Pre-Calculus
10/19/2006
polynomial function
degree n leadcoefficient
1
a
zero function f(x) = 0 undefined
constant function f(x) = 5 0
linear function f(x) = 2x + 5
quadratic function f(x) = x2 + 2x + 5 2
slope
linear constant non-zero
Pre-Calculus
10/19/2006
roots or solutions
x = -1 or 3.5
2y a(x h) k vertex: (h, k)
complete the square
vertex: (–4, –1)
axis of symmetry: x = –4
Pre-Calculus
10/19/2006
vertex: (1, 5)
vertex:x – intercepts:
Pre-Calculus
10/19/2006
constantspower constant of variation or proportion
varies as is proportional to
power functionpower: –4
not a power function: power isn’t a constantconstant of variation: 2
power function
power is 1, constant of variation is 2
independent variable: r
power is 2, constant of variation is 1
power: 2
direct variation
constant of variation:
g af(x) k x
f(x) x 3f(x) x 1
f(x)x
2f(x) x f(x) x
Pre-Calculus
10/19/2006
d = k F
d = k t 2
monomialdegree: 0lead coefficient: 4
not monomial
lead coefficient: 13
power is ½ (not an integer)
not monomial
monomialdegree: 3
non-negative
integer
power is a variable
Pre-Calculus
10/19/2006
vertical stretch / shrink
vertical stretch / shrink
reflection across the x-axis
domainrange
continuityincreasingdecreasingsymmetry
boundednessextrema
asymptotesend behavior
Pre-Calculus
10/19/2006
dividend divisor
quotient remainder
232 xx xx 52 2 xx 22
74 x24 x
5
2x x 2
Pre-Calculus
10/19/2006
k = 3(3)2 – 4(3) – 5 = 9 – 12 – 5 = –8
k = –2(–2)2 – 4(–2) – 5 = 4 + 8 – 5 = 7
k = 5(5)2 – 4(5) – 5 = 25 – 20 – 5 = 0
divides evenly
x - intercept zero
solution root
Pre-Calculus
10/19/2006
3(x + 4)(x – 3)(x + 1)
so factors are: x + 4, x – 3, x + 1
= 3x3 + 6x2 – 33x – 36
2(x + 3)(x + 2)(x – 5)
so factors are: x + 3, x + 2, x – 5
= 2x3 – 38x – 60
Pre-Calculus
10/19/2006
(x + 4)(x – 4) = 0
f(x) = x2 – 16
x = 4, x = –4
(x 3)(x 3) 0
2f(x) x 3
x 3,x 3
rational zeros
Pre-Calculus
10/19/2006
potential:
Use the rational zeros theorem to find the rational zeros of f(x) = 2x3 + 3x2 – 8x + 3
Use the rational zeros theorem to find the rational zeros of f(x) = 2x3 + 3x2 – 8x + 3
p = integer factors of the constant q = integer factors of the lead coefficient
p
q
1, 3
1, 2
1 3
1, 3, ,2 2
Pre-Calculus
10/19/2006
Pre-Calculus
10/19/2006
complex (real and non-real) zeros
* non-real zeros are not x – intercepts
zeros: 3i, – 3i, – 5
x-intercepts: – 5
5 i 23
4 4complex conjugate(a + bi and a – bi)
Pre-Calculus
10/19/2006
x4 – 14x3 + 78x2 – 206x + 221
Pre-Calculus
10/19/2006
denominator
the x – axis ( y = 0 )the line y = an / bm
there is no
quotient
output
input
Pre-Calculus
10/19/2006
vertical asymptote:
horizontal asymptote:
x – intercept
y – intercept
vertical asymptote:
horizontal asymptote:
x – intercept
y – intercept
none
y = 0
none
(0, 4)
x = –1
none
(0, 0) (1, 0)
(0, 0)slant asymptote: y = x – 2
Pre-Calculus
10/19/2006
(–3, 4) U (4, )
because the graph crosses the x-axis
because the graph does not cross the x-axis
[ –3, )
(– , –3)
(– , –3)
Pre-Calculus
10/19/2006
1, –3, 2
(– , –3) U (1, 2) U (2, )
(–3, 1)
–3 1 2
+++ –
Pre-Calculus
10/19/2006
Write a standard form polynomial function of degree 4 whose zeros include 1 + 2i and 3 – i.
Write a standard form polynomial function of degree 4 whose zeros include 1 + 2i and 3 – i.
4 3 2x 8x 27x 50x 50
quiz
Pre-Calculus
10/19/2006
Solve the following inequality using a sign chart:x3 + 2x2 – 11x – 12 < 0
Solve the following inequality using a sign chart:x3 + 2x2 – 11x – 12 < 0
( , 4 ] U [ 1, 3 ]
quiz
Pre-Calculus
10/19/2006
Write the following polynomial function in standard form. Then identify the zeros and the x – intercepts.
f(x) = (x – 3i) (x + 3i) (x + 4)
Write the following polynomial function in standard form. Then identify the zeros and the x – intercepts.
f(x) = (x – 3i) (x + 3i) (x + 4)
3i, 3i, 4
4zeros:
x – intercepts:
quiz
Pre-Calculus
10/19/2006
Without graphing, using a sign chart, find the values of x that cause f(x) = (x – 2) (x + 6) (x + 1) to be:
a.) zero ( f(x) = 0 )b.) positive ( f(x) > 0 )c.) negative ( f(x) < 0 )
Without graphing, using a sign chart, find the values of x that cause f(x) = (x – 2) (x + 6) (x + 1) to be:
a.) zero ( f(x) = 0 )b.) positive ( f(x) > 0 )c.) negative ( f(x) < 0 )
a.) 2, –1 , –6
b.) (–6, –1) U (2, )
c.) (–, –6) U (-1, 2)
quiz
Pre-Calculus
10/19/2006
Use the quadratic equation to find the zeros of f(x) = 5x2 – 2x + 5.
Your answer must be in exact simplified form.
Use the quadratic equation to find the zeros of f(x) = 5x2 – 2x + 5.
Your answer must be in exact simplified form.
1 2i 6
5
quiz
Pre-Calculus
10/19/2006
Find all zeros of f(x) = x4 + 3x3 – 5x2 – 21x + 22and write f(x) in its linear factorization form
Find all zeros of f(x) = x4 + 3x3 – 5x2 – 21x + 22and write f(x) in its linear factorization form
(x 1)(x 2)(x ( 3 i 2))(x ( 3 i 2))
Pre-Calculus
10/19/2006
2i is a zero of f(x) = 2x4 – x3 + 7x2 – 4x – 4. Find all remaining zeros and write f(x) in its linear factorization form.
2i is a zero of f(x) = 2x4 – x3 + 7x2 – 4x – 4. Find all remaining zeros and write f(x) in its linear factorization form.
(x 1)(2x 1)(x 2i)(x 2i)
quiz