Discussion:
f '(x) = +/- sqrt(-1)
(too old to reply)
spiny urchin
2015-05-28 14:30:27 UTC
Permalink
Raw Message
A random equation is deduced for the purpose of the problem:

f(x)=-10-2*x+2*(x^2)-x^3+2*(x^4)

The equation of its tangent line is determined:

g(x)=27.7006*x-27.7006*1.567318

And the equation of the normal is found:

h(x)=-27.7006*x+27.7006*1.567318

Precipitating the graph on the web page,

http://www.stonetabernacle.com/intersect-curve.html

* different points and lines on the graph are calculated*

A,B,C,D,E,F,X1,Xc,X2

From her the general case is determined. The roots to the 7th degree
polynomial are attempted.

This quantity apparently had to be true from the equations:

f '(x) = +/- sqrt(-1)
Skeezix LaRocca
2015-05-28 16:44:25 UTC
Permalink
Raw Message
Post by spiny urchin
f(x)=-10-2*x+2*(x^2)-x^3+2*(x^4)
g(x)=27.7006*x-27.7006*1.567318
h(x)=-27.7006*x+27.7006*1.567318
Precipitating the graph on the web page,
http://www.stonetabernacle.com/intersect-curve.html
* different points and lines on the graph are calculated*
A,B,C,D,E,F,X1,Xc,X2
From her the general case is determined. The roots to the 7th degree
polynomial are attempted.
f '(x) = +/- sqrt(-1)
Thats great, but can you make a good pot of chili ?
--
Freedom of religion is great, but I'll take freedom FROM it any day.
Loading...