Chapter 2 Notes

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HW: Pg. 175 #7-16

Polynomial Functions-
◦ Let n be a nonnegative integer and let a0, a1, a2,…,
an-1, an be real numbers with an≠0. The functions
given by
f(x)=anxn + an-1xn-1+…+a2x2+a1x+a0
Is a polynomial function of degree n. The leading
coefficient is an.
f(x)=0 is a polynomial function.
*it has no degree or leading coefficient.
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F(x) = 5x3-2x-3/4
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G(x) = √(25x4+4x2)
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H(x) = 4x-5+6x
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K(x)=4x3+7x7
Name
Form
Degree
Zero Function
F(x) = 0
Undefined
Constant Function
F(x)=a (a≠0)
0
Linear Function
F(x)=ax+b (a≠0)
1
Quadratic Function
F(x)=ax2+bx+c (a≠0)
2

F(x) = ax+b
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Slope-Intercept form of a line:
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Find an equation for the linear function f such
that f(-2) = 5 and f(3) = 6
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The average rate of change of a function
y=f(x) between x=a and x=b, a≠b, is
[F(b)-F(a)]/[b-a]
Weehawken High School bought a $50,000
building and for tax purposes are depreciating it
$2000 per year over a 25-yr period using
straight-line depreciation.
1.
2.
3.
4.
What is the rate of change of the value of the
building?
Write an equation for the value v(t) of the building
as a linear function of the time t since the
building was placed in services.
Evaluate v(0) and v(16)
Solve v(t)=39,000
Point of View
Characterization
Verbal
Polynomial of degree 1
Algebraic
F(x)=mx+b (m≠0)
Graphical
Slant line with slope m and yintercept b
Analytical
Function with constant nonzero
rate of change m, f is increasing if
m>0, decreasing if m<0
Sketch how to transform f(x)=x2 into:
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G(x)=-(1/2)x2+3
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H(x)=3(x+2) 2-1
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If g(x) and h(x) and in the form f(x)=ax2+bx+c,
what do you notice about g(x) and h(x) when a is
a certain value (negative or positive)?
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f(x)=ax2+bx+c
We want to find the axis of symmetry, which is
x=-b/(2a).
Then:
The graph of f is a parabola with vertex (x,y),
where x=-b/(2a). If a>0, the parabola opens
upward, and if a<0, it opens downward.

x=-b/(2a)
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F(x)=3x2+5x-4
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G(x)=4x2+12x+4
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H(x)=6x2+9x+3
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f(x)=5x2+10x+5
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Any Quadratic Function f(x)=ax2+bx+c, can
be written in the vertex form:
◦ F(x)=a(x-h)2+k
 Where (h,k) is your vertex
 h=-b/(2a) and k=is the y

F(x)=3x2+12x+11
=3(x2+4x)+11
=3(x2+4x+() - () )+11
=3(x2+4x+(2)2-(2)2)+11
=3(x2+4x+4)-3(4)+11
=3(x+2)2-1
f(x)=a(x-h)2+k
Factor 3 from the x term
Prepare to complete the square.
Complete the square.
Distribute the 3.
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F(x)=3x2+5x-4

F(x)=8x-x2+3

G(x)=5x2+4-6x
Point of View
Characterization
Verbal
Polynomial of degree ___
Algebraic
F(x)=______________ (a≠0)
Graphical
a>0
a<0
HW: Pg.189 #1-10

F(x)=k*xa
◦ a is the power, k is the constant of variation
EXAMPLES:
Formulas
Power
Constant of Variation
C=2∏r
1
2∏
A=∏r2
2
∏
F(x)=4x3
G(x)=1/2x6
H(x)=6x-2

F(x) = ∛x

1/(x2)
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What type of Polynomials are these functions?
(HINT: count the terms)
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6cx-5
h/x4
4∏r2
3*2x
ax
7x8/9
HW: Pg 203 #33-42e
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F(x)=x3+x
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G(x)=x3-x
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H(x)=x4-x2
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Find local extrema and zeros for each
polynomial
F(x)=2x3
F(x)=-x3
F(x)=-2x4
F(x)=4x4
What do you notice about the limits of each
function?
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F(x)=x3—2x2-15x
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What do these zeros tell us about our graph?
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F(x)=3x3 + 12x2 – 15x
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H(x)=x2 + 3x2 – 16
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G(x)=9x3 - 3x2 – 2x
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K(x)=2x3 - 8x2 + 8x
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F(x)=6x2 + 18x – 24
HW: Pg. 216 #1-6
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3587/32
(3x3+5x2+8x+7)/(3x+2)
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
F(x) = d(x)*q(x)+r(x)
F(x) and d(x) are polynomials where q(x) is
the quotient and r(x) is the remainder
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(3x3+5x2+8x+7)/(3x+2)
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Write (2x4+3x3-2)/(2x2+x+1) in fraction form
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
D(x)=x-k, degree is 1, so the remainder is a
real number
Divide f(x)=3x2+7x-20 by:
◦ (a) x+2
(b) x-3
(c) x+5
Remainder Theorem:
If a polynomial f(x) is divided by x-k, then the remainder
is r=f(k)
Ex: (x2+3x+5)/(x-2)
k=2
So, f(k)=f(2)=(2)2+3(2)+5=15=remainder
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Divide f(x)=3x2+7x-20 by:
◦ (a) x+2
(b) x-3
(c) x+5
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If d(x)=x-k, where f(x)=(x-k)q(x) + r
Then we can evaluate the polynomial f(x) at x=k:
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F(x)=2x2-3x+1; k=2
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F(x)=2x3+3x2+4x-7; k=2
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F(x)=x3-x2+2x-1; k=-3
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Now we can use this method to find both
remainders and quotients for division by x-k,
called synthetic division.
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(2x3-3x2-5x-12)/(x-3)
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K becomes zero of divisor
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3 | 2 -3 -5 -12
_____________
* Since the leading coefficient of
the dividend must be the leading
coefficient , copy the first “2” into
the first quotient position.
* Multiply the zero of the divisor
(3) by the most recent coefficient
of the quotient (2). Write the
product above the line under the
next term (-3).
* Add the next coefficient of the
dividend to the product just found
and record sum below the line in
the same column.
* Repeat the “multiply” and “add”
steps until the last row is
completed.
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(x3-5x2+3x-2)/(x+1)

(9x3+7x2-3x)/(x-10)
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(5x4-3x+1)/(4-x)
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Suppose f is a polynomial function of degree
n1 of the form f(x)=anxn+…+a0
with every coefficient an integer. If x=p/q is
a rational zero of f, where p and q have no
common integer factors other than 1, then
◦ P is an integer factor of the constant coefficient a0,
and
◦ Q is an integer factor of the leading coefficient an.
Example: Find rational zeros of f(x)=x3-3x2+1
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F(x)=3x3+4x2-5x-2
Potential Rational Zeros:
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F(x)=6x3-5x-1
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F(x)=2x3-x2-9x+9
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Let f be a polynomial function of degree n≥1
with a positive leading coefficient. Suppose
f(x) is divided by x-k using synthetic division.
If k≥0 and every number in the last line is
nonnegative (positive or zero), then k is an
upper bound for the real zeros of f.
If k≤0 and the numbers in the last line are
alternately nonnegative and nonpositive, then
k is a lower bound for the real zeros of f.

Lets establish that all the real zeros of
f(x)=2x4-7x3-8x2+14x+8 must lie in the
interval [-2,5]
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Establish bounds for real zeros
Find the real zeros of a polynomial functions
by using the rational zeros theorem to find
potential rational zeros
Use synthetic division to see which potential
rational zeros are a real zero
Complete the factoring of f(x) by using
synthetic division again or factor.
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F(x)=10x5-3x2+x-6
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F(x)=2x3-3x2-4x+6
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F(x)=x3+x2-8x-6
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F(x)=x4-3x3-6x2+6x+8
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F(x)=2x4-7x3-2x2-7x-4
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In the 17th century, mathematicians extended
the definition of √(a) to include negative real
numbers a.
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i =√(-1) is defined as a solution of (i )2 +1=0
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For any negative real number √(a) = √|a|*i
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a +bi , where a, b are real numbers
◦ a+bi is in standard form
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Sum: (a+bi) +(c+di) = (a+c) + (b+d)i
Difference: (a+bi) – (c+di) = (a-c) + (b-d)I
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EX: (a) (8 - 2i) + (5 + 4i)
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(b) (4 – i) – (5 + 2i)
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(2+4i)(5-i)
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Z=(1/2)+(√3/2)i, find Z2
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Z = a+bi = a – bi
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When do we need to use conjugates?
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(2+3i)/(1-5i)
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ax2+bx+c=0
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Try:
x2-5x+11=0
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Find all zeros:
f(x) = x4 + x3 + x2 + 3x - 6
HW: Pg. 234-235 #2-10e, 28-34e

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Fundamental Theorem of Algebra – A polynomial
function of degree n has n complex zeros (real
and nonreal).
Linear Factorization Theorem – If f(x) is a
polynomial function of degree n>0, then f(x) has
n linear factors and
F(x) = a(x-z1)(x-z2)…(x-zn)
Where a is the leading coefficient of f(x) and z1, z2, …, zn
are the complex zeros of the function.
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X=k is a…
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K is a
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Factor of f(x):
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F(x)=(x-2i)(x+2i)
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F(x)=(x-3)(x-3)(x-i)(x+i)
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Suppose that f(x) is a polynomial function
with real coefficients. If a+bi is a zero of f(x),
then the complex conjugate a-bi is also a
zero of f(x)
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1.
2.
3.
What can happen if the coefficients are not real?
Use substitution to verify that x=2i and x=-i are
zeros of f(x)=x2-ix+2. Are the conjugates of 2i and –i
also zeros of f(x)?
Use substitution to verify that x=i and x=1-i are
zeros of g(x)=x2-x+(1+i). Are the conjugates of i and
1-i also zeros of g(x)?
What conclusions can you draw from parts 1 and 2?
Do your results contradict the theorem about complex
conjugates?
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Given that -3, 4, and 2-i are zeros, find the
polynomial:
Given 1, 1+2i, 1-i, find the polynomial:
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The complex number z=1-2i is a zero of
f(x)=4x4+17x2+14x+65, find the remaining
zeros, and write it in its linear factorization.
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3x5-2x4+6x3-4x2-24x+16
HW: Pg. 246 #19-30
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F and g are polynomial functions with g(x)≠0.
the functions:
◦ R(x)=f(x)/g(x) is a rational function
◦ Find the domain of : f(x)=1/(x+2)
(a)
g(x)=2/(x+3)
(b)
H(x)=(3x-7)/(x-2)
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Find horizontal and vertical asymptotes of
f(x)=(x2+2)/(x2+1)
Find asymptotes and intercepts of the
function f(x)=x3/(x2-9)
HW: Pg. 254 #7-17
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X + 3/x = 4
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2/(x-1) + x = 5
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(2x)/(x-1) + 1/(x-3) = 2/(x2-4x+3)
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3x/(x+2) + 2/(x-1) = 5/(x2+x-2)
(x-3)/x + 3/(x+2) + 6/(x2 +2x) = 0
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Find the dimensions of the rectangle with
minimum perimeter if its area is 200 square
meters. Find the least perimeter:
A = 200
HW: Finish 2.9 WKSH
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F(x)=(x+3)(x2+1)(x-4)2
Determine the real number values of x that
cause the function to be zero, positive, or
negative:
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(x+3)(x2+1)(x-4)2 > 0
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(x+3)(x2+1)(x-4)2 ≥ 0

(x+3)(x2+1)(x-4)2 < 0
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(x+3)(x2+1)(x-4)2 ≤ 0
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2x3-7x2-10x+24>0
Solve Graphically: x3-6x2≤2-8x
*Plug function into your calculator*
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Section 2.9 #1-12 odd

Check yourself

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