software

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
------------------------------------------------------------------------
davesexcel's Profile: www.excelforum.com/member.php...oamp;userid=31708
View this thread: www.excelforum.com/showthread...hreadid=519106Just solve the quadratic equation:

0= -0.0881x^2 2.8089x-0.1615

x=31.82549

see:

mathworld.wolfram.com/QuadraticEquation.html
--
Gary''s Studentquot;PAUL GRAZIDEquot; wrote:

gt; I am setting up a calibration curve, where x= concentration and y=
gt; absorbance. The curve is polynomial.
gt;
gt; after plotting x and y values using a scattergraph and applying a trend
gt; line, i got the following regression line:
gt; y= -0.0881x^2 2.8089x 0.0505.
gt;
gt; If the absorbance reading (y) = 0.212, how do I calculate the value of x?
gt;
gt; Thanks in advance
gt;
gt;
gt;


Gary''s Student Wrote:
gt; Just solve the quadratic equation:
gt;
gt; 0= -0.0881x^2 2.8089x-0.1615
gt;
gt; x=31.82549
gt;
gt; see:
gt;
gt; mathworld.wolfram.com/QuadraticEquation.html
gt; --
gt; Gary''s Student
gt;
gt;
gt; quot;
gt; gt;[/color]
Well I got that wrong Hmmnn... I wonder why the goal seek didn't work?--
davesexcel
------------------------------------------------------------------------
davesexcel's Profile: www.excelforum.com/member.php...oamp;userid=31708
View this thread: www.excelforum.com/showthread...hreadid=519106You 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 Studentquot;davesexcelquot; wrote:

gt;
gt; 0.057672821
gt; y= -0.0881x^2 2.8089x 0.0505.
gt;
gt; If the absorbance reading (y) = 0.212, how do I calculate the value of
gt; x?
gt; I am just thinking
gt; A1=B2
gt; B2=-.0881*C1^2 2.8089*C1 .0505
gt; then I went to Tools, Goal seek
gt; set the value of A1 to .212 by changing Cell C1
gt;
gt;
gt; --
gt; davesexcel
gt; ------------------------------------------------------------------------
gt; davesexcel's Profile: www.excelforum.com/member.php...oamp;userid=31708
gt; View this thread: www.excelforum.com/showthread...hreadid=519106
gt;
gt;

Do NOT copy the polynomial coefficients from the trendline but use LINEST -
see
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 emailquot;PAUL GRAZIDEquot; gt; wrote in message
...
gt;I am setting up a calibration curve, where x= concentration and y=
gt;absorbance. The curve is polynomial.
gt;
gt; after plotting x and y values using a scattergraph and applying a trend
gt; line, i got the following regression line:
gt; y= -0.0881x^2 2.8089x 0.0505.
gt;
gt; If the absorbance reading (y) = 0.212, how do I calculate the value of x?
gt;
gt; Thanks in advance
gt;
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

quot;Bernard Liengmequot; wrote:

gt; Do NOT copy the polynomial coefficients from the trendline but use LINEST -
gt; see
gt; www.stfx.ca/people/bliengme/E...Polynomial.htm
gt; When you look at the quadratic solution, remember only positive x is
gt; meaningful in this situation
gt; best wishes from a chemist
gt; --
gt; Bernard V Liengme
gt; www.stfx.ca/people/bliengme
gt; remove caps from email
gt;
gt;
gt; quot;PAUL GRAZIDEquot; gt; wrote in message
gt; ...
gt; gt;I am setting up a calibration curve, where x= concentration and y=
gt; gt;absorbance. The curve is polynomial.
gt; gt;
gt; gt; after plotting x and y values using a scattergraph and applying a trend
gt; gt; line, i got the following regression line:
gt; gt; y= -0.0881x^2 2.8089x 0.0505.
gt; gt;
gt; gt; If the absorbance reading (y) = 0.212, how do I calculate the value of x?
gt; gt;
gt; gt; Thanks in advance
gt; gt;
gt;
gt;
gt;