Solution: Performing the operations on the specified function in .NET framework

Generation EAN 13 in .NET framework Solution: Performing the operations on the specified function

In visualizing problems it is very helpful to know what certain functions look like You should review the hctions described in this section until you can look at a h c t i o n and picture "in your mind's eye" what it looks like This skill will prove valuable to you as you progress through your calculus course

The linear algebraic function (see Fig 1-7 ) is y = mx +b, where m is the slope of the straight line and b is the intercept, the point where the line crosses the y-axis Th~sis not the only form for the linear hction, but it is the one that is used in graphing and is the one most easily visualized

Fig 1-7

1-15

Graph the function y = 2x - 3

Solution: This is a straight line, and it is in the correct form for grqhing Because the slope is positive, the curve rises with increasing x The coefficient 2 tells you that the curve is steeper than a slope I, (which has a 45" angle) The constant 3 is the intercept, the point where the line crosses the y-axis (See Fig 1-8)

You should go through this little visualization exercise wt every function you graph ih Knowing the general shape of the curve makes graphing much easier With a little experience you should look at this function and immediately visualize that (1) it is a straight line (first power), (2) it has a positive slope greater than 1 so it is a rather steep line rising to the right, and (3) the constant term means that the line crosses the y-axis at -3 Knowing generally what the line looks like, place the first (easiest) point at x = 0, y = -3 Again knowing that the line rises to the right, pick x = 2, y = 1, and as a check x = 3, y = 3

Fig 1-8

If you are not familiar wt visualizing the function before you start calculating points ih graph a few straight lines, but go through the exercise outlined above before you place any points on the graph

Quadratics The next most complicated function is the quadratic (see Fig 1-9), and the simplest quadratic is y = x2, a curve of increasing slope, symmetric about the y-axis (y has the same value for x = + or - 1, + or - 2, etc) This symmetry property is very uselid in graphing Quadratics are also called parabolas Adding a constant to obtain y = x2 + c serves to move the curve up or down the y-axis in the same way the constant term moves the straight line up and down the y-axis

y=x2-3

Fig 1-9

1-16

Graphy=x2-3

Solution: First note that the curve is a parabola with the symmetry attendant to parabolas and it is moved down on the y-axis by the -3 The point x = 0, y = -3 is the key point, being the apex, or lowest point for the curve, and the defining point for the symmetry line, which is the y-axis Now, knowing the general shape of the curve add the point x = +,2, y = 1 This is sufficient information to construct the graph as shown in Fig 1-9 Further points can be added if necessary

Adding a constant a in front of the x2 either sharpens (a > 1) or flattens (a < 1) the graph A negative value causes the curve to open down

Solution: Looking at the function, note that it is a parabola (x2 term), it is flatter than normal (05 coefficient), it opens up (positive coefficient of the x2 term), and it is moved up the axis one unit Now put in some numbers: x = 0, y = 1 is the apex, and the y-axis is the symmetry line Add the points x = +2, y = 3 and sketch the graph (Fig, 1-10)

1 1 1

Fig 1-10

1-18

Graphy=-2x2-2

Solution: Look at the function and veri& the following statement This is a parabola that opens down, is sharper than normal, and is displaced two units in the negative direction Put in the two points x = 0 and x = +1 and veriQ the graph shown in Fig 1-10

Adding a linear term, a constant times x, so that the h c t i o n has the form y = L& +bx + c produces the most complicated quadratic The addition of this constant term moves the curve both up and down and sideways If the quadratic fbnction is factorable then the places where it crosses the x-axis are obtained directly from the factored form

1- 19 Graph the function y = f ( x ) = x2 + 2x - 8

Solution: This is a parabola that opens up, and is displaced up or down and sideways This quadratic is factorable to y = (x + 4)( x - 2) The values x = 2 and x = -4 make y = 0 so these are the points where the curve crosses the x-axis Place these points on the graph

Now here is where the symmetry property of parabolas is used Because of the symmetry, the parabola must be symmetric about a line halfway between x = 2 and x=-4, or about the line x = -1 The apex of the parabola is on this x = -1 line so substitute to find the appropriate value of y: f(-1) = (-1 + 4)( -1 - 2) = -9 These three points are sufficient to sketch the curve (see Fig 1- 1 1)

+2~-8

Fig 1-11