EXAMPLE 1 Identify polynomial functions 4 Decide whether the function is a polynomial function. If so, write it in standard form and state its degree,

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Presentation transcript:

EXAMPLE 1 Identify polynomial functions 4 Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient. a. h (x) = x 4 – x SOLUTION a. The function is a polynomial function that is already written in standard form. It has degree 4 (quartic) and a leading coefficient of 1.

EXAMPLE 1 Identify polynomial functions Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient. b. g (x) = 7x – 3 + πx 2 SOLUTION b. The function is a polynomial function written as g(x) = πx 2 + 7x – 3 in standard form. It has degree 2 (quadratic) and a leading coefficient of π.

EXAMPLE 1 Identify polynomial functions Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient. c. f (x) = 5x 2 + 3x -1 – x SOLUTION c. The function is not a polynomial function because the term 3x – 1 has an exponent that is not a whole number.

EXAMPLE 1 Identify polynomial functions Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient. d. k (x) = x + 2 x – 0.6x 5 SOLUTION d. The function is not a polynomial function because the term 2 x does not have a variable base and an exponent that is a whole number.

EXAMPLE 2 Evaluate by direct substitution Use direct substitution to evaluate f (x) = 2x 4 – 5x 3 – 4x + 8 when x = 3. f (x) = 2x 4 – 5x 3 – 4x + 8 f (3) = 2(3) 4 – 5(3) 3 – 4(3) + 8 = 162 – 135 – = 23 Write original function. Substitute 3 for x. Evaluate powers and multiply. Simplify

GUIDED PRACTICE for Examples 1 and 2 Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient. 1. f (x) = 13 – 2x SOLUTION f (x) = – 2x + 13 It is a polynomial function. Standard form: – 2x + 13 Degree: 1 Type: linear Leading coefficient of – 2.

GUIDED PRACTICE for Examples 1 and 2 2. p (x) = 9x 4 – 5x – SOLUTION p (x) = 9x 4 – 5x – The function is not a polynomial function.

GUIDED PRACTICE for Examples 1 and 2 3. h (x) = 6x 2 + π – 3x SOLUTION h (x) = 6x 2 – 3x + π The function is a polynomial function that is already written in standard form will be 6x 2 – 3x + π. It has degree 2 ( linear ) and a leading coefficient of 6. It is a polynomial function. Standard form: 6x2– 3x + π Degree: 2 Type: quadratic Leading coefficient of 6

GUIDED PRACTICE for Examples 1 and 2 Use direct substitution to evaluate the polynomial function for the given value of x. 4. f (x) = x 4 + 2x 3 + 3x 2 – 7; x = –2 SOLUTION f (x) = x 4 + 2x 3 + 3x 2 – 7; x = – 2 Write original function. f (–2) = (–2) (–2) (–2 ) 2 – 7 = 16 – – 7 Substitute –2 for x. Evaluate powers and multiply. = 5 Simplify

GUIDED PRACTICE for Examples 1 and 2 5. g (x) = x 3 – 5x 2 + 6x + 1; x = 4 SOLUTION g (x) = x 3 – 5x 2 + 6x + 1; x = 4 Write original function. g (x) = 4 3 – 5(4) 2 + 6(4) + 1 Substitute 4 for x. = 64 – = 9 Evaluate powers and multiply. Simplify