Ch 3 review Quarter test 1 And Ch 3 TEST. Graphs of Quadratic Functions Where a, b, and c are real...
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Transcript of Ch 3 review Quarter test 1 And Ch 3 TEST. Graphs of Quadratic Functions Where a, b, and c are real...
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
EXAMPLE: Find the x and y intercepts of the rational functions.
b. No common factors…in lowest terms.
y-intercept: x-intercept:
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!
EXAMPLE: Find the x and y intercepts of the rational functions.
c. Cannot be factored…in lowest terms.
y-intercept: x-intercept:
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
Graphing Rational Functions Using Transformations
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
Graphing Rational Functions Using Transformations
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...
Graphing Rational Functions Using Transformations
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
Next, find the x & y-intercepts:
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
Graphing Rational Functions Using Transformations
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
Next, find the x & y-intercepts:
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
Three Types of Rational Functions
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
Three Types of Rational Functions
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
EXAMPLE: For each function, find the domain, x and y-intercepts, vertical asymptotes, horizontal asymptotes, slant asymptotes.
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
EXAMPLE: For each function, find the domain, x and y-intercepts, vertical asymptotes, horizontal asymptotes, slant asymptotes.
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
EXAMPLE: Factor and find the zeros of the polynomial function.
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
EXAMPLE: Factor and find the zeros of the polynomial function.
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
EXAMPLE: Factor and find the zeros of the polynomial function.
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
EXAMPLE: Factor and find the zeros of the polynomial function.
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…
EXAMPLE: Factor and find the zeros of the polynomial function.
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)
EXAMPLE: Factor and find the zeros of the polynomial function.
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.
EXAMPLE: Factor and find the zeros of the polynomial function.
c. 18452553)( 234 xxxxxf
)9)(13)(2()( 2 xxxxf
iix 3,3,31,2