Expert: Sherry Wallin - 10/4/2009

Question
Sherry,
Here is an exact copy/paste of my actual problem.  Let me know if you can help me with this.  Thanks again, Mikel

12. Let f (x) = 2x+16/x2. Then the equation of the
tangent line to the graph of f (x) at the point (2,8) is given by
y = mx+b for
m =
and
b =

Answer
Ok the tangent line at any point has slope f'(x). (This is one interpretation of the first derivative). First lets rewrite f(x)= 2x +16*x^-2, now calculate f'(x) = 2 +-2*16x^-3 = 2+ -32/x^3. At the point (2,8) x is 2 so f'(2) = 2-4 = -2. That's the slope of the tangent line on f(x) at the point (2,8) and call it m. Then use what you know about lines since what you are trying to do is write the equation of the tangent line. You have a point and you have the slope so use the point slope formula of a line: y-y1=m(x-x1). In your case y -8 =-2(x-2). Simplify by putting this equation into slope intercept form: y - 8 = -2x + 4 -> y = -2x + 12. This tells you that b = 12 while m = -2

Any questions?

Math Prof