Judy A.

# how to find point of inflection ?

i have 5 set of data
when i plotting these data , i got straight line but in the end it bend to the right side.
by calculating r2, i knew that i got straight line by removing the two points where ris 0.994.
Then in this case , can i use the value between the third point and fourth point  as  the value of inflection point?

By: Tutor
4.9 (135)

Patient MIT Grad For Math and Science Tutoring

Judy A.

After removing two point i got r=0.99, actually i want straight line to get the value of intercept. but i also need to find out the point of inflection to calculate the pka value. Actually this is drug stability kinetic problem. Therefore can i use the value between  the last point of the straight line and the adjacent point.these are the data
xvalue= 1.5,2.45,3.46,4.24,7.17
y value=-1.1737,-2.2823,-3.1739,-3.6989,-3.886.
Report

11/28/14

Judy A.

these are the data
X= 1.5,2.45,3.46,4.24,7.17
Y= -1.1737,-2.2823,-3.1739,-3.6989,-3.886
Actually , i need straight line to find out the intercept value and also need point of reflection . In this case , can i use the value between the last point of straight and the adjacent point.
Report

11/28/14 Russ P.

Judy,
Given your data below, I think you were originally correct, so you can determine the crossover point of the two lines in my "kink" above which occurs between x = 3 & x=4.

Your fitted regression line for the first 3 data points is  y = mx + b or approximately y = -1.019 x + 0.307, using the formulas (where I removed the subscripts for clarity) and N=3:
m = {N Σ(xy) - (Σx)(Σy)}/{N Σ(x2) - (Σx)2}   and  b = {Σy - m Σx}/N

And the straight line for data points 4 & 5 i  y = -0.06386 x - 3.4281  from m = Δy/Δx  and b = yi - mxi i = 5.

So set the Y's of these 2 lines equal, solve for x at the crossover point, and then the y from either line and you get:
x = 3.26 & y = 3.02  for the kink point. where the slope changes as line 1 is crossed over to line 2.

NOTE: Five data points is rather sparse, and the precision of the data is questionable, so realistically that point occurs somewhere in the neighborhood of the calculated kink point. Also, by looking at how the slope (of the line segment between 2 successive data points) changes (-1.167, -0.883, -0.673, -0.064), it is pretty clear that it is flattening out approaching zero so the drug becomes stable with x.  Hence, our "kink" approach is not a bad simplification and approximation.
Report

11/28/14 Russ P.

Judy,
I must have forgotten to save the long version of my comment, so here is the short version:

The "kink" approach provides a reasonable estimate of where the slope changes dramatically from the regression line based on the first 3 data points ( y = -1.019 x + 0.307) and the second line through data points 4 & 5
(y = -0.06386 x - 3.4281).  Setting the y's equal, solving for x, then y in either line, we get x = 3.26 & y = -3.02 as the approximate "kink" inflection point.  So the drug is stable as the slope approaches zero beyond that point.
Report

11/28/14

## Still looking for help? Get the right answer, fast.

Get a free answer to a quick problem.
Most questions answered within 4 hours.

#### OR

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.