Expert: Paul Klarreich - 11/21/2010

Question
Find the slope of a line tangent to the curve by=x^2 at the point P(3,9)by finding the limit of the secant lines PQ where Q has x- value 2.5,2.9,2.99,and2.999. can you explain this to me step by step.
I can not do the caculations for the table.for colume Q1,Q2,Q3,Q4

Answer
Questioner:   ken
Country:  United States
Category:  Calculus
Private:  No
 
Subject:  Differential calculus
Question:  Find the slope of a line tangent to the curve by=x^2 at the point P(3,9)by finding the limit of the secant lines PQ where Q has x- value 2.5,2.9,2.99,and2.999. can you explain this to me step by step.
I can not do the caculations for the table.for colume Q1,Q2,Q3,Q4
............................

The slope of the secant line is:
   f(x) - f(a)
m = ------------
      x - a

In your example, use:

f(x) = x^2

a = 3

x = 2.5,
x = 2.9,  << I'll do this one:
x = 2.99,
x = 2.999,

   f(2.9) - f(3)<< always f(3), which is 9
m = ------------
      2.9 - 3<< always 3


   2.9^2 - 9
m = ----------
      -.1

   8.41 - 9
m = --------
      -.1

   - .59
m = -------
    -.1

m = .5.9

You can do the others.