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?

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

Well, let me explain it to you in my own humorous way!

To find the slope of the secant line, we need to choose any two points on the curve. So, let's choose P(2,-1) and another point, say Q. But remember, Q can be any point on the curve.

Now, the slope of the secant line is determined by the change in y divided by the change in x between the two points. So, we need to calculate (yQ - yP) / (xQ - xP).

Since P(2,-1), we can use the coordinates of the other points you listed here to calculate the slopes.

But instead of going through all the calculations, let's have some fun with these numbers! It looks like a mathematical roller coaster ride!

As we move closer and closer to point P(2,-1), the slopes of the secant lines seem to be approaching a certain value, right? That value is the slope of the tangent line to the curve at P.

You've guessed it right! The slope of the tangent line turns out to be 1.

To find the equation of the tangent line, we can use the point-slope form.

With a slope of 1, and knowing that the point P(2,-1) lies on the line, we get the equation: y = x - 3.

So there you have it, the equation of the tangent line is y = x - 3. It means that the tangent line gently kisses the curve at point P, just like a caring grandma kissing her mischievous grandchildren.

Mathematics can be quite amusing, don't you think?

To find the slope of the secant line passing through the points P(2, -1) and Q(x, 1/(1-x)), you need to calculate the difference in the y-values and the difference in the x-values and then divide the y-difference by the x-difference.

Let's calculate the slope using the given points:

For (2, -1) and (1.5, 2):
y-difference = 2 - (-1) = 3
x-difference = 1.5 - 2 = -0.5
Slope = (y-difference)/(x-difference) = 3/(-0.5) = -6.

For (2, -1) and (1.9, 1.111111):
y-difference = 1.111111 - (-1) = 2.111111
x-difference = 1.9 - 2 = -0.1
Slope = (y-difference)/(x-difference) = 2.111111/(-0.1) = -21.111111.

Similarly, you can calculate the slope for the remaining points.

Now, since the points are getting closer to (2, -1), we can see that as x approaches 2, the slope of the secant line is approaching 1.

To find the equation of the tangent line at P(2, -1), we can use the point-slope equation of a line:

y - y1 = m(x - x1),

where (x1, y1) is the given point and m is the slope.

Using the point (2, -1) and the slope we found (1), the equation becomes:

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

So the equation of the tangent line to the curve at P(2, -1) is y = x - 3.

To find the slope of the secant line passing through the points P(2, -1) and Q(x, 1/(1-x)), you need to calculate the difference in y-coordinates divided by the difference in x-coordinates. Let's take the points Q(1.9, 1.111111) and P(2, -1) as an example:

Difference in y-coordinates: 1/(1 - 1.9) - (-1) = 1/(-0.9) + 1 = -10/9 + 1 = -10/9 + 9/9 = -1/9
Difference in x-coordinates: 1.9 - 2 = -0.1

Slope of the secant line: (-1/9) / (-0.1) = 1/9 * 10 = 10/9

You can follow the same process to find the slopes for the other points.

Now, to guess the value of the slope of the tangent line at point P(2, -1), you would ideally compute the slope of the secant line as the difference in y-coordinates approaches zero. However, from the given points, we can see that as the x-coordinate of Q approaches 2, the slope of the secant line approaches 1. Therefore, we can guess that the slope of the tangent line at point P is 1.

To find the equation of the tangent line at point P, we can use the point-slope form of a linear equation. With a point P(2, -1) and a slope of 1, the equation becomes:

y - y1 = m(x - x1)
y - (-1) = 1(x - 2)
y + 1 = x - 2
y = x - 2 - 1
y = x - 3

Hence, the equation of the tangent line at point P(2, -1) is y = x - 3.