Transcript CHAPTER 2 POLYNOMIAL & RATIONAL FUNCTIONS

CHAPTER 2
POLYNOMIAL & RATIONAL
FUNCTIONS
2.1
• COMPLEX NUMBERS
Objectives
•
•
•
•
Add & subtract complex numbers
Multiply complex numbers
Divide complex numbers
Perform operations with square roots of
negative numbers
i = the square root of negative 1
• In the real number system, we can’t take
the square root of negatives, therefore the
complex number system was created.
• Complex numbers are of the form, a+bi,
where a=real part & bi=imaginary part
• If b=0, a+bi = a, therefore a real number
(thus reals are a subset of complex #)
• If a=0, a+bi=bi, therefore an imaginary #
(imaginary # are a subset of complex #)
Adding & Subtracting Complex #
• Add real to real, add imaginary to
imaginary (same for subtraction)
• Example:
(6+7i) + (3-2i)
•
(6+3) + (7i-2i) = 9+5i
• When subtracting, DON’T FORGET to
distribute the negative sign!
• (3+2i) – (5 – i)
• (3 – 5) + (2i – (-i)) = -2 + 3i
Multiplying complex #
• Treat as a binomial x binomial, BUT what is i*i?
It’s -1!! Why??
• Let’s consider i raised to the following powers:
i  (  1)  1
2
2
i  i  i  (1)i  i
3
2
i  (i )  (1)  1
4
2 2
2
EXAMPLE
(2  3i )  (3  6i )
 6  12i  9i  18i
2
 6  3i  18 (1)
 6  3i  18  24  3i
Dividing Complex #
• It is not standard to have a complex # in a
denominator. To eliminate it, multiply be a wellchosen one: ( conjugate/conjugate)
• The conjugate of a+bi=a-bi
• We use the following fact:
(a  bi)  (a  bi)  a  abi  abi  b i
2
 a  b  (1)  a  b
2
2
2
2 2
2
EXAMPLE
3  8i (4  3i )
(3  8i )  (4  3i ) 

4  3i (4  3i )
12  9i  32i  24i

16  9
 12  41i  12 41i



25
25 25
2
2.2 Quadratic Functions
• Objectives
– Recognize characteristics of parabolas
– Graph parabolas
– Determine a quadratic function’s minimum or
maximum value.
– Solve problems involving a quadratic
function’s minimum or maximum value.
Quadratic functions,
2
f(x)= ax  bx  c
graph to be a parabola. The vertex
of the parabolas is at (h,k) and “a”
describes the “steepness” and
direction of the parabola given
f ( x)  a( x  h)  k
2
Minimum (or maximum) function
value for a quadratic occurs at the
vertex.
• If equation is not in standard form, you may have
to complete the square to determine the point
(h,k). If parabola opens up, f(x) has a min., if it
opens down, f(x) has a max.
f ( x)  2 x 2  4 x  3
f ( x)  2( x 2  2 x)  3
f ( x)  2( x  1) 2  2  3  2( x  1) 2  1
(h, k )  (1,1)
• This parabola opens up with a “steepness” of 2
and the minimum is at (1,1). (graph on next page)
Graph of
f ( x)  2x  4x  3  2( x 1)  1
2
2
2.3 Polynomial Functions & Their
Graphs
• Objectives
– Identify polynomial functions.
– Recognize characteristics of graphs of
polynomials.
– Determine end behavior.
– Use factoring to find zeros of polynomials.
– Identify zeros & their multiplicities.
– Use Intermediate Value Theorem.
– Understand relationship between degree &
turning points.
– Graph polynomial functions.
f ( x)  an x  an1x
n
n1
 ... a2 x  a1x  a0
2
• The highest degree in the polynomial is the
degree of the polynomial.
• The leading coefficient is the coefficient of the
highest degreed term.
• Even-degreed polynomials have both ends
opening up or opening down.
• Odd-degreed polynomials open up on one end
and down on the other end.
• WHY? (plug in large values for x and see!!)
Zeros of polynomials
• When f(x) crosses the x-axis.
• How can you find them?
– Let f(x)=0 and solve.
– Graph f(x) and see where it crosses the xaxis.
What if f(x) just touches the x-axis, doesn’t
cross it, then turns back up (or down) again?
This indicates f(x) did not change from pos.
or neg. (or vice versa), the zero therefore
exists from a square term (or some even
power). We say this has a multiplicity of 2 (if
squared) or 4 (if raised to the 4th power).
Intermediate Value Theorem
• If f(x) is positive (above the x-axis) at
some point and f(x) is negative (below the
x-axis) at another point, f (x) = 0 (on the xaxis) at some point between those 2 pts.
• True for any polynomial.
Turning points of a polynomial
• If a polynomial is of degree “n”, then it has
at most n-1 turning points.
• Graph changes direction at a turning point.
Graph
f ( x)  2 x  6 x  18x
3
2
f ( x)  2 x( x 2  3x  9)  2 x( x  3) 2
Graph, state zeros & end behavior
f ( x)  2 x  12x  18x  2 x( x  6 x  9)
3
2
2
f ( x)  2 x( x  3) 2
• END behavior: 3rd degree equation and the leading
coefficient is negative, so if x is a negative number such as
-1000, f(x) would be the negative of a negative number,
which is positive! (f(x) goes UP as you move to the left.)
and if x is a large positive number such as 1000, f(x) would
be the negative of a large positive number (f(x) goes
DOWN as you move to the right.)
• ZEROS: x = 0, x = 3 of multiplicity 2
• Graph on next page
Graph f(x)
f ( x)  2 x 3  12x 2  18x  2 x( x 2  6 x  9)
f ( x)  2 x( x  3) 2
Which function could possibly
coincide with this graph?
1)  7 x  5 x  1
5
2)9 x  5 x  7 x  1
5
2
3)3 x  2 x  1
4
2
4)  4 x  2 x  1
4
2
2.4 Dividing polynomials;
Remainder and Factor Theorems
• Objectives
– Use long division to divide polynomials.
– Use synthetic division to divide polynomials.
– Evaluate a polynomials using the Remainder
Theorem.
– Use the Factor Theorem to solve a polynomial
equation.
How do you divide a polynomial by
another polynomial?
• Perform long division, as you do with
numbers! Remember, division is repeated
subtraction, so each time you have a new
term, you must SUBTRACT it from the
previous term.
• Work from left to right, starting with the
highest degree term.
• Just as with numbers, there may be a
remainder left. The divisor may not go into
the dividend evenly.
Remainders can be useful!
• The remainder theorem states: If the
polynomial f(x) is divided by (x – c), then
the remainder is f(c).
• If you can quickly divide, this provides a
nice alternative to evaluating f(c).
Factor Theorem
• f(x) is a polynomial, therefore f(c) = 0 if
and only if x – c is a factor of f(x).
• If we know a factor, we know a zero!
• If we know a zero, we know a factor!
2.5 Zeros of Polynomial Functions
• Objectives
– Use Rational Zero Thm. to find possible zeros.
– Find zeros of a polynomial function.
– Solve polynomial equations.
– Use the Linear Factorization Theorem to find
polynomials, given the zeros.
– Use Descartes’s Rule of Signs.
Rational Root (Zero) Theorem
• If “a” is the leading coefficient and “c” is the
constant term of a polynomial, then the only
possible rational roots are  factors of “a” divided
by  factors of “c”.
• Example: f ( x)  6x5  4x3 12x  4
• To find the POSSIBLE rational roots of f(x), we
need the FACTORS of the leading coefficient and
the factors of the constant term. Possible rational
1 3 1 3
roots are  1,2,3,6

 1,2,3,6, , , , 
 1,2,4
2 2 4 4

How many zeros does a polynomial
with rational coefficients have?
• An nth degree polynomial has a total of n zeros.
Some may be rational, irrational or complex.
• Because all coefficients are RATIONAL, irrational
roots exist in pairs (both the irrational # and its
conjugate). Complex roots also exist in pairs (both
the complex # and its conjugate).
• If a + bi is a root, a – bi is a root
• If a  b is a root, a  b is a root.
Descartes’s Rule of Signs
• Depends on the number of sign changes
(switching from pos. to neg. or vice versa)
• Count sign changes in f(x). # of pos. real
zeros is the # of sign changes or an even
number less than # of sign changes.
• Count sign changes in f(x). # of neg. real
zeros is # of sign changes or an even
number less than the # of sign changes.
2.6 Rational Functions & Their
Graphs
• Objectives
– Find domain of rational functions.
– Use arrow notation.
– Identify vertical asymptotes.
– Identify horizontal asymptotes.
– Use transformations to graph rational functions.
– Graph rational functions.
– Identify slant (oblique) asymptotes.
– Solve applied problems with rational functions.
Vertical asymptotes
• Look for domain restrictions. If there are values
of x which result in a zero denominator, these
values would create EITHER a hole in the graph
or a vertical asymptote. Which? If the factor
that creates a zero denominator cancels with a
factor in the numerator, there is a hole. If you
cannot cancel the factor from the denominator, a
vertical asymptote exists.
• If you evaluate f(x) at values that get very, very
close to the x-value that creates a zero
denominator, you notice f(x) gets very, very, very
large! (approaching pos. or neg. infinity as you
get closer and closer to x)
•
3x  7
f ( x) 
x2
Example
• f(x) is undefined at x = 2
• As x  2 , f ( x)  
x  2 , f ( x)  
• Therefore, a vertical asymptote exists at x=2. The
graph extends down as you approach 2 from the
left, and it extends up as you approach 2 from the
right.
What is the end behavior of this
rational function?
• If you are interested in the end behavior, you are
concerned with very, very large values of x.
• As x gets very, very large, the highest degree term
becomes the only term of interest. (The other
terms become negligible in comparison.)
• SO, only examine the ratio of the highest degree
term in the numerator over the highest degree term
of the denominator (ignore all others!)
3x  7
3x
f
(
x
)

f
(
x
)

3
• As x gets large,
becomes
x2
x
• THEREFORE, a horizontal asymptote exists, y=3
What if end behavior follows a line
that is NOT horizontal?
8 x 2  3x  2
f ( x) 
2x  2
• Using only highest-degree terms, we are left with
• This indicates we don’t have a horizontal
asymptote. Rather, the function follows a slanted
line with a slope = 4. (becomes y=4x as we head
towards infinity!)
• To find the equation of the slant asymptote,
proceed with long division, as previously done.
The quotient is the slant (oblique) asymptote. For
this function, y = 4x – 11/2
• NOTE: f(x) also has a vertical asymptote at x=1.
Graph of this rational function
8 x  3x  2
f ( x) 
2x  6
2
What is the equation of the oblique
asymptote?
4 x  3x  2
f ( x) 
2x 1
2
1.
2.
3.
4.
y = 4x – 3
y = 2x – 5/2
y = 2x – ½
y = 4x + 1
2.7 Polynomial & Rational
Inequalities
• Objectives
– Solve polynomial inequalities.
– Solve rational inequalities.
– Solve problems modeled by polynomial or
rational inequalities.
Solving polynomial inequalities
• Always compare the polynomial to zero.
• Factor the polynomial. We are interested
in when factors are either pos. or neg., so
we must know when the factor equals
zero.
• The values of x for which the factors equal
zero provide the cut-offs for regions to
check if the polynomial is pos. or neg.
(x – 3)(x + 1)(x – 6) < 0
• In order for the product of 3 terms to be less than
zero (negative), either all 3 terms must be neg. or
exactly 1 of them be neg.
• The 3 “cut-off” values are x = 3,-1,6
• The 3 cut-off values create 4 intervals along the xaxis: (,1), (1,3), (3,6), (6, )
• Pick a point in each interval & determine if that
value for x would make all 3 factors neg. or exactly
1 negative. If so, the function is < 0 on that interval.
• x<-1, f(x) < 0
-1<x<3, f(x) > 0
• 3<x<6, f(x)< 0
x>6, f(x) > 0
• Solution: {x: x < -1 or 3 < x < 6}
Given the following graph of f(x),
give interval notation for x-values
such that f(x)>0.
1)(3,1)  (0,2)  (4, )
2)(,3)  (1,0)  (2,4)
3)(, )
4)(3,4)
Solving rational inequalities
• VERY similar to solving polynomial inequalites
EXCEPT if the denominator equals zero, there
is a domain restriction. The function COULD
change signs on either side of that point.
• Step 1: Compare inequality to zero. (add
constant to both sides and use a common
denominator to have a rational expression)
• Step 2: Factor both numerator & denominator to
find “cut-off” values for regions to check when
function becomes positive or negative.
2.8 Modeling Using Variation
• Objectives
– Solve direct variation problems.
– Solve inverse variation problems.
– Solve combined variation problems.
– Solve problems involving joint variation.
Direct variation
• y varies directly as x (y is directly
proportional to x) if y = kx.
• k is the constant of variation (constant of
proportionality)
• This is the graph of a linear function with
slope = m, crossing through the origin.
• Example: You are paid $8/hr. Thus, pay is
directly related to hours worked:
pay=8(hours worked)
Direct variation COULD involve an
nth power of x (no longer linear)
• Still involves a constant of proportionality
n
y  kx
• y is directly proportional to the nth power of x.
Inverse Variation
k
y
x
• As x gets bigger, y gets smaller
• As x gets smaller, y gets bigger
Combined Variation Problem
• y is impacted by TWO variables in TWO different
ways. One variable causes y to get bigger, while
the other variable causes it to become smaller.
kx
y
z
• As x gets bigger, y gets bigger, but as z gets
bigger, y gets smaller.
• k must take into account both influences