00:01
In this problem we are given the curve y is equal to 2 by 5 minus x and we are given that the point p with coordinate 6 negative 2 lies on this curve.
00:12
And we are also given that q is the point x 2 by 5 minus x.
00:20
We are asked to find the slop of the second line pq for different values of x.
00:26
The first value of x is x is equal to 5 .9.
00:31
So if m pq denotes the slop of the second line, then by definition for a point x1 y1 and another point x2 y2, the slope between them is defined as y2 minus y1 by x2 minus x2.
00:57
So here, p is the point x1 y1 and q is the point x2 y2 so that by definition this is y2 which is 2 by 5 minus x minus y 1 which is negative 2 divided by x which is divided by x2 which is x minus x1 is 6.
01:19
So this slop can be simplified as 2 by 5 minus x plus 2 divided by x minus 6.
01:27
This can be further simplified as 2 plus 10 minus 2x by 5 minus x into x minus 6.
01:39
That is this is negative 2 times x minus 6 divided by 5 minus x times x minus 6.
01:49
So the common factor x minus 6 can be cancelled from the numerator and denominator.
01:54
So that we get the slope as negative 2 by 5 minus 6.
01:58
X for any point pq so the slop m pq is negative 2 by 5 minus x so when x is equal to 5 .9 the slop m pq is negative 2 by 5 minus 5 .9 which is negative 2 by negative 0 .9 which can be approximated to a value of 2 .2 2 .2 so at the point 5 .8 5 .9 the slope is 2 .22.
02:33
The second point is x is equal to 5 .99 so that the slop mpq is determined as negative 2 by 5 minus x.
02:43
So here x is 5 .99 so that this is equal to negative 2 by negative 0 .99 which can be approximated to a value 2 .0 to 0 .02.
02:58
So the slop at this point at this point.
03:00
Point is 2 .0202.
03:03
The third point is the point x is equal to 5 .999 so that the slop mpq is negative 2 by 5 minus x.
03:14
Here x is 5 .999 so that this is negative 2 by negative point 999...
