Calculus of a Single Variable, Chapter 2, 2.2, Section 2.2, Problem 54

a) So we are given: y=x^3 - 3x
If we derive this function, then we will have the following:
y'=3x^2 -3
This will show us what the slope of the original line would be at any x value that we plug in. So if we want to know the equation of the line tangent to the graph at (2,2), then we have to plug in that x value:
f'(2)=3(2)^2 -3=9
So the slope of the graph at (2,2) should be 9, or 9/1
Since we have to have "m" and "b" for the slope-intercept(y=mx+b) form of an equation, we just need to find "b" now. We can do that by plugging in the slope (m), x value, and y value that we have now:
2=(9)2+b
b=-16
So the equation of the line should be:
y=9x-16