Sample Problems. Practice Problems

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1 Lecture Notes Graphs of Factored Polnomials page Sample Problems. Plot the graph of f () = ( + ) ( ).. Plot the graph of f () = ( ) ( + ) ( ) ( ) (6 ) ( + ). Practice Problems Plot the graph of each of the following functions.. f () = ( ) ( + ). f () = ( + ) ( + ) ( ). f () = ( + ) ( ) ( ) (6 ). f () = ( + 6) ( + ) ( + ) ( + ) ( ) ( 6) ( 7). f () = ( + ) ( ) ( ) (6 ) ( + ) 6. f () = ( + ) ( + ) ( ) ( ) 7. f () = ( + ) ( + ) ( + 8) ( + ) ( ) (7 ) 8. f () = ( + ) ( + ) ( ) ( ) 9. f () = ( + ) ( ) ( ) (8 ) ( + ). f () = ( ) ( + 6) ( + ) ( ). f () = ( + ) ( + ) ( ) ( ) c copright Hidegkuti, Powell, 8 Last revised: April 6, 9

2 Lecture Notes Graphs of Factored Polnomials page Sample Problems - Answers. f () = ( + ) ( ) - -. f () = ( ) ( + ) ( ) ( ) (6 ) ( + ). - - Practice Problems - Answers. f () = ( ) ( + ) f () = ( + ) ( + ) ( ) c copright Hidegkuti, Powell, 8 Last revised: April 6, 9

3 Lecture Notes Graphs of Factored Polnomials page. f () = ( + ) ( ) ( ) (6 ) - -. f () = ( + 6) ( + ) ( + ) ( + ) ( ) ( 6) ( 7) Solution: Since it is a polnomial, the functions domain is R; and f is continuous on R. The degree is 9; the leading coe cient is negative, and so the end-behavior is lim f () = and lim f () =!! intercepts: at = 6; ; ; ; ; 6 and 7, and nowhere else. There is a change in sign at = 6; ; ; 6; and 7: There is no change in sign at = and f () = ( + ) ( ) ( ) (6 ) ( + ) Solution: Since it is a polnomial, the functions domain is R; and f is continuous on R. We rst bring this epression to a form easier to graph b factoring out the leading coe cient from each factor. ( ) = ( ) since ( ) = [ ( )] = ( ) ( ) = ( ) The function s equation is then 6 = ( ) and ( + ) = ( ) ( + ) ( ) ( ) (6 ) ( + ) = factor leading coe cients ( + ) ( ) ( ) ( ) ( ) ( ) ( ) = bring numbers to front ( + ) ( ) ( ) ( ) ( ) = organize factors left-to right b zero = ( + ) ( ) ( ) ( ) We can now determine the degree of the polnomial and the sign of its leading coe cient. (degree: 9, positive leading coe cient.) Based on this, the end-behavior is: lim f () = and lim f () =!! intercepts: at = ; ; ; and, and nowhere else. There is a change in sign at = ; ; and : There is no change in sign at = and. c copright Hidegkuti, Powell, 8 Last revised: April 6, 9

4 Lecture Notes Graphs of Factored Polnomials page - 6. f () = ( + ) ( + ) ( ) ( ) Solution: We rst determine the degree: =. Sine the leading coe cient is positive, we have that lim f () = and lim f () =. The rest of the stor is given to us b continuit. The!! intercepts are ; ; ; ; and. Looking at the multiplicit of the roots, we can determine if f changes sign around the zero or not f () = ( + ) ( + ) ( + 8) ( + ) ( ) (7 ) f () = ( + ) ( + ) ( ) ( ). Solution: The degree is 9, the leading coe cient is negative Thus the end-behavior is: lim = and lim =!! There is sign change through the zeroes corresponding to linear factors with odd eponents: at ;, and. There is no change of sign through the zeroes corresponding to linear factors with even eponents: at. The graph is continuous, thus we have: c copright Hidegkuti, Powell, 8 Last revised: April 6, 9

5 Lecture Notes Graphs of Factored Polnomials page 9. f () = ( + ) ( ) ( ) (8 ) ( + ) -. f () = ( ) ( + 6) ( + ) ( ) f () = ( + ) ( + ) ( ) ( ). Solution: degree: (odd), leading coe cient negative c copright Hidegkuti, Powell, 8 Last revised: April 6, 9

6 Lecture Notes Graphs of Factored Polnomials page 6 Sample Problems - Solutions. Plot the graph of f () = ( + ) ( ) Solution: Since f is a polnomial, its domain is the set of all real numbers (in short: R) and f is continuous on the entire domain. We determine that f is of degree 6; and its leading coe cient is negative. Based on this, the end-behavior is and so our graph so far is lim f () = and lim f () =!! - - The intercepts are at = ; ; and ; and there are no other zeroes. We will determine the behavior of each zero, left to right. First, at =. At = ; the factor ( + ) is zero. Since this linear factor is raised to the rst power (odd power), there is a change in sign there. Thus our graph is now - - The net zero is at =. At = ; the factor is zero. Since this linear factor is raised to the second power (even power), there is no change in the sign there. Thus our graph is now - - c copright Hidegkuti, Powell, 8 Last revised: April 6, 9

7 Lecture Notes Graphs of Factored Polnomials page 7 The net zero is at =. At = ; the factor is zero. Since this linear factor is raised to the third power (odd), there is a change in the sign there. Thus our graph is now - - and so the nal graph is - - Note that we can not et determine how tall the "humps" are between the zeroes. that the eist because the function is continuous. We can onl conclude. Plot the graph of f () = ( ) ( + ) ( ) ( ) (6 ) ( + ) Solution: First we bring the polnomial to a form that is more convenient for graphing. First, we will factor out the leading coe cients from each of the linear factors. ( ) = ( ) = [ ( + )] = ( ) ( + ) = ( + ) In short, ( ) = ( + ). Similarl, ( ) = ( ) and (6 ) = ( ) and ( + ) = ( ). Thus the function f is f () = ( ) ( + ) ( ) ( ) (6 ) ( + ) = ( + ) ( + ) ( ) ( ) ( ) ( ) ( ) ( ) We now re-arrange the factors: we bring the numbers to the front, and oraganize the linear factors b their zeroes, left to right. f () = ( + ) ( + ) ( ) ( ) ( ) We will use this form to graph the function. The degree and the sign of the leading coe cient determines the end-behavior of f. The degree is 8; the leading coe cient is positive, and so lim f () = lim f () =.!! The polnomial has zeroes onl at = ; ; ; ; and ; and at no other points. Since all polnomials are continuous on R; change in sign can onl occur through a zero. c copright Hidegkuti, Powell, 8 Last revised: April 6, 9

8 Lecture Notes Graphs of Factored Polnomials page 8 We graph left to right, starting and ending with positive values as discussed above. - - The linear factor ( + ) has multiplicit ; (its eponent is ). Consequentl, f will not change sign between numbers less than and numbers larger than : (Ver much like the function g () = ( + ) at = ) - - The net zero is at ; caused b the linear factor ( + ). It is not repeated (eponent ) and thus f changes sign between numbers less than and greater than. - - The net zero is caused b the factor ( ), thus there is again a change in sign. - - c copright Hidegkuti, Powell, 8 Last revised: April 6, 9

9 Lecture Notes Graphs of Factored Polnomials page 9 The net zero is caused b ( ). Because this eponent is even, there is no change in sign. - - The net zero is caused b ( ). Because this eponent is even, there is no change in sign. - - Thus the graph is: - - Note that we can not et determine how tall the "humps" are between the zeroes. that the eist because the function is continuous. We can onl conclude For more documents like this, visit our page at and click on Lecture Notes. questions or comments to mhidegkuti@ccc.edu. c copright Hidegkuti, Powell, 8 Last revised: April 6, 9