Transcript PC Ch4

Jonathon Vanderhorst & Callum Gilchrist
Period 1
INCLUDES:
•Polynomial Functions and Models
•Properties of Rational Functions
•The Graph of a Rational Function
•Polynomial and Rational Inequalities
•Real Zeros
•Complex Zeros
Polynomial functions and
models
A polynomial function is of the form:
f(x) = anxn+an-1x n-1+…+a1x+a0
The degree is based on the largest power of x
A power function, however, is simply a monomial:
f(x) = axn
For any given function n must be a non-negative integer
A power function; y=x2
TECHNIQUES:
•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= anxn
Properties of Rational
Functions
A rational function is of the form:
R(x) = p(x)/q(x)
Where p and q are polynomial functions
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 an/bm 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
The Graph of a Rational
Function
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
Polynomial and Rational
Inequalities
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.
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
Factor
Step 3: Use the zeros to separate the number line into four intervals:
(-∞, -2)
(-2,0)
(0,2)
(2, ∞)
-2
0
2
∞
Interval
(-∞, -2)
(-2,0)
(0,2)
(2, ∞)
Number
-3
-1
1
3
Value
45
-3
-3
45
Conclusion
Positive
Negative
Negative
Positive
Conclusion {x|-2 ≤ X ≤ 2} or [-2,2]
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)
-3
(1,2)
2
(2,
1
∞)
∞
Interval
(-∞, -3)
(-3,1)
(1,2)
(2, ∞)
Number
-4
0
3/2
3
Value
-6/25
6
9
-3/2
Conclusion
Negative
Positive
Positive
Negative
Conclusion {x|-3 < X < 2, X ≠ 1} or (-3,1) U (1,2)
The real zeros of a polynomial
function
If f(x) is divided by x - c, then the remainder is
f(c)
Find the remainder if f(x) = x2 – 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
2x3 – x2 + 2x – 3
2(1)3 –(1)2 +2(1) -3 =0 (x-1) is a factor by the
Factor Theorem
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)= 3x6 – 4x4 + 3x3 + 2x2 – 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
EXAMPLE 4
List the potential rational zeros of:
f(x)= 2x3 + 11x2 – 7x - 6
First have the constant a0 (-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
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.
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)= x5 + 3x3 – 9x2 + 5
The absolute value of the largest number
is |-9|= 9. You add 1 to 9= 10.
Conclusion: Every zero of f(x)= x5 + 3x3 – 9x2
+ 5 lies between -10 and 10.
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.
y
1
Zero
x
2
Show that f(x) = x5 – x3 – 1 has a zero
between 1 and 2.
f(1) = -1 and f(2) = 23
Since f(1)<0 and f(2) > 0, it follows
from the Intermediate Value Theorem
that f has a zero between 1 and 2
Complex Zeros
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.
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(x2 – 2x + 1)(x2 – 4x – ix + 4x + ix
+16 + 4i - 4i - i2)
f(x)= a(x2 – 2x + 1)(x2 + 8x + 17)
f(x)= a(x4 + 6x3 + 2x2 – 26x + 17)
Find the Complex Zeros of the polynomial function: f(x)= 3x4 + 5x3
+ 25x2 +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: 3x3 – x2 + 27x – 9 = 0
x2(3x-1) + 9(3x-1) = 0 ▬► a) x2 + 9 =0 and b) 3x – 1 = 0
a) x2 = -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)
=)