1. Jonathon Vanderhorst & Callum Gilchrist Period 1

2. Polynomial Functions and Models Properties of Rational Functions The Graph of a Rational Function Polynomial and Rational Inequalities Real Zeros Complex Zeros INCLUDES:

3. Polynomial functions and models

4.  A polynomial function is of the form: f(x) = a n x n +a n-1 x n-1 +…+a 1 x+a 0  The degree is based on the largest power of x  A power function, however, is simply a monomial: f(x) = ax n For any given function n must be a non-negative integer A power function; y=x 2

5. T ECHNIQUES : Transformations include shifting, compression, stretching, and reflection (as seen in section 2.4) If f(r)= 0 then r is called a real zero of the function, and if it repeats then it’s called a multiplicity. X intercepts Behavior near a zero Turning points, or local maxima and minima End behavior; where for large values of a function, it resembles the graph of the power function y= a n x n

6. Properties of Rational Functions

7. A rational function is of the form: R(x) = p(x)/q(x) Where p and q are polynomial functions

8. 1.If the numerator is < the denominator, R will have the horizontal asymptote y=0 (the x-axis) 2.If the numerator is ≥ the denominator, long division is used… a) If the degrees are =, the line a n /b m is a horizontal asymptote b) If the numerator is one more, the quotient is of the form ax+b and that is the line of the asymptote c)If the numerator is 2or more, the quotient is degree 2 or higher and R has neither a horizontal or an oblique asymptote

9. The Graph of a Rational Function

10. Step1: Factor the numerator & denominator and find the domain Step2: Write R in lowest terms and find the real zeros of the numerator, which are the x-intercepts, as well as indicate the behavior near them Step3: Find the real zeros of the denominator, which determine the vertical asymptotes and graph each with a dotted line Step4: Locate horizontal & oblique asymptotes and graph them Step5: Use the real zeros to determine where the function is above/below the x-axis Step6: Analyze the behavior near each asymptote Step7: Use all the above information to graph R

11. Polynomial and Rational Inequalities

12. Step 1: Write the inequality so that a polynomial or rational expression f is on the left side and zero is on the right side: f (x) > 0 f (x)<0 f (x) ≥ 0 f (x) ≤ 0 Step 2: Determine the real numbers at which the expression f on the left side equals zero, and if the expression is rational, the real numbers at which the expression f on the left side is undefined. Step 3: Use the numbers found in Step 2 and separate the real number line into intervals. Step 4: Select a number in each interval and then evaluate f at the number. (a) If the value of f is positive, then f (x) > 0 for all numbers in the x interval. (b) If the value of f is negative, then f (x) < 0 for all numbers in the x interval.

13. 4 2 Solve the inequality X ≤ 4x, and graph the solution set. Step 1: Rearrange the inequality so that 0 is on the right side. 4 2 4 2 X ≤ 4x ▬► X - 4X ≤ 0 Subtract 4x^2 from both sides of the inequality Step 2: Find the real zeros by setting the equation equal to 0 2 2 2 X (x - 4) = 0 ▬► X (X+2)(X-2) = 0 ▬► X = 0 or X = -2 or X = 2 Factor Step 3: Use the zeros to separate the number line into four intervals: (- ∞, -2) (-2,0) (0,2) (2, ∞ ) Conclusion {x|-2 ≤ X ≤ 2} or [-2,2] -2 0 2 ∞ Interval(-∞, -2)(-2,0)(0,2)(2, ∞) Number-313 Value45-3 45 ConclusionPositiveNegative Positive

14. EXAMPLE: (X+3)(2-X) Solve the inequality (X-1) 2 > 0, and graph the solution set. Step 1: The domain of X is X ≠ 1. The inequality already has 0 on the right side. Step 2: The real zeros of the numerator are X=-3 and X=2; the real zero of the denominator is X=1 Step 3: Use the zeros to separate the number line into four intervals: (- ∞, -3)(-3,1) (1,2)(2, ∞ ) Conclusion {x|-3 < X < 2, X ≠ 1} or (-3,1) U (1,2) -3 2 1 ∞ Interval(-∞, -3)(-3,1)(1,2)(2, ∞) Number-403/23 Value-6/2569-3/2 ConclusionNegativePositive Negative

15. The real zeros of a polynomial function

16. If f(x) is divided by x - c, then the remainder is f(c) Find the remainder if f(x) = x 2 – 4x -5 f(3) = (3) 2 – 4(3) -5 = -8 The remainder is -8 by the Remainder Theorem Use substitution to find if x – 1 is a factor of 2x 3 – x 2 + 2x – 3 2(1) 3 –(1) 2 +2(1) -3 =0 (x-1) is a factor by the Factor Theorem

17. Let f denote a polynomial function written in standard form. The number of positive real zeroes of f either equals the number of variations in the sign of the coefficients of f(x) or equals that number minus an even integer. The number of negative real zeros of f either equals the number of variations in the sign of the coefficients of f(-x) or equals that number minus an even integer. f(x)= 3x 6 – 4x 4 + 3x 3 + 2x 2 – x – 3; there are 3 sign changes so there are either 3 or 1 positive real zeros f(-x)= 3x6 – 4x4 -3x3 + 2x2 – 3; there are 3 sign changes so there are either 3 or 1 negative real zeros

18. EXAMPLE 4 List the potential rational zeros of: f(x)= 2x 3 + 11x 2 – 7x - 6 First have the constant a 0 (-6) as p Then have the leading coefficient (2) as q List all the factors p: ±1, ±2, ±3, ±6 q: ±1, ±2 Then form all possible ratios of p/q p/q: ±1, ±2, ±3, ±6, ± ½, ±3/2

19. Steps for Finding the Real Zeros of a Polynomial Function Step 1: Use the degree of the polynomial to determine the maximum number of real zeros. Step 2: Use Descartes’ Rule of Signs to determine the number of positive and negative zeros. Step 3: (a) Use the Rational Zeros Theorem to find potential Rational Zeros. (b) Use substitution, synthetic division, or long division to test the zeros. Step 4: Each time you find a zero, use the depressed equation to find more potential zeros.

20. To find the Bounds on the zeros of a polynomial you add 1 to the absolute value of the largest coefficient in the polynomial. EXAMPLE 8 f(x)= x 5 + 3x 3 – 9x 2 + 5 The absolute value of the largest number is |-9|= 9. You add 1 to 9= 10. Conclusion: Every zero of f(x)= x 5 + 3x 3 – 9x 2 + 5 lies between -10 and 10.

21. Let f denote a polynomial function. If a<b and if f(a) and f(b) are of opposite sign, there is at least one real zero of f between a and b. 2 1 Zero Show that f(x) = x 5 – x 3 – 1 has a zero between 1 and 2. f(1) = -1 and f(2) = 23 Since f(1) 0, it follows from the Intermediate Value Theorem that f has a zero between 1 and 2 x y

22. Complex Zeros

23. Let f(x) be a polynomial whose coefficients are real numbers. If r (complex zero) = a + bi is a zero of f, then the complex conjugate, r = a – bi is also a zero of f. EXAMPLE Find the remaining two zeros of a polynomial of degree 5 which has the real zeros 1, 5i, and 1 + i. Conclusion: The remaining zeros are the conjugates of 5i and 1 + i: -5i and 1 – i.

24. EXAMPLE Find a polynomial of degree 4 whose coefficients are real numbers and has the zeros 1, 1, and -4 + i. Since -4 + i is a zero, -4 – i must be a zero too. f(x)= a(x-1)(x-1)[x-(-4+i)][x-(-4-i)] f(x)= a(x 2 – 2x + 1)(x 2 – 4x – ix + 4x + ix +16 + 4i - 4i - i 2 ) f(x)= a(x 2 – 2x + 1)(x 2 + 8x + 17) f(x)= a(x 4 + 6x 3 + 2x 2 – 26x + 17)

25. Find the Complex Zeros of the polynomial function: f(x)= 3x 4 + 5x 3 + 25x 2 +45x – 18 Step 1: The degree is 4 so f will have four complex zeros. Step 2: By Descartes’ Rule of Signs there is one positive zero and either three or one negative zeros. Step 3: Potential Rational Zeros are: ±1/3, ±2/3, ±1, ±2, ±3, ±6, ±9, ±18 Step 4: Test potential zeros. f(-2)= 0 so -2 is a factor. You can factor the depressed equation: 3x 3 – x 2 + 27x – 9 = 0 x 2 (3x-1) + 9(3x-1) = 0 ▬► a) x 2 + 9 =0 and b) 3x – 1 = 0 a) x 2 = -9 ▬► x= 3i, x = -3i b) 3x = 1 ▬► x = 1/3 The polynomial in factored form is: f(x)= 3(x+3i)(x-3i)(x+2)(x-1/3)

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