f(x) = x^2 + x + 3
I know f ' (x) = 2x + 1.
Now how do I find the equation of the tangent line to the curve at x = -2, or any value of x for that matter
Hopefully you understand where this comes from, Zabulius. In particular, you should be aware that the derivative gives a formula for finding the slope at any value of x for a particular function. the tangent line is a straight line, and is hence of the form y = mx + b where m is the slope (hence given by the derivative) and b is the y-intercept. once you find the slope and a point the line passes through (this is the point (a, f(a)) in the above formula) you can find the equaiton of the line