Explanation Procedure used: The graph of a polynomial can sketch as follows. 1. Use the leading term test to determine the end behavior. 2. Find the zeros of the function by solving f ( x ) = 0 . Any real zeros are the first coordinates of the x -intercepts. 3. Use the x -intercepts (zeros) to divide the x -axis into the intervals, and choose a test point in each interval to determine the sign of all function values in that interval. 4. Find f ( 0 ) . This gives the y -intercept of the function. 5. If necessary, find additional function values to determine the general shape of the graph and then draw the graph. 6. As a partial check, use the facts that the graph has at most n − x intercepts and at most n − 1 turning points. Multiplicity of zeros can also be considered in order to check where the graph crosses or is tangent to the x axis. We can also check with a graphing calculator. Calculation: The given polynomial function is f ( x ) = x 3 − 7 x + 6 . Use the leading term test to determine the end behavior. The leading term in the given polynomial is x 3 . So, the degree of the polynomial is 3, which is an odd number. The coefficient of the leading term is 1, which is a positive number. So, it is clear that the value of f ( x ) → ∞ as x → ∞ and f ( x ) → − ∞ as x → − ∞ . Obtain the zeros of the function by solving f ( x ) = 0 . Any real zeros are the first coordinates of the x -intercepts. f ( x ) = 0 x 3 − 7 x + 6 = 0 Find the zeros of the above polynomial by using synthetic division. Check x = 1 is the zero of the polynomial. 1 1 0 − 7 6 1 1 − 6 _ 1 1 − 6 0 The zero of the polynomial is 1. Find the remaining zeros of the polynomial as follows. x 3 − 7 x + 6 = 0 ( x − 1 ) ( x 2 + x − 6 ) = 0 ( x − 1 ) ( x + 3 x − 2 x − 6 ) = 0 On further simplifications, ( x − 1 ) ( ( x + 3 ) − 2 ( x + 3 ) ) = 0 ( x − 1 ) ( x + 3 ) ( x − 2 ) = 0 x − 1 = 0 or x + 3 = 0 or x − 2 = 0 x = 1 or x = − 3 or x = 2 Thus, the zeros of the functions are –3, 1 and 2

COLLEGE ALGEBRA: GRAPHS AND MODELS, LOOS

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ISBN: 9780136175636

Author: BITTINGER

Publisher: PEARSON

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Chapter 4.3, Problem 51E

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To graph: The polynomial function f(x)=x3+4x2+x−6.

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Chapter 4.3, Problem 50E

Chapter 4.3, Problem 52E

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Chapter 4 Solutions

COLLEGE ALGEBRA: GRAPHS AND MODELS, LOOS

Ch. 4.1 - Determine the leading term, the leading...Ch. 4.1 - Prob. 2E Ch. 4.1 - Determine the leading term, the leading...Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Determine the leading term, the leading...Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Determine the leading term, the leading...Ch. 4.1 - Prob. 10E

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To graph: The polynomial function f ( x ) = x 3 + 4 x 2 + x − 6 .

Solution Summary: The author explains how the graph of a polynomial can sketch as follows: the leading term test determines the end behavior; the zeros are the first coordinates of the x -intercepts.

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To graph: The polynomial function f(x)=x3+4x2+x−6.

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Chapter 4 Solutions