Expert: Paul Klarreich - 11/6/2006

Question
How do you find the equation of the tangent line to the graph of f at th given point.

f(s) = (s-5)(s^2-3) at(1,8)

I've tried to do this problem and the many other problems I have. But i don't think, I'm undertanding. I did however find the derivative of the problem which is 3s^2-10s-3

Answer
Questioner:  SimoneRenee
Category:  Calculus
 
Subject:  Tangent Lines
Question:  How do you find the equation of the tangent line to the graph of f at the given point.

f(s) = (s-5)(s^2-3) at(1,8)

I've tried to do this problem and the many other problems I have. But I don't think I'm undertanding. I did however find the derivative of the function which is 3s^2-10s-3
......................................
Hi, SimoneRenee,  [Which is your name?]

To find the equation of the tangent line, use the Point-slope Form of the equation of a line:

y - y0 = m(x - x0), where

1. (x0,y0) is a point on the line, and you get y0 = f(x0) if you don't already have it, and

2. m is the slope, and you get it as  f'(x0)

In this case, you already have x0 = 1, y0 = 8, and you need m.  [OK, it's  s0, not x0, but who's looking?]

f'(s) = 2s^2 - 10s - 3, which you already got.

f'(s0) = f'(1) = 2 - 10 - 3 = - 11

So your equation is:

y - 8 = -11(x - 1), which you can simplify a bit:

y - 8 = -11x + 11
11x + y = 19