Calculus

A) The point p(3,1) lies on the curve y= square root x-2. If Q is the point (x, square root x-2) use your calculator to find the slope of the decant line PQ for the following values of x
1) 2.5
2.) 2.9
3.) 2.99
4.) 2.999
5.) 3.5
6.) 3.1
7.) 3.01
8.) 3.001

B) using the results of part a, guess the value of the slope of the tangent line to the curve at P(3,1)
C) using the results from part b, find an equation of the tangent line to the curve at P(3,1)
D) sketch the curve, two of the scant lines and the tangent line.

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  1. I will do one of the points, you do the others in the same way
    Slope of the secant = change in y/change in x
    using x = 3.01 , Q is (3.01, √(3.01 - 2)) = Q(3,1.004987...)
    slope PQ = (1.004987.. - 1)/(3.01 - 3) = .4987 , looks close to .5 or 1/2

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