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4.1 Polynomial Functions and Models
Essential Question(s):
• How do you graph polynomial functions?
• How do you divide polynomials?
|Polynomial Function | |
| |[pic] where n is a nonnegative integer and [pic] are real numbers such that [pic] |
| | |
|leading Coefficient – |The first term of a function in correct algebraic order |
|Turning points | |
| |Points where the graph changes from increasing to decreasing or vice versa. If f is a polynomial function of|
| |degree n, then the graph of f has at most n – 1 turning points |
A number r is said to be a zero or root of a function [pic] if [pic]
Properties of graphs of polynomial functions:
Let [pic] be a polynomial of degree [pic] with real coefficients. Then the graph of [pic]:
1. Is continuous for all real numbers
2. Has no sharp corners
3. Has at most [pic] real zeros
4. Has at most [pic] turning points
5. Increases or decreases without bound as [pic] and as [pic]
The Leading Coefficient Test
If [pic] is the leading term of a polynomial, then the behavior of the graph as [pic] or as [pic] can be described in one of the four following ways.
|If n is even and [pic]: |If n is even and [pic]: |
|[pic] |[pic] |
|If n is odd and [pic]: |If n is odd and [pic]: |
|[pic] |[pic] |
Examples
Use the following to answer questions 1-2:
P(x) = (x – 8)(x2 – 1)(x2 + 81)
|1. |List all zeros of the polynomial function. |
[pic]
8, -1, 1, -9i, 9i
|2. |Which zeros are x-intercepts? |
8, 1, –1
|3. |Graph the polynomial function. |
| |P(x) = x3 – 3x – 2 |
Example: Long Divide to see if [pic] is a factor of [pic]. Identify the dividend, quotient, divisor and remainder.
Synthetic Division: may only be used when dividing a polynomial by a first degree divisor of the form x – r, where r is a constant
1. Write the value of r in a box.
2. Write the coefficients of the dividend to the right of the box.
3. Skip a line and draw a horizontal line below the list of coefficients.
4. Bring down the leading coefficient and write it below the line.
5. Multiply the value of r by the number below the line. Write the result in the next column above the line.
6. Add the numbers in the column above the line and write the result below the line.
Repeat 5 & 6 until all columns are completed.
7. To get the final result, we use the numbers below the line. The number in the last column is the remainder. The other numbers are the coefficients of the quotient.
Use synthetic division to see if [pic] is a factor of [pic]. Identify the dividend, quotient, divisor and remainder.
Check the synthetic division.
[pic]
Define or describe the following:
| | |
|The Remainder Theorem |If a number c is substituted for x in the polynomial f(x), then the result f(c) is the |
| |remainder that would be obtained by dividing f(x) by x – c |
| | |
|The Factor Theorem |For a polynomial f(x), if f(c) = 0, then x – c is a factor of f(x) |
Example: Show three different ways to obtain a remainder when [pic]is divided by [pic].
|Long Division |Synthetic Division |
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|[pic] |[pic] |
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|Remainder Theorem |
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|[pic] |
| |
4. Use synthetic division to compute the quotient and remainder.
(x3 – 4x2 + x – 1) ÷ (x + i)
[pic]
4.2 Real Zeros and Polynomial Inequalities
Essential Question(s):
• How do you find the real zeros of polynomial equations?
• How do you solve polynomial inequalities?
Upper and Lower Bounds for Real Zeros
Upper and Lower Bound Theorem:
Let [pic] be a polynomial of degree [pic] with real coefficients [pic];
1. Upper bound: A number [pic] is an upper bound for the real zeros of [pic] if, when [pic] is divided by [pic] using synthetic division, all numbers in the quotient row, including the remainder are nonnegative.
2. Lower bound: A number [pic] is a lower bound for the real zeros of [pic] if, when [pic] is divided by [pic] using synthetic division, all numbers in the quotient row, including the remainder, alternate in sign. (note, the sign on 0 could be considered + or - , but not both)
The Intermediate Value Theorem (Location Theorem)
For any polynomial function [pic]with real coefficients, suppose that for [pic], [pic] and [pic] are of opposite signs. Then the function has a real zero between a and b.
Two sketches that demonstrate visually the meaning of the Intermediate Value Theorem:
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Examples
|1. |Use the location theorem to explain why the polynomial function has a zero on the indicated interval. |
| | |
| |P(x) = 3x3 + x2 – 4x + 9; (–2, –1). |
P(–2) < 0 and P(–1) > 0
Use the following to answer questions 2-3:
P(x) = x3 – 2x2 + 3x – 7
|2. |Find the smallest positive integer and largest negative integer that, by the Upper and Lower Bound Theorem, are upper and lower |
| |bounds, respectively for the real zeros of P(x). |
[pic]
Upper bound: 3; lower bound: –1
|3. |Use the bisection method to approximate a real zero of the polynomial to one decimal place. |
| |Sign of P |
|Sign Change Interval |Midpoint |P(a) |P(m) |P(b) |
|(a, b) |m | | | |
|(-1, 3) |1 |- |- |+ |
|(1, 3) |2 |- |- |+ |
|(2, 3) |2.5 |- |+ |+ |
|(2, 2.5) |2.25 |- |+ |+ |
|(2, 2.25) |2.125 |- |- |+ |
|(2.125, 2.25) |2.1875 |- |+ |+ |
|(2.125, 2.1875) |2.15625 |- |+ |+ |
|(2.125, 2.15625) |2.140625 |- |+ |+ |
|(2.125, 2.140625) |stop here |- | |+ |
2.1
Approximating Zeros at Turning Points ( Use graphing calculator
|4. |Approximate the zeros of the polynomial function to two decimal places, using maximum or minimum commands to approximate any |
| |zeros at turning points. |
| |P(x) = x4 – 4x3 – 4x2 + 16x + 16 |
[pic] [pic]
–1.24, 3.24
Procedure for solving a polynomial inequality:
1. Transform the inequality so that one side is zero.
2. Solve the related equation to find boundary points.
3. Mark the boundary points on a number line to divide the number line into intervals.
4. Check a value in each interval to see whether or not it solves the inequality.
5. Graph your solution on a number line and state your solution in interval notation.
Example: [pic]
[pic] ( Intervals are [pic]
|Interval |Test Value |Substitute into [pic] |Conclusion |
|[pic] |[pic] |[pic] Positive |[pic] for all x in [pic] |
|[pic] |[pic] |[pic] Negative |[pic] for all x in [pic] |
|[pic] |[pic] |[pic] Positive |[pic] for all x in [pic] |
Solution: [pic]
|Example: [pic] |Example: [pic] |
|[pic] |[pic] |
| | |
|Intervals are [pic] |( Intervals are [pic] |
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|Solution: [pic] |Solution: [pic] |
4.3 Complex Zeros and Rational Zeros of Polynomials
Essential Question(s):
• How do you find complex zeros and rational zeros of polynomial equations?
|Multiplicity and x-intercepts | |
| |The multiplicity of a zero tells how many times the zero occurs as a root of the polynomial. If r is a zero |
| |of even multiplicity, then the graph touches the x-axis and turns around at r. If r is a zero of odd |
| |multiplicity, then the graph crosses the x-axis at r. |
| | |
|Fundamental Theorem of Algebra: |If f(x) is a polynomial of degree n, where [pic], then the equation [pic] has at least one complex root. |
|The Linear Factorization Theorem | |
| |If [pic] where [pic] and [pic], then [pic], where [pic] are complex numbers (An nth degree polynomial can be |
| |expressed as the product of a nonzero constant and n linear factors) |
|Complex Conjugate Theorem | |
| |If [pic] is a zero of a polynomial, then its conjugate, [pic], is also a root |
|Linear and Quadratic Factors Theorem | |
| |If [pic] is a polynomial of degree [pic] with real coefficients, then [pic] can be factored as a product of |
| |linear factors (with real coefficients) and quadratic factors (with real coefficients and imaginary zeros). |
|Real Zeros and Polynomials of Odd Degree | |
| |Every polynomial of odd degree with real coefficients has at least one real zero. |
|Zeros of Even or Odd Multiplicity | |
| |Let [pic] be a polynomial with real coefficients |
| |If r is a real zero of [pic] of even multiplicity, then [pic] has a turning point at r and does not change |
| |sign at r. |
| |If r is a real zero of [pic] of odd multiplicity, then [pic] does not have a turning point at r and changes |
| |sign at r. |
|Rational Root Theorem | |
| |If a polynomial has a rational root it will be a |
| | |
| |[pic] or [pic] |
Examples
|1. |List all zeros of the polynomial function P(x) = (x – 6)(x2 – 64)(x2 + 81). |
[pic] [pic] [pic]
[pic]
|2. |Find the zeros of the polynomial and indicate the multiplicity of each. |
| |P(x) = (x2 – 9)4(x2 + 1)4(x – 6i)2 |
[pic] [pic] [pic]
The zeros are: - 3, 3, -i, i, 6i
-3, 3, -i, and i all have multiplicity of 4 (from the exponent on their corresponding factors)
6i has multiplicity of 2 (from the corresponding exponent on its factor)
|3. |Find a polynomial P(x) of lowest degree, with leading coefficient 1, that has the indicated zeros. Write P(x) as a product of |
| |linear factors. |
| |–3 (multiplicity 1), 8 (multiplicity 5) |
[pic]
(x + 3) comes from the fact that – 3 is a zero with multiplicity 1
(x – 8 ) comes from the fact that 8 is a zero with multiplicity 5
|4. |Find a polynomial of lowest degree, with leading coefficient 1, that has the indicated graph. Assume all zeros are integers. |
| |[pic] |
[pic]
Graph is negative from our discussion about “end behavior”
(x + 1)2 from the fact that -1 is a zero and the graph touches and turns around at this point
(x – 2) from the fact that graph crosses x – axis at 2.
|5. |List all possible rational zeros for the polynomial. |
| |f(x) = 5x3 – 2x2 + x – 35 |
Factors of the constant term (-35) are: [pic]
Factors of the leading term (5) are: [pic]
Possible rational factors [pic] are:
[pic]
|6. |Factor the polynomial in two ways. |
| |(a) As a product of linear factors (with real coefficients) and quadratic |
| |factors (with real coefficients and imaginary zeros) |
| |(b) As a product of linear factors with complex coefficients. |
| |P(x) = x3 – x2 + 9x – 9 |
a) [pic] b) [pic]
|7. |Find all zeros exactly (rational, irrational, and imaginary). |
| |P(x) = 4x4 + x3 + 33x2 +9x – 27 |
List possible rational roots: [pic]
Use synthetic division to determine a root (you can find the roots that work by graphing the function on your calculator and finding the zeros). You must use the synthetic division to reduce the polynomial.
Synthetically dividing the P(x) by x + 1 yields the quotient: [pic] which can be factored as: [pic]
So, P(x), our original function, factors as [pic]
The zeros are [pic]
|8. |Find a polynomial P(x) having root 1 + 3i, degree 2, leading coefficient 1, and real coefficients. Write the polynomial in |
| |expanded form. |
[pic] Remember, complex numbers come in pairs that are conjugates
Foil this to get:
[pic]
4.4 Rational Functions and Inequalities
Essential Question(s):
• How do you find horizontal, vertical, and oblique asymptotes of functions?
• How do you graph rational functions?
Rational means … fractional
A. Domain of a Rational Function ( all real values except for those that make the denominator equal to zero
B. Vertical Asymptotes & Holes
[pic]
• Let f be a rational function defined by where n(x) and d(x) are polynomials with no common factors. If a is a real number such that d(a) = 0, then x = a is a vertical asymptote of f.
• A Hole occurs when the denominator is zero and the factor can be reduced with a common factor in the numerator.
• A graph may cross a horizontal or oblique asymptote but never a vertical asymptote!
Locate any vertical asymptotes of the rational function.
C. Horizontal Asymptotes
[pic]
•
• The graph of a rational function may or may not cross a horizontal asymptote.
[pic]
D. Oblique or Slant Asymptotes
[pic]
Find the oblique asymptote of the rational function.
NOTE: Asymptotes should be shown on graphs as dashed lines!
Example:
Find the asymptotes for [pic].
[pic], and the function cannot be reduced ( Vertical asymptote:[pic]
[pic] is a horizontal asymptote since degree of numerator = degree of denominator
Graphing Rational Functions:
[pic]
Example:
Make a sketch of [pic]. Show all asymptotes, zeros and three points
in each region.
[pic]
Procedure for solving a rational inequality:
1. Transform the inequality so that one side is zero and the other side is a single quotient.
2. Set the numerator and denominator equal to zero and solve to locate the boundary points.
3. Mark the boundary points on a number line to divide the number line into intervals.
4. Check a value in each interval to see whether or not it solves the inequality.
5. Graph your solution on a number line and state your solution in interval notation.
Example: [pic]
|Numerator |Denominator |
| |[pic] |
|[pic] | |
Intervals are [pic]
Solution: [pic]
4.5 Variation and Modeling
Essential Question(s):
• What is direct variation, inverse variation, joint variation, and combined variation?
|Direct Variation |
| | |
|Whenever a real-life situation produces pairs of numbers in which the |Example: |
|ratio of the second coordinate to the first (the slope) is the same, | |
|there is direct variation. |Find the constant of proportionality and an equation when y varies directly with x|
| |and y = 64 when x = 12. |
|If a situation gives rise to a linear function [pic], where k is a | |
|constant, we have |[pic] |
|direct variation . | |
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|The number k is called the constant of proportionality | |
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|Key phrases identifying direct variation: | |
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|1. y varies directly as x | |
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|2. y is directly proportional to | |
|Inverse Variation |
| | |
|Whenever a situation produces pairs of numbers in which the product is|Example: |
|constant, we say there is inverse variation . | |
| |Find the variation constant and an equation of variation in which y varies |
|If a situation gives rise to a function [pic], where k is a positive |inversely as x, and y = 32 when x = .04. |
|constant, we say that we have inverse variation . | |
| |[pic] |
|The number k is called the constant of proportionality | |
| | |
|Key phrases identifying inverse variation: | |
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|1. y varies inversely as x | |
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|2. y is inversely proportional to | |
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|Combined Variation |
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|Direct combined variation |Example: |
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|If a situation gives rise to a linear function |Find an equation of variation in which y varies directly as the square of x and y |
|[pic], where k is a constant, we have direct combined variation. |= 24 when x = 4. |
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| |[pic]( So, [pic] |
|The number k is called the constant of proportionality | |
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|Key phrase identifying direct combined variation: | |
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|y varies directly as the nth power of x | |
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|Inverse combined variation |Example: |
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|If a situation gives rise to a linear function [pic], where k is a |Find an equation of variation in which y varies inversely as the square of x and y|
|constant, we have |= 24 when x = 4. |
|inverse combined variation. | |
| |[pic] ( So, [pic] |
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|The number k is called the constant of proportionality | |
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|Key phrase identifying inverse combined variation: | |
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|y varies inversely as the nth power of x | |
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|Joint variation |Example: |
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|If a situation gives rise to a linear function [pic], where k is a |Find an equation of variation in which y varies jointly as the square of x and the|
|constant, we have joint variation. |cube of z and y = 24 when x = 4 when z = 3. |
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| |[pic] ( So, [pic] |
|The number k is called the constant of proportionality. | |
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|Key phrase identifying joint variation: | |
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|y varies jointly as x and z | |
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-----------------------
( Dividend
(Remainder
(Quotient
Divisor (
[pic]
( Dividend
(Remainder
Quotient (
Divisor (
[pic]
[pic]
y
x
a
b
c
P(a) ( +
P(b) ( –
P(c) = 0
[pic]
y
x
a
b
c
f(b) ( +
f(a) ( –
f(c) = 0
[pic]
– 6
6
– 3
3
– 3
4
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
Vertical asymptotes: [pic], [pic]
Horizontal asymptote: [pic]
Zeros: [pic] (from numerator)
– 3
4
5
................
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