Calculus
Lessons 29 - 44

Calculus
Translations
lesson 29

Translations

As we have seen with the graphs of the parabola, circle, ellipse and hyperbola if we replace x by x+ a constant it will shift the graph to the left or right.  If we add a constant the graph will be shifted up or down that number of units. Try this with some different equations.  Graph this function by highlighting the cell and pressing enter.

Plot[f[x] = Abs[x], {x, -4, 4}];

Change the following equation and see if you can shift the graph 2 units left and 2 units down.

Plot[f[x] = Abs[x], {x, -4, 4}];

Try this one

Plot[f[x] = ((x^2)/2), {x, -4, 4}];

Then see if you can shift the graph 3 units right and 2 units up.  Use the following equation.

Plot[f[x] = ((x^2)/2), {x, -4, 4}];

Start with the graph of this function then using the second equation see if you can shift it down 2 units and Pi/2 units to the left.

Clear[f, x]Plot[f[x] = Abs[4 Sinx], {x, -π, 4 π}];

Clear[f, x]Plot[f[x] = Abs[4 Sinx], {x, -π, 4 π}];

Rational functions I

If a fraction has a polynomial for both numerator and denominator we say that the fraction is a rational polynomial expression.  Rational polynomial expressions are called rational functions.
If we graph the rational function f[x]= 1/x we note that as x gets larger positively the value of f[x] gets smaller and smaller and as x gets smaller positively then f[x] gets larger and larger.  As the |x| gets smaller f[x] increases negatively and as |x| increases f[x] gets smaller.  Here is what the graph looks like.  Highlight the following cell and press enter.

Clear[f, x]Plot[f[x_] = (1/x), {x, -4, 4}, 
      PlotStyle -> RGBColor[1, 0, 0], AspectRatio -> 1];

We can see that the graph approaches the y axes but never touches.  The y axes is called a vertical asymptote (a line that the graph approaches but never touches).  It is important to note the graph goes up on one side of the asymptote and reappears from the down direction on the other side.  This always happens when the expression in the denominator is a linear expression.  When we replace the x in the previous equation by -x see what happens.  Highlight the cell to the right of the equation and press the enter button.

Clear[f, x]Plot[f[x_] = -(1/x), {x, -4, 4}, 
      PlotStyle -> RGBColor[1, 0, 1], AspectRatio -> 1];

You  can see the graph is flipped upside down about the x axes.  What do you think will happen if we graph f[x]= 1/(x-3) +2?  Try it:

Clear[f, x]Plot[f[x] = (1/(x - 3)) + 2, {x, -8, 8}, 
      PlotStyle -> RGBColor {0, 0, 1}];

As you can see this shifted the graph to the right and up.  The vertical asymptote is now the line x=3.  If we place a negative in front of the fraction the graph should flip up side down about the line y=2.  Try it to see if it actually happens.

Rational Functions II
Lesson 39

Poles and Zeroes

The graph of a function touches the x axes at every value of x for which the function f[x] equals zero.  These values of x are called the zeros of the function.  The value of a function can be zero only for values of x that cause the numerator to be equal to zero.  When the denominator of a function equals zero we call this a pole.  There is a vertical asymptote at every pole.  Remember the graph of a function approaches but never touches a vertical asymptote.

Zeros of the numerator are zeros of the function.
Zeros off the denominator are poles of the function.

We will begin by investigating graphs of rational functions by considering the special case of functions that are factored into linear factors that occur only once each and that have more  factors in the denominator than in the numerator.  This last stipulation will ensure that the x axes will be a horizontal asymptote.
A rational function that is composed of unique nonrepeating linear factors changes signs at every zero of the numerator and denominator, this means the graph must cross the x axes at every zero and jump across the x axes at every pole.
The graphs can be sketched quickly if we draw vertical dashed lines at every pole and place dots at every zero.  We then start at the right with a large positive value for x and substitute it in the equation and see if f[x] is positive or negative.  Then we start above or below the x axes and follow the above rules.
Lets try it with the function f[x]=-[(x)(x-7)]/[(x+5)(x+2)(x-2)(x-5)]
First we plot the zeros and vertical asymptotes.  Highlight the cell to the right and press enter.

Clear[f, x]my = 
      Plot[f[x] = x, {x, -8, 8}, PlotStyle -> RGBColor[1, 1, 1], 
        DisplayFunction -> Identity]; Show[my, 
      Graphics[{{Dashing[{0.05, 0.05}], Line[{{5, 8}, {5, -8}}], 
            Line[{{-5, 8}, {-5, -8}}], 
            Line[{{2, 8}, {2, -8}}], 
            Line[{{-2, 8}, {-2, -8}}]}, {PointSize[ .03], 
            Point[{7, 0}], Point[{0, 0}]}}], 
      DisplayFunction -> $DisplayFunction];

Then a large positive value for x yields a negative value for f[x] so we begin below the x axes and cross the x axes at zeros and jump across it at the poles.  This is what it should look like.  Highlight the cell and press enter.

Clear[f, x, p, l]pl = 
      Plot[f[x_] = 
          -((x (x - 7))/((x + 5) (x + 2) (x - 2) 
                  (x - 5))), {x, -8, 8}, 
        PlotStyle -> RGBColor[1, 0, 0], 
        DisplayFunction -> Identity]; Show[pl, 
      Graphics[{{Dashing[{0.05, 0.05}], Line[{{5, 8}, {5, -8}}], 
            Line[{{-5, 8}, {-5, -8}}], 
            Line[{{2, 8}, {2, -8}}], 
            Line[{{-2, 8}, {-2, -8}}]}, {PointSize[ .03], 
            Point[{7, 0}], Point[{0, 0}]}}], 
      DisplayFunction -> $DisplayFunction];

Try this one: f[x]= (x+5)(x-3)(x-1)/(x-4)(x+2)(x+4)(x-1)

We can begin by canceling out the (x-1) factors but you must remember to put a hole in the graph at x=1.

Clear[f, x, p, l]pl = 
      Plot[f[x_] = 
          (((x + 5) (x - 3))/((x - 4) (x + 2) 
              (x + 4))), {x, -8, 8}, 
        PlotStyle -> RGBColor[1, 0, 0], 
        DisplayFunction -> Identity]; Show[pl, 
      Graphics[{{Dashing[{0.05, 0.05}], Line[{{4, 8}, {4, -8}}], 
            Line[{{-2, 8}, {-2, -8}}], 
            Line[{{-4, 8}, {-4, -8}}], 
            Line[{{1, 8}, {1, -8}}]}, {PointSize[ .03], 
            Point[{-5, 0}], Point[{3, 0}]}}], 
      DisplayFunction -> $DisplayFunction];

Check your understanding

In the following rational functions tell what the zeros would be, where the poles would be drawn and for a large value of x will you start graphing above or below the x axes.

1.  f[x]= (x-2)(x+3)/(x-5)(x+2)(x-3)

2.  f[x]= -X(x-3)/(x-1)(x+2)(x+3)

Factors of polynomial functions
Lesson 44

Factors of polynomial functions

In lesson 39 we considered the graphs of rational polynomial functions.  In this lesson we will investigate factors of polynomials and graphing second,third and fourth degree polynomial equations.
If all coefficients of a polynomial are real numbers, the polynomial is called a real polynomial.  The degree is the value of the greatest exponent.  Carl Friedrich Gauss proved that every real polynomial of degree N has exactly N linear factors.  Some factors may be complex linear factors but must occur in conjugate pairs (x+2i)(x-2i).  A factor of the form (x\(TraditionalForm\`\^2\)+4)  is called and irreducible quadratic factor because it can't be factored into linear real factors.   This factor can never equal zero for any real number value for x.  Thus irreducible quadratic factors never cause a polynomial to equal zero.  
Only linear real factors can cause a polynomial to equal zero.  The graph will cross the x axes at a zero if the factor occurs an odd number of times and will touch but not cross the x axes if it occurs an even number of times. 
If you have the function f[x]= (x + 4) (((x + 2))^4) ((x^2)+3)(((x - 5))^3)(((x - 7))^2) at what values of x will the graph touch the x axes and at what values of x will it cross the x axes?  After answering the question check it out by highlighting the following cell and press enter.

Clear[f, x]Plot[
      f[x_] = (x + 4) (((x + 2))^4) ((x^2) + 3) (((x - 5))^3) 
          (((x - 7))^2), {x, 8, -8}];

The turning point theorem tells us that the graph of a polynomial function has fewer turning points than the degree of the polynomial.

Thus the graph of a third degree polynomial function has at most two turning points; the graph of a fourth-degree polynomial function has at most three turning points;etc.
The term of highest degree in a polynomial is the dominant term because, for large absolute values of x, the value of the highest degree term will be greater than the absolute value of the sum of all the other terms in the equation.  
We can tell by the dominant term what the ends of the graph of a polynomial function will do.  The graph will increase or decrease according to the sign of the highest degree term evaluated when x is a large negative and a large positive number.  If the value of the term is negative it will decrease if positive it will increase.  
Example: f[x] = (x^4) - 3 (x^3) + 2 x - 9   For positive values of x,  (x^4) would be a positive number.  For negative values of x,  (x^4) would be a positive number.  So both ends of the graph will increase.  Take a look.

Clear[f, x]Plot[f[x_] = (x^4) - 3 (x^3) + 2 x - 9, {x, -8, 8}];

What do you think would happen to the ends of this polynomial function:
f[x]=(x^3) + 2 (x^2) - x + 4 Will the negative side increase or decrease?  How about the positive side?  Try it and see.

Clear[f, x]Plot[f[x_] = (x^3) + 2 (x^2) - x + 4, {x, -8, 8}];

Just as is the case with linear equations the constant shifts the graph up or down.  The same is true of a polynomial function.  The general form of an n-th degree polynomial function is f[x]= (ax^n)+(bx^(n - 1))+(cx^(n - 2))+...+k
The sum of all real and complex roots is -b/a and the product of all real and complex roots is k/a if the degree of the polynomial is even and -k/a if the degree is odd.  The average value of all roots is

Average value of all roots = -(b/(n a))

This will give us a good idea of the x value of the center of the graph.  Using this fact find the x coordinate of the center of the following polynomial function.  f[x]= --3 (x^3) + 3 (x^2) - 2 x + 3
What does the graph do to the right of this center point and to the left?  Try it and see if you were correct.

Clear[f, x]Plot[f[x] = -3 (x^3) + 3 (x^2) - 2 x + 3, {x, -4, 4}];

Try another polynomial function and do the same as the one above then check. f
[x]= (x^4) - 3  (x^2)+2x+5

Clear[f, x]Plot[f[x] = (x^4) - 3 (x^2) + 2 x + 5, {x, -4, 4}];