x1) and get y2-y1 = m(x2-x1).

For a general equation y2 and x2 are just y and x because thats the next y and x value you're solving for so write as Y - y1 = m(X - x1), m is the slope at tht point which we found to be 16.

Plug in your values and get Y - f(-2) = m(x- -2) = Y - f(-2) = -16 *(X + 2)

f(-2) = 4(-2)²-4 = 12 so plug that in and you get Y - 12 = -16*(x+2) and get Y = -16*(x+2) +12 =

Y = -16*x - 20