# calculus 1

The point P(2,-1) lies on the curve y=1/(1-x)
If Q is the point (x, 1/(1-x) find slope of secant line.

these are the points
2, -1
1.5,2
1.9,1.111111
1.99,1.010101
1.999,001001
2.5,0.666667
2.1,0.909091
2.01,0.990099
2.001,0.999001
using the results from the points guess the value of the slope of tangent line to the curve at P(2,-1)

then find equation

the slope answer was 1
would you set it up like this
(1/(1-2)+1)/(2-1) the answer comes to zero

then the equation was y=x-3
how?

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2. 👎 0
3. 👁 1,050
1. Now that you have the slope and a point on the line, recall how you found the equation of a straight line , (the secant), from earlier grades

in the form y = mx + b
y = x + b , since m = 1
plug in the point (2,-1)
-1 = 2 + b
b = -3

so y = x - 3

or

y-(-1) = 1(x-2)
y + 1 = x - 2
y = x - 3

or

x - y - 3 = 0

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2. 👎 0

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