Question

?Given f(x)= x^2 - 2x and the derivative is f'(x)= 2x - 2. The slope of the tangent line at x = - 3 is -8. The slope of the tangent line at x= -1 is -4. The slope of the tangent line at x= 4 is 6.

How to graph f, and sketch in the tangent lines at x = -3,-1,4?

Answer 1

See below

Explanation:

The graph of `f(x)` is a parabola with `x` intercepts `0` and `2`.

graph{x^2-2x [-24.38, 26.94, -3.49, 22.16]}

Now go to the point where `x=-3` (and `y = f(-3) = 15`) and draw in a line that just touches the graph at that point. Since the slope of that line is -8, it should also contain the point `(-2,7)`.

The tangent at `x=-1` (and `y = 3`) has slope `-4`, so it also goes through the point `(0,-1)`.

The tangent at `(4,8)` also contains the point `(5,14)`.