I am setting up a calibration curve, where x= concentration and y=
absorbance. The curve is polynomial.

after plotting x and y values using a scattergraph and applying a trend
line, i got the following regression line:
y= -0.0881x^2+2.8089x+0.0505.

If the absorbance reading (y) = 0.212, how do I calculate the value of x?

Thanks in advance

0.057672821
y= -0.0881x^2+2.8089x+0.0505.

If the absorbance reading (y) = 0.212, how do I calculate the value of
x?
I am just thinking
A1=B2
B2=-.0881*C1^2+2.8089*C1+.0505
then I went to Tools, Goal seek
set the value of A1 to .212 by changing Cell C1


--
davesexcel
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You are correct davesexcel:

This quadratic has two separate solutions:
one around .0576 and the other around 31.82548752

=(-2.8089+(2.8089^2-4*(-0.0881)*(-0.1615))^0.5)/(2*(-0.0881))
=(-2.8089- (2.8089^2-4*(-0.0881)*(-0.1615))^0.5)/(2*(-0.0881))

--
Gary's Student


"davesexcel" wrote:


0.057672821
y= -0.0881x^2+2.8089x+0.0505.

If the absorbance reading (y) = 0.212, how do I calculate the value of
x?
I am just thinking
A1=B2
B2=-.0881*C1^2+2.8089*C1+.0505
then I went to Tools, Goal seek
set the value of A1 to .212 by changing Cell C1


--
davesexcel
------------------------------------------------------------------------
davesexcel's Profile: http://www.excelforum.com/member.php...o&userid=31708
View this thread: http://www.excelforum.com/showthread...hreadid=519106

Just solve the quadratic equation:

0= -0.0881x^2+2.8089x-0.1615

x=31.82549

see:

http://mathworld.wolfram.com/QuadraticEquation.html
--
Gary''s Student


"PAUL GRAZIDE" wrote:

I am setting up a calibration curve, where x= concentration and y=
absorbance. The curve is polynomial.

after plotting x and y values using a scattergraph and applying a trend
line, i got the following regression line:
y= -0.0881x^2+2.8089x+0.0505.

If the absorbance reading (y) = 0.212, how do I calculate the value of x?

Thanks in advance

Gary''s Student Wrote:
Just solve the quadratic equation:

0= -0.0881x^2+2.8089x-0.1615

x=31.82549

see:

http://mathworld.wolfram.com/QuadraticEquation.html
--
Gary''s Student


"
[/color]
Well I got that wrong Hmmnn... I wonder why the goal seek didn't work?


--
davesexcel
------------------------------------------------------------------------
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View this thread: http://www.excelforum.com/showthread...hreadid=519106

Do NOT copy the polynomial coefficients from the trendline but use LINEST -
see
http://www.stfx.ca/people/bliengme/E...Polynomial.htm
When you look at the quadratic solution, remember only positive x is
meaningful in this situation
best wishes from a chemist
--
Bernard V Liengme
www.stfx.ca/people/bliengme
remove caps from email


"PAUL GRAZIDE" wrote in message
...
I am setting up a calibration curve, where x= concentration and y=
absorbance. The curve is polynomial.

after plotting x and y values using a scattergraph and applying a trend
line, i got the following regression line:
y= -0.0881x^2+2.8089x+0.0505.

If the absorbance reading (y) = 0.212, how do I calculate the value of x?

Thanks in advance

Or if you do use the chart equation, format it to display scientific notation
with 14 decimal places to avoid rounding issues.

More fundamentally, are you sure that a 4- or 5-parameter logistic would not
be a better model for your data than a polynomial?

Jerry

"Bernard Liengme" wrote:

Do NOT copy the polynomial coefficients from the trendline but use LINEST -
see
http://www.stfx.ca/people/bliengme/E...Polynomial.htm
When you look at the quadratic solution, remember only positive x is
meaningful in this situation
best wishes from a chemist
--
Bernard V Liengme
www.stfx.ca/people/bliengme
remove caps from email


"PAUL GRAZIDE" wrote in message
...
I am setting up a calibration curve, where x= concentration and y=
absorbance. The curve is polynomial.

after plotting x and y values using a scattergraph and applying a trend
line, i got the following regression line:
y= -0.0881x^2+2.8089x+0.0505.

If the absorbance reading (y) = 0.212, how do I calculate the value of x?

Thanks in advance