The point P(2,-1) lies on the curve y=1/(1-x)

Question

If Q is the point (x, 1/(1-x) find slope of secant line.

these are the points
1.5 1.9 1.99 1.999 2.5 2.1 2.01 2.01 2.001

when i calculated it i got a negative number
1/(1-1.5)=-2 but the answer is 2. also would the points be (1.5, 2) 0r (2,2)?

Answer 1

first let's look at the derivative (slope at a point)
y = 1/(1-x)
dy/dx = slope = [-1(-1)]/(1-x)^2

bottom always +
so the slope is always positive

now secant from (2,-1) to
( x , 1/(1-x) )

secant slope P to Q =
(1/(1-x) +1) / (x-2)

(1 + 1 - x )/ [(1-x)(x-2)]

= (x-2) / (x^2 - 3x + 2)

if x = 1.5 then
= (-.5 /(2.25 - 4.5 + 2
= -.5 / (-.25)
= + 2 sure enough

Answer 2

it says to round to 6 decimal numbers

so it would for example the first point
(1.5, 2)
then it would be

(1.9,1.111111) right?