Polynomials and Rational Functions (2.1) The shape of the graph of a polynomial function is related to the degree of the polynomial Shapes of Polynomials Look at the shape of the odd degree polynomials fx()=− x3 27 x fx()= x53−++ 5 x 4 x 1 Graph of Odd polynomial y=x^5-5x^3+4x+1 5 4 3 2 1 0 -3 -2 -1-1 0 1 2 3 -2 -3 Graph of Odd Polynomial fx()=− x3 27 x y=x^3-27x 60 40 20 0 0 102030405060 -20 -40 -60 Graphs of even degree polynomials Now, look at the y=x^4-6x^2 shape of the even 30 degree polynomial 25 20 fx()=− x42 6 x 15 10 5 0 -4 -3 -2 -1 0 1 2 3 4 -5 -10 -15 Graph of even degree polynomial Here is another f(x)=3x^2+6x-1 example of an 50 even degree polynomial : 40 30 2 fx()=+− 3 x 6 x 1 20 10 0 -6 -4 -2 0 2 4 -10 Generalization: The graphs of odd-degree polynomials start negative, end positive and cross the x-axis at least once. The even-degree polynomial graphs start positive, end positive, and may not cross the x axis at all Characteristics of polynomials: Graphs of polynomials are continuous. One can sketch the graph without lifting up the pencil. 2. Graphs of polynomials have no sharp corners. 3. Graphs of polynomials usually have turning points, which is a point that separates an increasing portion of the graph from a decreasing portion. Turning points and x intercepts Theorem 1 : Turning points and x Intercepts of Polynomials The graph of a polynomial function of positive degree n can have at most n-1 turning points and can cross the x axis at most n times. Largest value of the roots of a polynomial Theorem 2: Maximum value of an x- intercept of a polynomial. If r is a zero of the polynomial P(x) this means that P(r) = 0. For example, px()= x2 − 4 x is a second degree polynomial . and p(4)= 42 −= 4(4) 0 , so r = 4 is a zero of the polynomial as well as being an x-intercept of the graph of p(x). Cauchy’s Theorem A theorem by a French mathematician named Cauchy allows one to determine the maximum value of a zero of a polynomial (maximum value of the x- intercept). Let’s take an example: the polynomial p()xx=−2 4 x Cauchy’s Theorem According to this theorem r < 1 + maximum value of {1,− 4} = 1 + 4 = 5 The numbers within the absolute value symbols are the coefficients of the polynomial p(x). p()xx= 2 − 4 x Result of application of Cauchy’s theorem From this result we have , which means -5 < r < 5 . This tells us that we should look for any potential x intercepts within the range of -5 and 5 on the x –axis. In other words, no intercepts (roots) will be found that are greater than 5 nor less than -5. Conclusion , From the graph of px()=− x2 4x we find that the other zero is located at (0,0). Thus, the two zeros , 0 , -4, are within the range of -5 to 5 on the x-axis. Now, let’s try another example: An Example: Example: Approximate the real zeros of Px()=+ 3 x32 12 x ++ 9 x 4 First step: Coefficient of cubic term must equal one, so divide each term by three to get a new polynomial Q(x)= 4 xxx32+++43 3 Roots of new polynomial are the same as the roots of P(x). Example, continued Step 2: Use the theorem: ⎧ 1 ⎫ r <+1max4,3,⎨ ⎬ ⎩⎭3 r <+14 = 5 r < 5 Example, continued Step3: We know that all possible x intercepts (roots) are found along the x-axis between -5 and 5. So we set our viewing rectangle on our calculator to this window and graph the polynomial function. Step 4. Use the zero command on our calculator to determine that the root is approximately -3.19 (there is only one root). Rational Functions Definition: Rational function: a quotient of two polynomials, P(x) and Q(x), is a rational function for all x such that Q(x) is not equal to zero. Example: Let P(x) = x + 5 and Q(x) = x – 2 then R(x)= x + 5 x − 2 is a rational function that is defined for all real values of x with the exception of 2 (Why?) Domain of rational functions Domain : {xx≠ 2} and x is a real number. This is read as “the set of all numbers, x , such that x is not equal to 2. X intercepts of a rational function: To determine the x- intercepts of the graph of any function, we find the values of x for which y = 0 . In our case y = 0 implies that 0 = x + 5 x − 2 This implies that x + 5 = 0 or x = -5 . Y-intercept of a rational function Y intercept: The y intercept of a function is the value of y for which x = 0 . Setting x = 0 in the equation we have y = , or -5/2. So, the y- intercept is located at ( 0, -2.5). Notice that the y-intercept is a point described by an ordered pair, not just a single number. Also, remember that a function can have only one y intercept but more than one x-intercept ( Why?) Graph of a Rational function: 1. Plot points near the value at which the function is undefined. In our case, that would be near x = 2. Plot values such as 1.5, 1.7. 1.9 and 2.1, 2.3, 2.5. Use your calculator to evaluate function values and make a table. 2. Determine what happens to the graph of f(x) if x increases or decreases without bound. That is, for x approaching positive infinity or x approaching negative infinity. 3. Sketch a graph of a function through these points. 4. Confirm the results using a calculator and a proper viewing rectangle. Graph of rational function 50 40 30 20 y=(x+5)/(x-2) approaches zero as x gets small 10 approaches zero as x gets large 0 Series1 -2-101234567891011-10 -20 undefined at -30 x = 2 -40 -50 Conclusions: From the graph we see that there is a vertical asymptote at x = 2 because the graph approaches extremely large numbers as x approaches 2 from either side. We also see that y = 0 is a horizontal asymptote of the function since y tends to go to zero as x tends to either a very large positive number or very small negative number. Exponential functions The equation f ()xb= x defines the exponential function with base b . The domain is the set of all real numbers, while the range is the set of all positive real numbers ( y > 0). Note y cannot equal to zero. Riddle Here is a problem related to exponential functions: Suppose you received a penny on the first day of December, two pennies on the second day of December, four pennies on the third day, eight pennies on the fourth day and so on. How many pennies would you receive on December 31 if this pattern continues? 2) Would you rather take this amount of money or receive a lump sum payment of $10,000,000? Solution (Complete the table) Day No. pennies 1 1 2 2 2^1 3 4 2^2 4 8 2^3 5 16 6 32 7 64 Generalization Now, if this pattern continued, how many pennies would you have on Dec. 31? Your answer should be 2^30 ( two raised to the thirtieth power). The exponent on two is one less than the day of the month. See the preceding slide. What is 2^30? 1,073,741,824 pennies!!! Move the decimal point two places to the left to find the amount in dollars. You should get: $10,737,418.24 Solution, continued The obvious answer to question two is to take the number of pennies on December 31 and not a lump sum payment of $10,000,000 (although, I would not mind having either amount!) This example shows how an exponential function grows extremely rapidly. In this case, the exponential function fx()= 2x is used to model this problem. x Graph of fx()= 2 Use a table to graph the exponential function above. Note: x is a real number and can be replaced with numbers such as 2 as well as other irrational numbers. We will use integer values for x in the table: Table of values x y -4 −4 11 2 = 4 = 1216 -3 2−3 = 8 -2 1 2−2 = 4 -1 1 2−1 = 2 0 210 = 1 221 = 2 242 = Graph of y = fx()= 2x Characteristics of the graphs of fx()= bx where b> 1 1. all graphs will approach the x-axis as x gets large. 2. all graphs will pass through (0,1) (y-intercept) 3. There are no x – intercepts. 4. Domain is all real numbers 5. Range is all positive real numbers. 6. The graph is always increasing on its domain. 7. All graphs are continuous curves. Graphs of f () xb = x if 0 < b < 1 1. all graphs will approach the x-axis as x gets large. 2. all graphs will pass through (0,1) (y-intercept) 3. There are no x – intercepts. 4. Domain is all real numbers 5.