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Feb 8th, 2004, 09:59 AM
#1
Thread Starter
New Member
a vb math problem
Hi, i'm new here. I making a ecuation plotter without mscript and i have the following problems:
1º I want to know how to get te convex and concav intervals. I'm trying with the second derivative (f''(x)).
2º How to get the maximums and the minimums (with derivative)
3º And finally, how to get the asymptotic vertical and oblicuos.
I have f'(x) and f''(x) done and i tried things but doesn't work.
This is the main part of the program (in a module):
[Highlight=VB]
Public Sub Dibujar()
Dim X As Single, Y As Single
Dim XA, YA As Single
' Sistema de coordenadas
Form1.Picture1.Scale (-(Form1.Slider1.Value), Form1.Slider1.Value)-(Form1.Slider1.Value, -(Form1.Slider1.Value))
' Eje x
Form1.Picture1.Line (-(Form1.Slider1.Value), 0)-(Form1.Picture1.ScaleWidth, 0)
' Eje y
Form1.Picture1.Line (0, (Form1.Slider1.Value))-(0, Form1.Picture1.ScaleHeight)
'Dibujar curva
For X = -(Form1.Slider1.Value) To Form1.Slider1.Value Step 0.001
Y = (Form1.Text1.Text * X ^ 3 + Form1.Text2.Text * X ^ 2 + Form1.Text3.Text * X + Form1.Text4.Text) / _
(Form1.Text5.Text * X ^ 3 + Form1.Text6.Text * X ^ 2 + Form1.Text7.Text * X + Form1.Text8.Text)
Form1.Picture1.PSet (X, Y)
Next X
Form1.Cuadricula.Enabled = True
End Sub
[code]
The way to solve the asymptotic vertical is fyinding the value of x with this ecuation y=(ax^3+bx^2+cx+d)/(ex^3+fx^2+gx+h), next 0 = ex^3+fx^2+gx+h to found when the division is by 0. But i don't know how i can solve this equation (maybe x= root((fx^2+gx+h)/a) but with the root ^3). I don't know, thanks for all.
And sorry for my english.
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