EVALUATING POLYNOMIAL FUNCTIONS A polynomial function is a

EVALUATING POLYNOMIAL FUNCTIONS A polynomial function is a function of the form f (x) = an x nn + an – 1 x nn – 11 +· · · + a 1 x + a 00 Where ann ≠≠ 00 and the exponents are all whole numbers. coefficient For this polynomial function, aann is the leading coefficient, constant term, term and n is the degree aa 00 is the constant A polynomial function is in standard form if its terms are descending order of of exponents from left to to right. written in descending

EVALUATING POLYNOMIAL FUNCTIONS You are already familiar with some types of polynomial functions. Here is a summary of common types of polynomial functions. Degree Type Standard Form 0 Constant f (x) = a 0 1 Linear f (x) = a 1 x + a 0 2 Quadratic f (x) = a 2 x 2 + a 1 x + a 0 3 Cubic f (x) = a 3 x 3 + a 2 x 2 + a 1 x + a 0 4 Quartic f (x) = a 4 x 4 + a 3 x 3 + a 2 x 2 + a 1 x + a 0

Identifying Polynomial Functions Decide whether the function is a polynomial function. If it is, write the function in standard form and state its degree, type and leading coefficient. f (x) = 1 x 2 – 3 x 4 – 7 2 SOLUTION The function is a polynomial function. Its standard form is f (x) = – 3 x 4 + 1 2 x – 7. 2 It has degree 4, so it is a quartic function. The leading coefficient is – 3.

Identifying Polynomial Functions Decide whether the function is a polynomial function. If it is, write the function in standard form and state its degree, type and leading coefficient. f (x) = x 3 + 3 x SOLUTION The function is not a polynomial function because the x term 3 does not have a variable base and an exponent that is a whole number.

Identifying Polynomial Functions Decide whether the function is a polynomial function. If it is, write the function in standard form and state its degree, type and leading coefficient. – f (x) = 6 x 2 + 2 x 1 + x SOLUTION The function is not a polynomial function because the term 2 x – 1 has an exponent that is not a whole number.

Identifying Polynomial Functions Decide whether the function is a polynomial function. If it is, write the function in standard form and state its degree, type and leading coefficient. f (x) = – 0. 5 x + π x 2 – 2 SOLUTION The function is a polynomial function. Its standard form is f (x) = π x 2 – 0. 5 x – 2. It has degree 2, so it is a quadratic function. The leading coefficient is π.

Identifying Polynomial Functions Polynomial function? f (x) = 12 x 2 – 3 x 4 – 7 f (x) = x 3 + 3 x f (x) = 6 x 2 + 2 x– 1 + x f (x) = – 0. 5 x + π x 2 – 2

Using Synthetic Substitution One way to evaluate polynomial functions is to use direct substitution. Another way to evaluate a polynomial is to use synthetic substitution. Use synthetic division to evaluate f (x) = 2 x 4 + -8 x 2 + 5 x - 7 when x = 3.

Using Synthetic Substitution SOLUTION 2 x 4 + 0 x 3 + (– 8 x 2) + 5 x + (– 7) Polynomial in standard form 33 • 2 0 – 8 5 – 7 Coefficients 6 18 30 105 6 10 35 98 x-value 2 The value of f (3) is the last number you write, In the bottom right-hand corner.

GRAPHING POLYNOMIAL FUNCTIONS The end behavior of a polynomial function’s graph is the behavior of the graph as x approaches infinity (+ ) or negative infinity (– ). The expression x + is read as “x approaches positive infinity. ”

GRAPHING POLYNOMIAL FUNCTIONS END BEHAVIOR

GRAPHING POLYNOMIAL FUNCTIONS CONCEPT SUMMARY END BEHAVIOR FOR POLYNOMIAL FUNCTIONS an n x – x + >0 even f (x) + >0 odd f (x) – f (x) + <0 even f (x) – <0 odd f (x) + f (x) –

Graphing Polynomial Functions Graph f (x) = x 3 + x 2 – 4 x – 1. SOLUTION To graph the function, make a table of values and plot the corresponding points. Connect the points with a smooth curve and check the end behavior. – 2 the leading – 1 0 1 is positive, 2 3 x The degree is– 3 odd and coefficient – 3 23 + + (x) as 3 x so f (x)f(x) – – 7 as x 3 – 3 and f– 1 .

Graphing Polynomial Functions Graph f (x) = –x 4 – 2 x 3 + 2 x 2 + 4 x. SOLUTION To graph the function, make a table of values and plot the corresponding points. Connect the points with a smooth curve and check the end behavior. – 3 0 coefficient 1 3 x The degree is even– 2 and the– 1 leading is 2 negative, f (x) –– 21 as x 0 –– 1 and 0 f (x) 3 – – 16 + as x – 105 so f (x) .