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