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
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)?
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
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?
To find the slope of the secant line passing through the points P(2, -1) and Q(x, 1/(1-x)), we can use the slope formula:
slope = (y2 - y1) / (x2 - x1)
In this case, P(2, -1) is the first point (x1, y1), and Q(x, 1/(1-x)) is the second point (x2, y2).
Substituting the values into the formula, we have:
slope = (1/(1-x) - (-1)) / (x - 2)
Now, let's calculate the slope for some of the given values of x:
1. For x = 1.5:
slope = (1/(1-1.5) - (-1)) / (1.5 - 2)
= (-2 - (-1)) / (-0.5)
= (-2 + 1) / (-0.5)
= -1 / (-0.5)
= 2
Therefore, the slope for x = 1.5 is 2.
2. For x = 2:
slope = (1/(1-2) - (-1)) / (2 - 2)
= (1/(-1) + 1) / 0
= (-1 + 1) / 0
= 0 / 0
When calculating the slope for x = 2, we encounter the division by zero, which is undefined. We cannot determine the slope for this particular point.
Regarding the points, the numbers you provided seem to be the x-values of the points. To find the corresponding y-values, substitute each x-value into the equation y = 1/(1-x). For example, for x = 1.5, the point would be (1.5, 2), and for x = 2.01, the point would be (2.01, 2).