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Warm-up: Simplify. Warm-up: Answers. Check Homework. p. 533 # 15-43 odd Quiz 5.1 tomorrow Objective(5.2A) : Classify polynomials and evaluate polynomials using synthetic substitution. A Polynomial Function is a function of the form. f(x) = a n x n + a n-1 x n-1 + …+a 1 x + a 0
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Check Homework p. 533 # 15-43 odd Quiz 5.1 tomorrow Objective(5.2A): Classify polynomials and evaluate polynomials using synthetic substitution.
A Polynomial Functionis a function of the form • f(x) = anxn + an-1xn-1 + …+a1x + a0 The exponents are all whole numbers, and the coefficients are all real numbers. anis the leading coefficient n is the degree
Degree Type Standard Form Example You are already familiar with some types of polynomial functions. Here is a summary of common types ofpolynomial functions. 0 Constant f (x) = 4 1 Linear f (x) = 2x -10 2 Quadratic f (x) = 5x2+2x + 7 3 Cubic f (x) = 5x3+ 3x2- x + 6 4 Quartic f (x) = 5x4 + 4x3+ 7x2+9x + 1
1 f(x) = x2– 3x4– 7 2 1 Its standard form is f(x) = –3x4+x2 – 7. 2 Decide whether the function is a polynomial function. If it is, write the function in standard form and state its degree, typeand leading coefficient. SOLUTION The function is a polynomial function. It has degree 4, so it is a quartic function. The leading coefficient is – 3.
f(x) = x3+ 3x Decide whether the function is a polynomial function. If it is, write the function in standard form and state its degree, typeand leading coefficient. SOLUTION The function is not a polynomial function because the term 3xdoes not have a variable base and an exponentthat is a whole number.
f(x) = 6x2+ 2x–1+ x Decide whether the function is a polynomial function. If it is, write the function in standard form and state its degree, typeand leading coefficient. SOLUTION The function is not a polynomial function because the term2x–1has an exponent that is not a whole number.
f(x) = –0.5x+x2– 2 Its standard form is f(x) = x2– 0.5x– 2. Decide whether the function is a polynomial function. If it is, write the function in standard form and state its degree, typeand leading coefficient. SOLUTION The function is a polynomial function. It has degree 2, so it is a quadratic function. The leading coefficient is .
1 f(x) = x2– 3x4– 7 2 f(x) = –0.5x+ x2– 2 Polynomial function? f(x) = x3+ 3x f(x) = 6x2+ 2x–1+ x
Degree of a Monomial What is the degree of the monomial? • The degree of a monomial is the sum of the exponents of the variables in the monomial. • The exponents of each variable are 4 and 2. 4+2 = 6. • The degree of the monomial is 6. • The monomial can be referred to as a sixth degree monomial.
Example 1 Classify the polynomials by degree and number of terms. Classify by number of terms Classify by degree Polynomial Degree a. b. c. d. Zero Constant Monomial First Linear Binomial Second Quadratic Binomial Trinomial Third Cubic
Example 2 Write the polynomials in standard form and identify the polynomial by degree and number of terms. a) b) c)
Practice • Page 338 (1-5)
Use synthetic division to evaluate f(x) = 2x4 + -8x2 + 5x- 7 when x = 3. Just Watch-- One way to evaluate polynomial functions is to usedirect substitution. Another way to evaluate a polynomialis to use synthetic substitution.
Polynomial in standard form 3 Coefficients x-value SOLUTION 2x4 + 0x3 + (–8x2) + 5x + (–7) Polynomial in standard form 2 0 –8 5 –7 3• Coefficients 6 18 30 105 35 10 98 2 6 The value of f(3) is the last number you write, In the bottom right-hand corner.
Write the coefficients of f (x) in order of descending exponents. Write the value at which f (x) is being evaluated to the left. STEP 1 EXAMPLE 3 Evaluate by synthetic substitution a) Use synthetic substitution to evaluate f (x) from Example2 when x = 3.f (x) = 2x4 – 5x3 –4x + 8 SOLUTION
STEP 2 Bring down the leading coefficient. Multiply the leading coefficient by the x-value. Write the product under the second coefficient. Add. STEP 3 Multiply the previous sum by the x-value. Write the product under the third coefficient. Add. Repeat for all of the remaining coefficients. The final sum is the value of f(x) at the given x-value. EXAMPLE 3 Evaluate by synthetic substitution
Synthetic substitution gives f(3) = 23, which matches the result in Example 2. ANSWER EXAMPLE 3 Evaluate by synthetic substitution
for Examples 3 and 4 GUIDED PRACTICE Use synthetic substitution to evaluate the polynomial function for the given value of x. b)f (x) = 5x3 + 3x2–x + 7; x = 2 Write the coefficients of f (x) in order of descending exponents. Write the value at which f (x) is being evaluated to the left.
Synthetic substitution gives f(2) = 57 ANSWER for Examples 3 and 4 GUIDED PRACTICE
– 1 – 2 – 1 0 4 – 5 for Examples 3 and 4 GUIDED PRACTICE c) g (x) = – 2x4–x3 + 4x – 5; x = – 1 Write the coefficients of g(x) in order of descending exponents. Write the value at which g (x) is being evaluated to the left.
2 –1 1 –5 – 2 1 –1 5 – 10 Synthetic substitution gives f(– 1) = – 10 ANSWER – 1 – 2 – 1 0 4 – 5 for Examples 3 and 4 GUIDED PRACTICE
Assignments Classwork: Practice 5.2 # 1-7 Homework (5.2A): p. 341 # 3-8, 15-23 (15 pts) Closure: Review exponent rules for quiz tomorrow