Ch 3 review Quarter test 1 And Ch 3 TEST. Graphs of Quadratic Functions Where a, b, and c are real numbers and a 0 Standard Form Domain: all real numbers

Ch 3 reviewQuarter test 1

And Ch 3 TEST

Graphs of Quadratic Functions

cbxaxxf 2)(Where a, b, and c are real numbers and a 0

Standard Form

•Domain: all real numbers

•Range: depends on the minimum and maximum

•The graph is a parabola.

if 0a

The graph of x2 is shifted “h” units horizontally and “k” units vertically.

opens:

axis of symmetry:

vertex:

k is the

range:

V(h, k) / minimum

x = h

up

x = h

(h, k)

minimum

ky

positive

if

The graph of x2 is shifted “h” units horizontally, “k” units vertically, and reflected over x-axis.

opens:

axis of symmetry:

vertex:

k is the

range:

V(h, k) / maximum

x = h

down

x = h

(h, k)

maximum

negative0a

ky

cbxaxxf 2)(Standard form: Vertex form: khxaxf 2)()(

can “see” the transformations…

The vertex form is easier to graph…to change from standard form to vertex form, either complete the square (YUCK!) or memorize this formula:

a

bh

2

and

a

bfk

2 h

Therefore, the vertex is at

a

bf

a

b

2,

2

and the axis of symmetry is . a

bx

2

k = f(h)

Example: Let . Find the vertex, axis of symmetry, the minimum or maximum value, and the intercepts. Use these to graph f(x). State the domain and range and give the intervals of increase and decrease. Then write the equation in vertex form and list the transformations that were made to the parent function, f(x) = x2.

163)( 2 xxxf

1st identify a, b, and c: a = 3 b = 6 c = 1

Next find h and k: 16

6

)3(2

6

2

a

bh

2)1()( fhfk

So, the vertex is (-1, -2) and the axis of symmetry

is x = -1. Since a > 0, then the graph opens up

and has a minimum value at -2.

To find y-intercepts evaluate f(0): 1)0( f

To find x-intercepts (roots/zeros) use the quadratic formula:

a

acbbx

2

42

)3(2

)1)(3(4)6()6( 2

6

246

-0.184and

-1.816

The intercepts are at (0, 1), (-0.184, 0), and

(-1.816, 0).

To graph, plot the vertex, intercepts, utilize the axis of symmetry.

V(-1, -2) y-int: (0,1) axis of sym: x = -1

So, to be symmetrical, another point will be at (-2, 1).

Check using your graphing calculator!

Domain: all real numbers

Range:

2y

Decreasing:

Increasing: ),1( )1,(

Vertex form: f(x) = a(x - h)2 + k

a = 3 h = -1 k = -2

2)1(3)( 2 xxf

It is the graph of x2 shifted left 1, vertically stretched by

3 and shifted down 2.

Example: Find the standard form equation of the quadratic function whose vertex is (1, -5) and whose y-intercept is -3.

h k (0, -3)

Vertex form: f(x) = a(x - h)2 + k

Fill in the information that was given:

)5()10(3 2 a Solve for a…

5)1(3 2 a53 a

2a

Write the equation in vertex form then simplify to standard form:

5)1(2)( 2 xxf

5)12(2)( 2 xxxf

5242)( 2 xxxf

342)( 2 xxxf

Power Functions

naxxf )(The polynomial that the graph resembles (the end behavior model)…

EX: The power function of the polynomial is…

33xy 153)( 23 xxxxf

22 )5(2)( xxxg 42xy

)35(2)( 2 xxxf 36xy

Properties of Power Functions with Even Degrees

1. f is an even function

a. The graph is symmetric with respect to the y-axis

b. f(-x) = f(x)

2. Domain: all real numbers

3. The graph always contains the points (0, 0), (1, 1) and (-1, 1)

4. As the exponent increases in magnitude, the graph becomes more vertical when x < -1 or x > 1; but for x near the origin, the graph tends to flatten out and be closer to the x-axis.

The graph always contains the points (0, 0), (1, 1) and (-1, 1)*

*Points used to make transformations

EX: Graph y = x4, y = x8 and y = x12 all on the same screen.

Let and be your viewing window.

What do you notice?

11 x 10 y

Properties of Power Functions with Odd Degrees

1. f is an odd function

a. The graph is symmetric with respect to the origin

b. f(-x) = -f(x)

2. Domain: all real numbers

Range: all real numbers

3. The graph always contains the points (0, 0), (1, 1) and (-1, -1)

4. As the exponent increases in magnitude, the graph becomes more vertical when x < -1 or x > 1; but for x near the origin, the graph tends to flatten out and be closer to the x-axis.

The graph always contains the points (0, 0), (1, 1) and (-1, -1)*

*Points used to make transformations

EX: Graph y = x3, y = x7 and y = x11 all on the same screen.

Let and be your viewing window.

What do you notice?

11 x 11 y

Graphs of polynomial functions are smooth (no sharp corners or cusps) and continuous (no gaps or holes…it can be

drawn without lifting your pencil)…

Is a polynomial graph Is not a polynomial graph

We can apply what we learned about transformations in Chapter 2

and what we just learned about power functions to graph

polynomials…

EXAMPLE: Graph f(x) = 1 – (x – 2)4 using transformations.

Step 1: y = x4 Step 2: y = (x – 2)4

Step 3: y = - (x – 2)4 Step 4: y = 1 – (x – 2)4

Start with (0, 0), (1, 1) & (-1, 1)

Shift right 2 units

Reflect over x-axis

Shift up 1 unit

EXAMPLE: Graph f(x) = 2(x + 1)5 using transformations. Check your work with your graphing calculator.

x5…(0, 0), (1, 1) & (-1, -1)

(x + 1)5…shift left 1 unit

2(x + 1)5…vertical stretch by factor 2

multiply the y-values by 2

Zeros and the Equation of a Polynomial Function

If f is a polynomial function and r is a real number for which f(r) = 0, then r is called a real zero of f, or a root of f. If r is a real zero/root of f then:

a. r is an x-intercept of the graph of f, and

b. (x – r) is a factor of f

In other words…if you know a zero/root, then you know a factor…if you know a factor, then you know a zero/root

EX: If (x – 4) is a factor, then 4 is a zero/root…

If -3 is a zero/root, then (x + 3) is a factor…

EXAMPLE: Find a polynomial of degree 3 with zeros -4, 1, and 3. (Let a = 1)

If x = -4, then the factor that solves to that is…

)4())4(( xx

If x = 1, then the factor that solves to that is… )1( x

If x = 3, then the factor that solves to that is… )3( x

)3)(1)(4()( xxxxf

)3)(1)(4()( xxxxf

)3)(43()( 2 xxxxf

)(xf 3x 23x 23x x9 x4 12

1213)( 3 xxxf

Rational Function

A function of the form , where p and q are

polynomial functions and q is not the zero

polynomial.

The domain is the set of all real numbers EXCEPT those for which the denominator q is zero.

)(

)(

xq

xp

2

1)(

x

xxfEX:

* Enter in calculator as (x + 1)/(x – 2)...MUST put parentheses!

*

Domain and Vertical Asymptotes

To find the domain of a rational function, find the zeros of the denominator…this is where the denominator would be zero…this is where x cannot exist.

The vertical asymptote(s) of a rational function are where x cannot exist…it is the virtual boundary line on the graph. Vertical asymptotes are defined by the equation ‘x =‘

How the graph reacts on either side of a vertical asymptote:

Goes in opposite directions as it approaches the asymptote:

Goes in the same direction as it approaches the asymptote:

THE GRAPH WILL NEVER CROSS THE VERTICAL ASYMPTOTE!!!

EXAMPLE: Find the domain and vertical asymptotes of the rational functions.

a.5

42)(

2

x

xxf

The graph will not exist where the denominator equals zero!

x + 5 = 0

x = -5 When x = -5, the graph will not exist!

5xThe domain is and the VA is 5x5xThe domain is and the VA is 5x5xThe domain is and the VA is 5x

EXAMPLE: Find the domain and vertical asymptotes of the rational functions.

2x

b.4

1)(

2

xxf c.

1)(

2

3

x

xxf

x2 – 4 = 0

(x + 2)(x – 2) = 0

x = -2 x = 2

Domain:

VA: 2,2 xx

x2 + 1 = 0

x2 = -1

x = not real

Domain: all real #’s

VA: none

Intercepts on the x and y axes

To find the y-intercepts of a rational function, that is where x = 0, evaluate f(0).

To find the x-intercepts of a rational function, first make sure the function is in lowest terms, that is the numerator and denominator have no common factors. Then, find the zeros of the numerator.

factor top & bottom first!!

The zeros of the numerator are the x-intercepts (zeros) of the rational function.

EXAMPLE: Find the x and y intercepts of the rational functions.

a.5

42)(

2

x

xxf

5

)2(2 2

x

xNo common factors…in

lowest terms.

y-intercept: x-intercept:

5

4

50

4)0(2)0(

2

f042 2 x

42 2 x22 x2x

The y-intercept is at and the x-intercepts

are at and 5

42 2


b. No common factors…in lowest terms.


The y-intercept is at and there are no x-intercepts 41

4

1)(

2

xxf

)2)(2(

1

xx

4

1

4)0(

1)0(

2

f none…the numerator has

no x in it to solve for!


c. Cannot be factored…in lowest terms.


01

0

10

)0()0(

2

3

f03 x0x

The y-intercept is at 0 and the x-intercept is at 0.

1)(

2

3

x

xxf

Graphing Rational Functions Using Transformations

a. Analyze the graph of 21)(

xxR

1st find the domain and any vertical asymptotes:

x2 = 0

x = 0

Domain: 0xVA: x = 0

Next, find the x & y-intercepts:

x-intercepts: none

y-intercepts: none

undefinedf 0

1

)0(

1)0(

2


a. Analyze the graph of 21)(

xxR

Is it even?

To graph without using a calculator, identify a few points on the graph by plugging in x-values:

)(1

)(

1)(

22xR

xxxR

It is an even function, so it is symmetric to the y-axis.

f(1) = 1 f(-1) = 1

f(2) = ¼ f(-2) = ¼


b. Use transformations to graph

Check the domain and any vertical asymptotes:

(x – 2)2 = 0

x – 2 =0

x = 2

Domain: 2xVA: x = 2

Next, check the y-intercept:

y-intercepts: 1.25 25.114

11

)20(

1)0(

2

f

1)2(

1)(

2

xxR

It is the graph of shifted right 2 and up 1.21

x

Shift the VA and points right 2 and up 1...


c. Analyze the graph of and use it to graph

Find the domain and any vertical asymptotes:

x - 2 = 0

x = 2

Domain: 2x

VA: x = 2


x-intercepts: 3

)2(

11)(

xxfxxg 1)(

It is the graph of shifted right 2, reflected over the x-axis, and shifted up 1.

x1

2

1

2

2

2

11

xx

x

x 2

12

x

x2

3

x

x x – 3 = 0

x = 3


c. Analyze the graph of and use it to graph

Find the domain and any vertical asymptotes:

x - 2 = 0

x = 2

Domain: 2x

VA: x = 2


x-intercepts: 3

y-intercepts: 1.52

3

2

11

)20(

11)0(

f

)2(

11)(

xxfxxg 1)(

It is the graph of shifted right 2, reflected over the x-axis, and shifted up 1.

x1

Properties of Rational Functions

Holes (Points of Discontinuity)

x-values for a rational function that cannot exist, BUT are not asymptotes. These

occur whenever the numerator and denominator have a common factor.

Must factor both top and bottom first!!

EXAMPLE: Find the domain and vertical asymptote(s).

34

1)(

2

xx

xxf

4

2)(

2

x

xxf

)3)(1(

1

xx

x a hole occurs at x + 1 = 0

Domain:

VA: x = -3

A hole occurs at x = -1

3,1 x

34

1)(

2

xx

xxf

EXAMPLE: Find the domain and vertical asymptote(s).

34

1)(

2

xx

xxf

4

2)(

2

x

xxf

)3)(1(

1

xx

x)2)(2(

2

xx

x

Domain:

VA: x = -3

A hole occurs at x = -1

a hole occurs at x - 2 = 0

A hole occurs at x = 2

Domain:

VA: x = -2

3,1 x 2x

Horizontal Asymptotesdescribe a certain behavior of the graph as

or as , that is its end behavior.

How the graph behaves on the far ends of the x-axis.

x x

The graph of a function may intersect a horizontal asymptote.

The graph of a function will never intersect a vertical asymptote.

Always written as y =

Three Types of Rational Functions

Balanced…the degree of the numerator and denominator are equal

4

132)(

2

2

x

xxxf Horizontal Asymptote: b

ay

2

22

x

xH.A.

The horizontal asymptote is where y = 2.

2


Bottom Heavy…the degree of the denominator is larger than the degree of the numerator.

Horizontal Asymptote: 0y

The horizontal asymptote is where y = 0.

4

3)(

2

x

xxf


Top Heavy…the degree of the numerator is larger than the degree of the denominator

Has NO HORIZONTAL ASYMPTOTE

There is no horizontal asymptote.

3

2)(

2

x

xxf

has an oblique asymptote instead…

Oblique (Slant) Asymptotean asymptote that is neither vertical nor

horizontal, but also describes the end behavior of a graph. Has the equation “y =“ and has an x in it. It is found by dividing the polynomial: top bottom (quotient only)

Top Heavy rational functions have oblique or

slant asymptotes instead of a horizontal asymptote.

EXAMPLE: Find the oblique asymptote of

1)(

2

4

x

xxxf

Note: The textbook considers only linear equations oblique asymptotes…

divide the polynomials using long division…

00010 2342 xxxxxx

2x

4x 30x 2x2x

1

x 02x x0 1

x 1 ignore remainder

The oblique asymptote is y = x2 + 1.

EXAMPLE: For each function, find the domain, x and y-intercepts, vertical asymptotes, horizontal asymptotes, and slant asymptotes.

a.4

132)(

2

2

x

xxxf

)2)(2(

)1)(12(

xx

xx

D: ___________________

x-int: _________________

y-int: _________________

VA: _________________

HA: _________________

Now find domain and vertical asymptotes

x + 2 = 0 x – 2 = 0

x = -2 x = 2

x = 2, x = -2

2x

Balanced

1st find the horizontal asymptote

22

2

2

x

xy

y = 2

EXAMPLE: For each function, find the domain, x and y-intercepts, vertical asymptotes, horizontal asymptotes, and slant asymptotes.

a.4

132)(

2

2

x

xxxf

)2)(2(

)1)(12(

xx

xx

D: ___________________

x-int: _________________

y-int: _________________

VA: _________________

HA: _________________

x = 2, x = -2

2x

y = 2

Find the x-intercepts:

Find the y-intercepts:

4

1

4)0(

1)0(3)0(2)0(

2

2

f

41

2x + 1 = 0 x + 1 = 0

x = -1/2 x = -1 -1/2, -1

EXAMPLE: For each function, find the domain, x and y-intercepts, vertical asymptotes, horizontal asymptotes, slant asymptotes.

b.

D: ___________________

x-int: _________________

y-int: _________________

VA: _________________

HA: _________________

Find domain and vertical asymptotes

x + 1 = 0 x – 1 = 0

x = -1 x = 1

x = 1, x = -1

1x

Bottom-heavy

y = 0

1

3)(

2

x

xxf

)1)(1(

3

xx

x

Find the x-intercepts:

Find the y-intercepts:

31

3

10

30)0(

2

f

x - 3 = 0

x = 3

3

3


c.D: ___________________

x-int: _________________

y-int: _________________

VA: _________________

HA: _________________None

Find the oblique asymptote:

3

2)(

2

x

xxf Top-heavy

oblique asymptote: __________

0023 2 xxx

3 2 0 0

2

66

18

18y = 2x - 6

y = 2x - 6


c.D: ___________________

x-int: _________________

y-int: _________________

VA: _________________

HA: _________________

x = -3

3x

None

Find the x-intercepts: Find the y-intercepts:

030

)0(2)0(

2

f2x2 = 0

x = 0

03

2)(

2

x

xxf

oblique asymptote: __________

Find domain and vertical asymptotes

x + 3 = 0

x = -3

0

y = 2x - 6

Real and Non-real Zeros of Polynomial Functions

The zeros of a polynomial function can be found by finding its factors.

The real zeros (roots) are the x-values where the graph crosses the x-axis. In this section, you will be finding both real and non-real

(imaginary) roots.

Can “SEE” Real roots x-intercepts

Cannot “see” imaginary roots...must use algebraic

method, such as the quadratic formula, to find them

NON-REAL

REAL

Remainder and Factor Theorems

Recall: Division Algorithm for Polynomials

)(

)()(

)(

)(

xg

xrxq

xg

xf

dividend

divisor

quotient

remainder

If the remainder is zero (0) then, g(x) divides evenly into f(x) and )()()( xqxgxf

EX: 12/4 = 3 with remainder 0, so 4 x 3 = 12

Remainder Theorem

Let f be a polynomial function. If f(x) is divided by x – c, then the remainder is f(c). f(c) = the remainder!

EXAMPLE: Find the remainder if f(x) = x3 – 4x2 – 5 is divided by x – 3.

THIS CAN BE DONE IN ONE OF 3 WAYS!!

Using Synthetic Division: 3 1 4 0 5

1

3

1

3

39

14 Remainder

Using the Remainder Theorem: (3)3 – 4(3)2 – 5 = -14f(3) =

Using Graphing: Let y1 = x3 – 4x2 - 5

When x = 3, y = -14 Look at table

The remainder

is -14.

Factor Theorem

Let f be a polynomial function. Then x – c is a factor of f(x) iff f(c) = 0.

1. If f(c) = 0, then x – c is a factor of f(x).

2. If x – c is a factor of f(x), then f(c) = 0.

Basically, if the remainder is zero, you have a factor and a zero/root…and if you have a factor or zero/root, the remainder will be zero…

EXAMPLE: Determine whether x – 1 is a factor of f(x) = 2x3 – x2 + 2x – 3. If so, then factor f(x).

1st check for a remainder of 0… 2(1)3 – (1)2 + 2(1) – 3 = 0f(1) =

Since the remainder is 0, then (x – 1) is a factor of f(x).

Now factor f(x) using synthetic division…

1 2 1 2 3

2

2

1

1

3

3

02x2 + x + 3

cannot be factored any further...

)32)(1()( 2 xxxxf

Complex Zeros (Roots) of a Polynomial Function

Fundamental Theorem of Algebra

Every polynomial function f(x) of degree n has exactly n numbers of real + imaginary zeros...that is, there are exactly n complex zeros.

Furthermore, a polynomial of odd degree has at least one real zero. WHY?? Goes in opposite directions, so it

MUST go through the x-axis!

Complex Roots (Conjugate Pairs) Theorem

Let f be a polynomial function. If a + bi is a complex zero of f, then a – bi is also a zero of the function.

Irrational AND Imaginary zeros must come in pairs!

3 are listed, so there are 2 more...3 + 2 = degree 5

imaginary and irrational zeros must come in conjugate pairs...

Steps for finding zeros (roots) of polynomial functions:

1. Determine the number of real and non-real roots the function will have by graphing.

2. Find the real zeros (x-intercepts) on your graph. If no real zeroes, then polynomial WILL BE FACTORABLE.

3. Factor the function using synthetic division. Continue to factor until you get a quadratic factor.

4. Solve each of the factors for the roots. Answer in exact form (not decimals).

where it crosses x-axis

real

TOTAL ZEROS = DEGREE

f(c) = 0

or get a polynomial that is factorable...

exact form

First check to see if the polynomial can be factored by “normal” means!!

EXAMPLE: Factor and find the zeros of the polynomial function.

a. 67112)( 23 xxxxf

Step 1: Graph and find # of real & non-real zeros

crosses the x-axis 3 times, so there are 3 real and 0 non-real


a. 67112)( 23 xxxxf

Step 2: Find the real zeros on your graph.

Is 1 a zero? Does f(1) = 0?

Is -6 a zero? Does f(-6) = 0? Yes

All three are real and can be found using your calculator!

Yes

Is -1/2 a zero? Does f(-1/2) = 0? Yes

So, f(x) factored is (x – 1)(2x + 1)(x + 6) and the zeros are x = -6, -1/2, 1


b. 483284)( 2345 xxxxxxf

crosses the x-axis once and touches once...

What are the possibilities?

Odd multiplicity Even multiplicity

3 real: multiplicity of 1 and multiplicity of 2 + 2 non-real


According to the graph, the real zeros are at x = -3 and x = 2

We must use synthetic division since we cannot factor by grouping. Choose one of the real zeros to use for the division.

Let’s start with x = -3…

b. 483284)( 2345 xxxxxxf


b. 483284)( 2345 xxxxxxf

If x = -3, then... 3 1 1 4 8

1

34

12

8

2416

161684 234 xxxx

32 4848

16

48

0

Not a quadratic and not factorable by grouping, so divide/factor again using synthetic division and another real zero…


b. 483284)( 2345 xxxxxxf

2

1

2

2 4

14

4 8 16 842 23 xxx16816

8 0

Use x = 2 to factor x4 – 4x3 + 8x2 – 16x + 16 further…

)2(2 xx )2(4 x

)4)(2( 2 xx

So, f(x) factored is (x + 3)(x - 2)(x - 2)(x2 + 4) =

(x + 3)(x – 2)2(x2 + 4)


b. 483284)( 2345 xxxxxxf

)4()2)(3()( 22 xxxxf

3x 2xmult. 2

042 x

42 x

ix 24

The zeros are -3, 2 (mult 2), 2i and -2i.


c. 18452553)( 234 xxxxxf

)9)(13)(2()( 2 xxxxf

iix 3,3,31,2

Documents

Ch 3 review Quarter test 1 And Ch 3 TEST. Graphs of Quadratic Functions Where a, b, and c are real numbers and a 0 Standard Form Domain: all real numbers