 Home
 Products
 Integration
 Tutorial
 Barcode FAQ
 Purchase
 Company
print barcode image c# \I& ye in .NET framework
\I& ye GTIN  13 Creation In Visual Studio .NET Using Barcode printer for .NET framework Control to generate, create UPC  13 image in Visual Studio .NET applications. Scanning EAN13 In .NET Framework Using Barcode recognizer for VS .NET Control to read, scan read, scan image in .NET applications. Watch
Barcode Encoder In Visual Studio .NET Using Barcode maker for .NET Control to generate, create bar code image in Visual Studio .NET applications. Bar Code Recognizer In VS .NET Using Barcode reader for Visual Studio .NET Control to read, scan read, scan image in Visual Studio .NET applications. GRAPHING
Making EAN13 Supplement 5 In C# Using Barcode creation for VS .NET Control to generate, create EAN / UCC  13 image in Visual Studio .NET applications. Encode UPC  13 In VS .NET Using Barcode creation for ASP.NET Control to generate, create EAN13 image in ASP.NET applications. again at the curve in Fig 47 and see that this is a very reasonable point for the curve to
Print European Article Number 13 In Visual Basic .NET Using Barcode creator for .NET Control to generate, create UPC  13 image in Visual Studio .NET applications. Encoding UPC Symbol In .NET Using Barcode drawer for .NET Control to generate, create UCC  12 image in VS .NET applications. change slope from becoming more and more negative to becoming more and more positive This is another graphing tool involving calculus Create Matrix 2D Barcode In .NET Framework Using Barcode maker for VS .NET Control to generate, create 2D Barcode image in Visual Studio .NET applications. Make EAN 128 In VS .NET Using Barcode encoder for Visual Studio .NET Control to generate, create EAN / UCC  14 image in .NET framework applications. Sketchthegraphof y = 8 x 5 5x4 20x3
1D Generator In .NET Using Barcode encoder for Visual Studio .NET Control to generate, create 1D Barcode image in .NET applications. Making Leitcode In Visual Studio .NET Using Barcode generation for VS .NET Control to generate, create Leitcode image in Visual Studio .NET applications. Solution: This is a Sth degree curve so it increases (rises) rapidly with large positive x and goes rapidly negative for large negative values of x The function factors to y = x3(8x2  5x  20) producing the points x=o, y = o EAN13 Supplement 5 Generator In VB.NET Using Barcode generation for VS .NET Control to generate, create European Article Number 13 image in Visual Studio .NET applications. Printing ANSI/AIM Code 128 In None Using Barcode generator for Online Control to generate, create Code 128 Code Set C image in Online applications. The first derivative is y'= 40x4  20x3  60x2 y'=20x2(2x2 33) y'= 20x2(2x3)(x+1) Code 3 Of 9 Reader In C# Using Barcode scanner for .NET Control to read, scan read, scan image in Visual Studio .NET applications. Draw UPC A In Java Using Barcode creation for Java Control to generate, create UPCA Supplement 2 image in Java applications. Fig 48 Paint Code 3/9 In ObjectiveC Using Barcode creation for iPhone Control to generate, create ANSI/AIM Code 39 image in iPhone applications. EAN 128 Creator In VB.NET Using Barcode generator for .NET Control to generate, create EAN128 image in .NET applications. Setting the frrst derivative equal to zero (y'=O) produces three points x = 0, 3/2, and  1 These are the points where the curve has zero slope The second derivative is y"=16Ox3  6 0 ~ 1~ 0 ~  2 y"= 2Ox(8x2  3 ~  6 ) Code39 Reader In .NET Using Barcode recognizer for .NET Control to read, scan read, scan image in .NET framework applications. Data Matrix ECC200 Generation In None Using Barcode maker for Word Control to generate, create DataMatrix image in Office Word applications. To determine the shape of the curve where the slope equals zero, find y" at each point: y"(0) = 0 , horizontal point of inflection; y"(3/2) = 225, U shape; y"(1) = 100, n shape The value of the function at each turning point is found by putting the values of x in the function At x = O , y(O)=O At x = 3/2, ~ ( 3 1 2= (3/2)3 [8(3/2)2  5(3/2)  201 = 64 ) At x = 1, y(1) = 1(8+5 20) = 7 CHAPTER^
All the inflection points, both horizontal and vertical, are found by setting y"= 0 : 20x(8x2  3 x  6 ) = 0 The inflection points are at x = 0 and the solutions to 8 x 2  3 x  6 = 0 are
3 f J94(8)(6)  3 k142 = 070,l 1 16
These are most reasonable points, being where we expect the points of inflection to occur The function is sketched in Fig 48 So far we have looked at polynomials This is the type of function you will encounter most often Your course may or may not include the graphing of rational functions (polynomial fiactions) Polynomial fiactions introduce one more interesting twist to the use of derivatives in curve sketching, what happens to a curve when the derivatives are undefined This is best illustrated by example Sketch the graph of y = x2/x1 Solution: At x = l this h c t i o n is undefrned (l/O) Therefore draw a dashed vertical line on the coordinate axes at x = l indicating that the curve may exist to the right or left of this line, but not on the line There is no dominant term in the same manner as for polynomials but, applying similar reasoning, look what happens when x is a large positive or negative number When x is large the x1 in the denominator looks like x and the h c t i o n looks like y = x In the language of the chapter on limits: As x+fao, y = x
Add a dashed line, y = x , to the coordinate axes remembering that this is an asymptote line
Fig 49 Now apply some calculus analysis The first derivative of the function is, using the quotient rule: GRAPHING y' = (x1)(2x)x2 (xl)2  2x2 2xx2  x 2 2x (X1)2 (x  1)2 Before setting y'= 0 , note that the derivative does not exist at x = 1 But we already knew that because the function does not exist at x = l so it is not surprising that the derivative does not exist there Note, however, that as x approaches 1 from either the positive or negative side, the slope of the curve is negative This information may be helpfbl in sketching the graph (See Fig 49) Setting y'= 0 produces x(x2) = 0 , and the two points where the slope equals zero, x = 0 , and x = 2 The values of the firnction for these two points are: y(O)=O and y(2)=L = 4 so the slopeof the curve is zero at (0,O) and (2,4) If you are unsure of the shape of the curve in certain regions, check a point With the information generated from the calculus and the concepts of limits you should get very close to the correct curve As you gain more confidence you will not resort to checking specific areas of the curve by testing a point The previous problem is typical of the more difficult ones you will encounter in your course It is probably beyond what you will encounter on a test because of the complexity of the analysis and the potential for confusion Sketches of the graphs of polynomials are much more popular as test problems Know how to graph polynomials and you will be well along toward a good test score in graphng Having gone through examples of what you can expect to encounter in graphing problems, it is now time to write down some procedural guidelines for graphing curves of the general form y = f ( x )

