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Polynomial Functions

Polynomial Functions. AFM Objective 1.01. Polynomial Functions. Polynomials have more than one term: 2x + 1 3x 2 + 2x + 1 The fancy explanation: f(x) = a n x n + a n-1 x n-1 + …+ a 2 x 2 + a 1 x + a Ex: 4x 3 + 3x 2 + 2x + 1

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Polynomial Functions

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  1. Polynomial Functions AFM Objective 1.01

  2. Polynomial Functions • Polynomials have more than one term: • 2x + 1 • 3x2 + 2x + 1 • The fancy explanation: • f(x) = anxn + an-1xn-1 + …+ a2x2 + a1x + a • Ex: 4x3 + 3x2 + 2x + 1 • Simply put: the addition and subtraction of terms with different exponents going from largest to smallest

  3. Polynomial Functions • Polynomial functions are named by their degrees • The largest exponent is the degree of the function 4x3 + 3x2 + 2x + 1 This has a degree of 3

  4. Polynomial Functions • Leading Term: • The term with the highest exponent • This function has a degree of 3 • By knowing the leading term we can analyze a function without graphing it *The highest exponent may not be written first in the function!! 4x3 + 3x2 + 2x + 1

  5. Quick Quiz 1 • What are the degrees of the following polynomial functions? • 3x5 – 2x4 + 9x – 2 • -6x3 + 2x2 – 7x + 10 • 2x11 + 5x6 – 3x2 • 4x2 + 5x4 – 2x3 + x – 5

  6. Quick Quiz 1 • What are the degrees of the following polynomial functions? • 3x5 – 2x4 + 9x – 2 • -6x3 + 2x2 – 7x + 10 • 2x11 + 5x6 – 3x2 • 4x2 + 5x4 – 2x3 + x – 5 Degree of 5 Degree of 3 Degree of 11 Degree of 4

  7. Polynomial Functions • Certain polynomial functions have similar qualities. • We will use the Leading Term to analyze the behavior of a function • For simplicity sake we will just use the function f(x) = axn

  8. axn • If a is positive and n is an odd number what kind of equations will you get? • 2x • 4x3 • 7x5 • Here’s an example:

  9. axn • Here’s another example: • Notice how the straight line and this line both point down on the left and up on the right • The have the same “end behavior”

  10. axn • All equations that have a positive a and odd n have the same end behavior

  11. axn • If a is negative and n is odd it will change the end behavior (ex: -2x3)

  12. axn • If a is positive and n is an even number what kind of equations will you get? • 2x2 • 4x4 • 7x6 • Here’s an example: • Notice that both ends point the same way, up!

  13. axn • All equations where a is positive and n is an even number will have the same end characteristics:

  14. axn • If a is negative and n is an even number it will change the end behavior • Here’s an example: • Notice that both ends still point the same way but down

  15. axn • All equations where a is negative and n is an even number the end behavior will be:

  16. Quick Quiz 2 • What is the end behavior of the following functions? • f(x) = 3x3 • f(x) = 4x6 – 2x2 + 1 • f(x) = -3x2 + 2x – 5 • f(x) = 4x – 5x5 + 3x3

  17. Quick Quiz 2 • What is the end behavior of the following functions? • f(x) = 3x3 • f(x) = 4x6 – 2x2 + 1 • f(x) = -3x2 + 2x – 5 • f(x) = 4x – 5x5 + 3x3

  18. Turning Points • Turning Points: • Where the graph changes direction • It goes from increasing to decreasing or the other way around Turning Point

  19. Turning Points • This is a cubic (x3) function and it has 2 turning points Turning Point

  20. Turning Points • This is also a cubic (x3) function • Notice there are noTurning Points! • This function never changes it’s direction • *Cubic (x3) functions can have up to 2 Turning Points

  21. Turning Points • This is a quartic (x4) function Turning Point

  22. Turning Points • This is also a quartic (x4) function • Notice there is just one Turning Point! • This function only changes direction once • *Quartic (x4) functions can have up to 3 Turning Points

  23. Turning Points • If you look at the exponent on the Leading Term you can analyze up to how many Turning Points there can be axn Has up to n – 1 turning points

  24. Quick Quiz 3 • Up to how many possible turning points do the following functions have? • 3x5 – 2x4 + 9x – 2 • -6x3 + 2x2 – 7x + 10 • 2x11 + 5x6 – 3x2 • 4x2 + 5x4 – 2x3 + x – 5

  25. Quick Quiz 3 • Up to how many possible turning points do the following functions have? • 3x5 – 2x4 + 9x – 2 • -6x3 + 2x2 – 7x + 10 • 2x11 + 5x6 – 3x2 • 4x2 + 5x4 – 2x3 + x – 5 4 2 10 3

  26. Zeros • A zero is the x value when the function crosses the x-axis Zeros • This is a quartic (x4) function • Notice it has 4 zeros

  27. Zeros • This is also a quartic (x4) function Zeros • Notice it has only 2 zeros • The others are “imaginary” • *Quartic functions can have up to 4 real zeros

  28. Zeros • If you look at the exponent on the Leading Term you can analyze up to how many Zeros there can be axn Has up to n real zeros

  29. Zeros • Remember that when a line crosses the x-axis, the y value is zero. • Any point that is (x, 0) is a “zero” or x-intercept • You can find some of the zeros by solving the equations

  30. Zeros • Example: Determine all the real zeros for • This is a cubic function so it can have up to 3 real zeros • First set the function equal to zero (y is zero) • We need to factor to find the solutions x3 – 5x2 + 6x

  31. Zeros • Find a common factor and pull it out • Factor what’s left inside the parenthesis x3 – 5x2 + 6x = 0 x(x2 – 5x + 6) = 0 x(x – 2)(x - 3) = 0

  32. Zeros • Set each part of the function equal to zero • Solve each one for x x(x– 2)(x - 3) = 0 x - 3 = 0 x – 2 = 0 x = 0 x = 0, 2, 3

  33. Zeros • Example 2: Determine all the real zeros for • We know this will have up to 4 real zeros • Also, this equation is in quadratic form since we can write it like: x4 – 8x2 + 15 (x2)2 – 8(x2) + 15 Two x2’s One x2 A #

  34. Zeros • In order to factor this properly we can replace the x2’s with the letter u • We now have an equation we can factor! (x2)2 – 8(x2) + 15 (u)2 – 8(u) + 15

  35. Zeros • Factor u2 – 8u + 15 • Remember, u is equal to x2 • Plug x2 in for u (u)2 – 8(u) + 15 (u – 3)(u - 5) (x2– 3)(x2- 5)

  36. Zeros • Solve each one for x x2– 5 = 0 x2– 3 = 0 x2= 5 x2= 3 x= ± x= ±

  37. Zeros • Determine all the real zeros for • This can have up to 4 real zeros • Factor out the GCF first (always take the negative with the first number) -x4 – x3 + 2x2 -x2(x2 + x – 2)

  38. Zeros • Factor inside the parenthesis • Set everything equal to zero -x2(x2 + x – 2) -x2(x + 2)(x – 1)

  39. Zeros • Solve each part for x -x2(x + 2)(x – 1) -x2 = 0 -1 -1 x2= 0 x= ± 0

  40. Zeros • Notice that you have two answers of zero • This is called “Multiplicity” • Multiplicity- when you have a repeated zero (answer) x= ± 0

  41. Zeros • Solve each part for x x – 1= 0 x + 2= 0 x = 1 x= -2

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