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3.1
3.2
3.3
3.4
3.5
3.6
3.1 Quadratic Functions and Models
Quadratic Functions ▪ Graphing Techniques ▪ Completing the
Square ▪ The Vertex Formula ▪ Quadratic Models and Curve
Fitting
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
3-1
3.1
Example 1(a) Graphing Quadratic Functions (page 304)
Graph f(x) = x2 + 4x – 4 by plotting points.
Give the domain and range.
X-coordinate
of vertex
-b/(2a)
Vertex:
X = -4/2 = -2
X
Y
-3
-2
-1
-7
-8
-7
Y = (-2)2 + 4(-2) – 4
(-2,-8)
Domain:
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Range:
3-2
3.1
add on: Quadratic graph w/ fraction value for a
Graph f(x) = 2/3 (x – 5)2 + 4 by plotting points.
Give the domain and range.
Vertex:
X
Y
(5,4)
2
5
8
10
4
10
Since the value of a = 2/3 choose x-values
around the vertex which are 3 units larger
and smaller than the x-coor of the vertex
Domain:
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Range: [4, ∞ )
3-3
Add on: Find the equation from the graph
Given the graph of a parabola with a vertex at (1,3)
passing through the point (-1,-2). Find the
equation of the parabola in vertex form:
y = a(x – h)2 + k
h=1
k=3
y = a (x – 1)2 + 3
Substitute (-1,-2) for (x,y) solve for a
-2 = a (-1 – 1)2 + 3
-5 = a (4)
- 5/4 = a
y = -5/4 (x – 1)2 + 3
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3-4
Add on
Find equation of parabola with x-intercepts of 2 and 5 and a
y-intercept of 4. Simplify your answer.
Roots:
Factors:
Equation:
Substitute
x = 2 and x = 5
(x – 2)(x – 5)
y = a(x – 2)(x – 5)
(0,4) for the x and y and solve for a
4 = a(0 – 2)(0 – 5)
4 = a(10)
2/ = a
5
Equation:
Answer:
y = 2/5 (x – 2)(x – 5)
y = 2/5 x2 – 2/5 x + 4
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
3-5
3.1
Example 2 Graphing a Parabola by Completing the
Square (page 305)
Graph f(x) = x2 + 2x – 5 by completing the square
and locating the vertex.
X
Y
I prefer: find vertex 1st
X coor = –2/2 = -1
Y coor = (-1)2 + 2(-1) – 5 = -6
Vertex (–1, –6)
Axis of Sym: x = –1
-2
-1
0
-5
-6
-5
3.1
Example 3 Graphing a Parabola by Completing the
Square (page 305)
Graph f(x) = 2x2 + x – 6 by completing the square
X
and locating the vertex.
I prefer: find vertex 1st
X coor = –1/4
Y coor = 2( –1/4)2 + (–1/4) – 6 = –49/8
-1
- 1/ 4
0
Y
-3
- 49/8
-6
Vertex is (–1/4, –49/8)
Axis of sym: X = –1/4
Their way:
3-7
3.1
Example 4 Finding the Axis and the Vertex of a Parabola
Using the Vertex Formula (page 308)
Find the axis and vertex of the parabola
f(x) = –3x2 + 12x – 8 using the vertex formula.
a = –3, b = 12, c = –8
Axis: x = 2
Vertex: (2, f(2))
Vertex: (2, 4)
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3-8
3.1
Example 5(a) Solving a Problem Involving Projectile
Motion (page 308)
A ball is thrown directly upward from an initial height of 75 ft
with an initial velocity of 112 ft per sec. Give the function that
describes the height of the ball in terms of time t.
The projectile height function is
After how many seconds does the ball reach its maximum
height? What is the maximum height?
The maximum height occurs at the vertex.
(3.5sec,271ft)
The ball reaches its maximum height, 271 ft, after 3.5 seconds.
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3-9
Additional Notes
Discriminant: b2 – 4ac
• Positive: then the graph crosses the x-axis twice
• Zero: then the graph is tangent to the x-axis
(meaning it touches the x-axis)
• Negative: then the graph never crosses the x-axis
because there are two imaginary roots
Disc: positive
a: positive
Disc: zero
a: positive
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Disc: positive Disc: negative
a: negative
a: negative
3-10
3.1 additional notes
The number of mosquitos, M(x), in millions, in a certain area of
Florida depends on the June rainfall, x, in inches. The function
M(x) = 10x – x2 models this phenomenon.
Find the amount of rainfall that will maximize the number of
mosquitos. What is the maximum number of mosquitos?
Vertex:
x = -b/2a
M(5) = 10(5) – 52
= 25
Vertex is
(5,25)
x = -10/-2 = 5
Since the parabola opens down the max # of mosquitos in
Flordia is 25 million which will occur when 5 inches of rain
has fallen in June
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3-11
3.1 Summary
Quadratics: y = ax2 + bx + c
y = a(x – h)2 + k, vertex (h,k)
Axis of symmetry of a quadratic equation: x = -b/2a
Vertex of a quadratic equation: (-b/2a, f(-b/2a ))
Quadratic formula:
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3-12
3.2 Synthetic Division
Synthetic Division ▪ Evaluating Polynomial Functions Using the
Remainder Theorem ▪ Testing Potential Zeros
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3-13
3.2
Example 1 Using Synthetic Division (page 323)
Use synthetic division to divide
.
x – 3 is in the form x – k.
Bring down the 1st #, then mult, add, mult, add, …
12
-9
6
-3
2
-4
3.2
Example 2 Applying the Remainder Theorem (page 324)
Let
theorem to find f(4).
. Use the remainder
Use synthetic division with k = 4.
f(4) = 17
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3-15
3.2
Example 3(a) Deciding Whether a Number is a Zero
(page 325)
Let
. Is k = 2 a zero?
Use synthetic division with k = 2.
2 is not a zero.
Let
. Is k = –3 a zero?
–3 is a zero.
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Additional Example
Express f(x) in the form (x – k)q(x) + r for the given
value of k
f(x) = 2x3 + 4x2 – 5x + 2 and k = 4
4 2
4
8
2 12
-5 2
48 172
43 174
Answer:
(x – 4)(2x2 + 12x + 43) + 174
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What if the 2 was missing in
the original f(x) equation?
We would need to put a zero
place holder in the last
position of synthetic division
3-17
3.2
Example 3(c) Deciding Whether a Number is a Zero
(page 325)
Let
. Is k = 1 + 3i a zero?
Use synthetic division with k = 1 + 3i.
1 + 3i is a zero.
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3-18
3.2 Summary
How to tell if a number is a zero of a polynomial or not
Synthetic Division:
Always rewrite the original problem in the correct order
and fill in zeros for missing exponents
The number to the left of the division bar comes from
setting the divisor = 0 and solving
Bring the first value down, then multiply, add, multiply,
add, …
When writing the answer, start with one degree less
than the original problem
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3-19
3.3 Zeros of Polynomial Functions
Factor Theorem ▪ Rational Zeros Theorem ▪ Number of Zeros ▪
Conjugate Zeros Theorem ▪ Zeros of a Polynomial Function ▪
Descartes’ Rule of Signs
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3-20
3.3
Example 1(a) Deciding Whether x – k is a Factor of f(x)
(page 328)
Let
. Is x + 4 a factor?
By the factor theorem x + 4 is a factor of f(x) if and
only if f(–4) = 0.
Insert 0 as the
coefficient for the
missing x3-term.
Since f(–4) = 0, x + 4 is a factor of f(x).
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3-21
3.3
Example 1(b) Deciding Whether x – k is a Factor of f(x)
(page 328)
Let
Is x + 4 a factor?
.
By the factor theorem x + 4 is a factor of f(x) if and
only if f(–4) = 0.
Since f(–4) ≠ 0, x + 4 is not a factor of f(x).
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3-22
3.3
Example 2 Factoring a Polynomial Given a Zero (page 329)
Factor
5 is a zero of f.
into linear factors if
Since 5 is a zero of f, x – 5 is a factor. Divide f(x) by x – 5.
The quotient is
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.
Side note:
6x2 – 7x – 3 =
6x2 – 9x + 2x – 3 =
3x(2x – 3) + 1(2x – 3) =
(3x + 1) (2x – 3)
3-23
Additional example
Factor f(x) into linear factors given that k is a
zero of f(x).
f(x) = x3 + (7 – 3i)x2 + (12 – 21i)x – 36i; k = 3i
3i
1
1
7 – 3i
12 – 21i
– 36i
3i
21i
36i
7
12
0
f(x) = (x – 3i)(x2 + 7x + 12)
Answer: f(x) = (x – 3i)(x + 3)(x + 4)
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3-24
3.3
Example 3(a) Using the Rational Zeros Theorem
(page 330)
List all possible rational zeros.
Note:
•from Algebra II “factors of last over factors of first”
•from College Alg book “factors of p/q”
p must be a factor of a0 = 4: ±1, ±2, ±4
q must be a factor of a4 = 8: ±1, ±2, ±4, ±8
Possible rational zeros, :
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3-25
3.3
Example 3(b) Using the Rational Zeros Theorem
(page 330)
Find all rational zeros and factor f(x) into linear factors.
Is x – 1 a factor?
Possible rational factors:
± {1, ½, ¼, 1/8, 2, 4, 8}
x – 1 is not a factor.
Is x + 1 a factor?
x + 1 is a factor.
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3-26
3.3 Example 3(b) Using the Rational Zeros Theorem
Possible rational factors: ± {1, ½, ¼, 1/8, 2, 4, 8}
(cont.)
Use the quotient polynomial
find the next factor. Is x – 4 a factor?
to
x – 4 is a factor.
The rational zeros are
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.
Side note:
8x2 – 2x – 1 =
8x2 – 4x + 2x – 1 =
4x(2x – 1) + 1(2x – 1) =
(4x + 1)(2x – 1)
3-27
3.3
Example 4(a) Finding a Polynomial Function that
Satisfies Given Conditions (Real Zeros)
(page 332)
Find a function f defined by a polynomial of degree 3
that has zeros of –3, –2, and 5, and f(–1) = 6.
The three zeros are: x = -3, x = -2, x = 5
The factors are:
(x + 3) (x + 2) (x – 5)
Since f(–1) = 6, we can solve for a:
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3-28
3.3
Example 4(b) Finding a Polynomial Function that
Satisfies Given Conditions (Real Zeros)
(page 332)
Find a function f defined by a polynomial of degree 3
that has zero 4 of multiplicity 3, and f(2) = –24.
The function f has form
Since f(2) = –24, we can solve for a:
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3-29
3.3
Example 5 Finding a Polynomial Function that Satisfies
Given Conditions (Complex Zeros) (page 334)
Find a polynomial function of least degree having
only real coefficients and zeros –4 and 3 – i.
By the conjugate zeros theorem, 3 + i is also a zero.
Zeros are:
x = -4 x = 3 – i x = 3 + i
Factors are: (x + 4) (x – 3 + i) (x – 3 – i)
F(x)
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Have the
students make a
3x3 box to
multiply
Have the
students make a
2x3 box to
multiply
3-30
3.3
Example 6 Finding All Zeros of a Polynomial Function
Given One Zero (page 335)
Find all zeros of
given that 2 + i is a zero.
Since 2 + i is a zeros of f(x), then 2 – i is also a zero.
2–i
2–i
1
3
6 – 3i
-10
-20 + 10i
0
F(x) = (x – 2 – i)(x – 2 + i)(x2 + 3x – 10)
F(x) = (x – 2 – i)(x – 2 + i)(x +5)(x – 2)
The zeros of f(x) are 2 + i, 2 – i, –5, and 2.
3.3
Example 7 Applying Descartes’ Rule of Signs (page 336)
Determine the possible number of positive real zeros and
negative real zeros of
f(x) = – + – + –
Thus, f(x) has 4 sign changes,
so f(x) has 4, or 2, or 0 positive real zeros.
For negative zeros, consider the variations in signs for f(–x).
f(-x)= – – – – –
There are no changes in sign, so there are no
negative zeros.
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3-32
3.3
Example 7 Applying Descartes’ Rule of Signs (page 336)
Cont. from previous slide to include complex imaginary roots
Determine the possible number of positive real zeros and
negative real zeros of
f(x) = – + – + –
Thus, f(x) has 4 sign changes,
so f(x) has 4, or 2, or 0 positive
real zeros.
+
–
i
4
0
0
2
0
2
0
0
4
f(-x)= – – – – –
There are no changes in sign,
so there are 0 negative zeros.
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There are 0, 2, or 4 complex
imaginary zeros.
3-33
3.3
Example 8 Applying Descartes’ Rule of Signs (page 336)
Determine the possible number of positive real zeros and
negative real zeros of
–
f(x) = – – – + –
Thus, f(x) has 2 sign changes,
so f(x) has 2 or 0 positive real zeros.
+
–
i
2
2
0
0
2
0
2
0
2
4
f(-x)= – + – – –
Thus, f(x) has 2 sign changes,
so f(x) has 2 or 0 negative real zeros.
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There are 0, 2, or 4 complex
imaginary zeros.
3-34
3.3 Summary
Given a polynomial, determine if a binomial is a factor or not
Factor a polynomial given a factor or a zero
List the possible rational zeros of a polynomial
Find all the rational zeros of a polynomial
Given a zeros of a polynomial find the polynomial
Multiplicity is the number of times the zeros is counted as a factor
If a polynomials with integral coefficients has an imaginary root, then
its conjugate must also be a root.
To find the number of possible positive and negative zeros count the
number of sign changes for f(x) and f(-x) respectfully, and subtract 2
at a time
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3-35
3.4 Polynomial Functions: Graphs,
Applications, and Models
Graphs of f(x) = axn ▪ Graphs of General Polynomial Functions ▪
Turning Points and End Behavior ▪ Graphing Techniques ▪
Intermediate Value and Boundedness Theorems ▪ Approximating
Real Zeros ▪ Polynomial Models and Curve Fitting
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3-36
3.4
Example 2(a) Examining Vertical and Horizontal
Translations (page 341)
Graph
X
Y
-1
0
0
1
1
2
2
9
The graph of
is
the same as the graph of
, but translated 1 unit
up.
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3-37
3.4
Example 2(b) Examining Vertical and Horizontal
Translations (page 341)
Graph
X
Y
1
-1
2
0
3
1
4
32
The graph of
is
the same as the graph of
, but translated 2
units right.
The multiplicity of 5 for the zero at 2 makes the graph flatter at that point
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3-38
3.4
Example 2(c) Examining Vertical and Horizontal
Translations (page 341)
Graph
X
Y
-5
-3
-3
5
-1
-3
1
-123
The graph of f(x) = -½ (x + 3)4 + 5 is the
same as the graph of f(x) = x4,but translated
3 units left, reflected across the x-axis,
stretched vertically by a factor of ½, and then
translated 5 units up.
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3-39
3.4
Example 3 Determining End Behavior Given The
Defining Polynomial (page 344)
Use the symbols for end behavior to describe the
end behavior of the graph of each function.
(a)
Since a = −1 < 0 and f has even degree, the end
behavior is shaped
(b)
Since a = 1 > 0 and g has odd degree, the end
behavior is shaped
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3-40
3.4
Example 3 Determining End Behavior Given The
Defining Polynomial (cont.)
(c)
Since a = −1 < 0 and h has odd degree, the end
behavior is shaped
(d)
Since a = 1 > 0 and f has even degree, the end
behavior is shaped
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3-41
3.4
Example 4 Graphing a Polynomial Function (page 345)
Graph
X
Y
-1
6
0
-6
1
-12
2
0
Have girls do synthetic
division for -1 and boys
synthetic division for 2
Setting each factor equal to zero gives
the zeros of f as 2, -1/2, and –3.
Add on
Graph (for the notes we will just complete the table but the homework you would graph these)
y = 2x(x – 5)(x + 2)
y = 1/5 (x – 2)2 + 3
X
Y
X
Y
0
0
-3
8
5
0
2
3
-2
0
7
8
1
-24
12
23
This equation is in factored
form so I would start with the
zeros of the polynomial
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
The coefficient has a 5 in the
denominator so I choose numbers
+ 5 units from the x-coordinate of
the vertex
3-43
3.4
Example 5 Locating a Zero (page 346)
Use synthetic division and a graph to show that
has a real zero between 1 and 2
Use synthetic division to find f(1) and f(2).
f(1) = 2
f(2) = –3
By the Intermediate Value Theorem,
there is a real zero between 1 and 2.
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3.4
Example 6(a) Using the Boundedness Theorem (page 348)
Show that
greater than 1.
has no real zero
f(x) has real coefficients and the leading coefficient, 1, is
positive, so the Boundedness Theorem applies. Divide
f(x) synthetically by x – 1.
Since 1 is positive and all the numbers in the last row of
the synthetic division are nonnegative, f(x) has no real
zeros greater than 1.
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3-45
3.4
Example 6(b) Using the Boundedness Theorem (page 348)
Show that
less than –2.
has no real zero
f(x) has real coefficients and the leading coefficient, 1, is
positive, so the Boundedness Theorem applies. Divide
f(x) synthetically by x + 2.
Since the 2 is negative and the numbers in the last row of
the synthetic division alternate in sign, f(x) has no real
zeros less than –2.
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3-46
Additional example
Given f(x) = x2(x – 4)(x + 5)2
Find the zeros, state the multiplicity for each, state if
graph crosses or is tangent at each zero, the degree
of the polynomial, the end behavior and then make
a rough sketch of the graph without the use of a
graphing calculator
Zeros:
Mult:
Cross/tangent:
X=0
2
tangent
X = 4 X = -5
1
2
cross tangent
Degree of poly: 5
End behavior: Down, Up
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
The graph should have a smooth curve
NOT a sharp point at (-5,0)
3-47
Additional example
Find the polynomial function with the given graph.
Assume the leading coefficient is 1.
Zeros:
Mult:
(0,-128)
X = -4 X = -1 X = 2
1
2
2
X=4
1
F(x) = a(x + 4)2(x + 1)(x – 2)2(x – 4)
F(0) = a(0 + 4)2(0 + 1)(0 – 2)2(0 – 4)
-128 = a(16)(1)(4)(–4)
-1/2 = a
F(x) = -1/2(x + 4)2(x + 1)(x – 2)2(x – 4)
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3-48
Additional example:
Make a table points used to graph the following:
y = 2x(x – 1)(x + 2)
Degree =
3
End behavior is
down,up
The zeros of the function are
0, 1, -2
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
X
Y
0
1
-2
0
0
0
-1
4
You will need one more point than the
degree for the homework system to
graph the polynomial for you
3-49
Additional example:
Make a table points used to graph the following:
y = x3 + 9x2 + 11x – 21
Degree =
3
End behavior is
down,up
X
Y
0
1
-1
-21
0
-24
2
45
You will need one more point than the
degree for the homework system to
graph the polynomial for you
Use synthetic division to obtain each yvalue of the table after x = 1 … I will
show you on the whiteboard if needed
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3-50
3.4 Summary
End behavior of polynomials
Steepness and shifting of graphs from parent functions
Reflecting polynomials over x- or y-axis
Multiplicity:
Odd multiplicity, the graph crosses the x-axis at that zero
Even multiplicity, the graph is tangent to the x-axis at that zero
Larger the value of the multiplicity the flatter the graph at that zero
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3-51
3.5 Rational Functions: Graphs,
Applications, and Models
The Reciprocal Function
▪ The Function
Asympototes ▪ Steps for Graphing Rational Functions ▪
Rational Function Models
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
▪
3-52
3.5
Example 1 Graphing a Rational Function (page 360)
Graph
. Give the domain and range.
The expression can be written as
, indicating
that the graph can be obtained by stretching the
graph of
vertically by a factor of 4.
Domain:
Range:
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3-53
3.5
Example 2 Graphing a Rational Function (page 361)
Graph
. Give the domain and range.
Shift the graph of
three units to the right, then
reflect the graph across the x-axis to obtain the graph
of
.
The horizontal asymptote is y = 0.
The vertical asymptote is x = 3.
Domain:
Range:
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3-54
3.5
Example 3 Graphing a Rational Function (page 363)
Graph
. Give the domain and range.
To obtain the graph shift the graph of
two
units to the right, stretch the graph vertically by a factor
of two, and then shift the graph four units down.
The vertical asymptote is x = 2.
The horizontal asymptote is y = –4.
Domain:
Range:
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3-55
3.5
Example 4(a) Finding Asymptotes of Rational Functions
(page 364)
Find all asymptotes of the function
To find the vertical asymptotes, set the denominator
equal to 0 and solve.
The equations of the vertical asymptotes are
x = –4 and x = 4.
The denominator has a bigger degree than the
numerator, so the horizontal asymptote is y = 0.
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3-56
3.5
Example 4(b) Finding Asymptotes of Rational Functions
(page 364)
Find all asymptotes of the function
To find the vertical asymptotes, set the denominator
equal to 0 and solve.
The equation of the vertical asymptote is
.
The numerator and the denominator have the equal
degrees, so the equation of the horizontal asymptote is
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3-57
3.5
Example 4(c) Finding Asymptotes of Rational Functions
(page 364)
Find all asymptotes of the function
To find the vertical asymptotes, set the den = 0 and solve.
Vert asy is x = 3.
Since the degree of the numerator is exactly one more
than the denominator, there is an oblique asymptote.
Divide the numerator by the
denominator and,
disregarding the remainder,
set the rest of the quotient
equal to y to obtain the
equation of the asymptote.
I would have used
Synthetic Division
Oblique asy
is y = 2x + 6.
3-58
3.5
Example 5 Graphing a Rational Function with the x-axis
as Horizontal Asymptote (page 366)
Graph
To find the vertical asymptotes, set the den = 0 and solve.
Find the horizontal asymptote.
The degree of den > degree of num,
so the horizontal asy is y = 0.
Find the y-intercept
Find the x-intercept
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3.5
Example 6 Graphing a Rational Function That Does Not
Intersect its Horizontal Asymptote (page 367)
Graph
To find the vertical asymptotes, set the den = 0 and solve.
Find the horizontal asymptote.
The degree of num = degree of den,
so the horizontal asymptote is y = 4.
Find the y-int
Find the x-int
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3.5
Example 8 Graphing a Rational Function With an
Oblique Asymptote (page 369)
Find the y-int
Graph
Find the x-int
Find the vertical asy
x(x+1) = 0
Find the horizontal asy.
Since the deg num > deg den
there isn’t a horiz asy. Divide
to find oblique asy.
I would have used
Synthetic Division
oblique asy is
y=x+2
x = 0 or -1
3.5
Example 9 Graphing a Rational Function Defined by an
Expression That is not in Lowest Terms
(page 370)
Graph
The domain of f cannot include 3.
In lowest terms, the function becomes
The graph of f is the same as the
graph of y = x + 2, with the exception
of the point with x-value 3. A “hole”
appears in the graph at (3, 5).
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3.5
Ex 7 Graphing a Rational Function That Intersects its
Horizontal Asymptote (page 368) Calculus bound
students only
Graph
Find the x-int
Find the vertical asy
Find the horizontal asy.
Degree of num = degree of denom,
so the horizontal asy is y = 2.
Find the y-int
3.5 Summary
Rational Graphs
X-intercept: solve for x, when y = 0
Y-intercept: solve for y, when x = 0
Vertical asymptote: set the denominator = 0
Horizontal asymptote:
degree of bottom = top,
y = coef
(BETC)
degree is bigger on bottom,
y=0
(BOBO)
Oblique asymptote:
(BOTD)
degree bigger on top, divide
asy: y = quotient
Holes occur in the graph when a factor reduces from the
numerator and denominator
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3.6 Variation
Direct Variation ▪ Inverse Variation ▪ Combined and Joint
Variation
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3.6
Example 1 Solving a Direct Variation Problem (page 380)
At a given average speed, the distance traveled by a
vehicle varies directly as the time. If a vehicle travels
156 miles in 3 hours, find the distance it will travel in 5
hours at the same average speed.
Equation: d = kt.
Find k, substitute d = 156 and t = 3
Final equation: d = 52t.
Solve the equation for d with t = 5.
The vehicle will travel 260 miles in 5 hours.
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3.6
Example 2 Solving an Inverse Variation Problem
(page 380)
In a certain manufacturing process, the cost of
producing a single item varies inversely as the
square of the number of items produced. If 100
items are produced, each costs $1.50. Find the cost
per item if 250 items are produced.
Let n = # of items produced and
c = cost per item
Solve for c with n = 250
Find k, sub c = 1.50 and n = 100
The cost per item will be $0.24.
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
3.6
Example 3 Solving a Joint Variation Problem (page 382)
The volume of a cylinder varies directly as the
height and the square of the radius. A cylinder
with radius 5 cm and height 10 cm has volume 785
cm3. Find the volume of a cylinder with radius 10
cm and height 15 cm.
V = khr2
Find k, sub V = 785, h =10
and r = 5
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Solve for V with h = 15 and
r = 10.
The volume of the
cylinder is 4710 cm3
3.6
Example 4 Solving a Combined Variation Problem
(page 383)
Vanessa Jones has determined that the number of children
enrolled in her day care center varies directly with the amount
she spends on advertising per year and inversely with her
weekly charge for each child. This year she spent $900 on
advertising, charged $150 per child per week, and had 51
children enrolled. Next year, if she spend $1000 on
advertising and charges $170 per week, how many children
should she expect to enroll?
Let n = the number of children,
a = $ spent on advertising/yr,
and c = the weekly charge.
She should expect to enroll 50 children.
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3.6 Summary
Varies Directly:
y = kx
Varies Inversely:
y = k/x
Varies jointly:
y = kxw
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3-70
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3.1
Ex 5(c) Solving a Problem Involving Projectile Motion
(pg 308)
For what interval of time is the height of the ball
greater than 200 ft?
Use the quadratic formula
a = –16, b = 112, c = –125
The ball will be greater than 200 ft
above ground level between 1.39
and 5.61 seconds after it is thrown.
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
3.1
Example 5(d) Solving a Problem Involving Projectile
Motion (page 308)
After how many seconds will the ball hit the ground?
Use the quadratic formula to find the positive solution of
Reject
The ball hits the ground after about 7.62 sec.
Verify with a graphing
calculator.
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3.1 Additional Story Problem
A farmer has 500 feet of fencing and is planning to
enclose a rectangular lot using the barn for one
side of the lot. What should the dimensions of the
lot be for the lamb to have the maximum amount of
area to graze on?
A = x(500 – 2x)
A = 500x – 2x2
A = -2x2 + 500x
barn
x
x
500-2x
Max area occurs when
x = – 500/–4 = 125
Max area is 500(125) – 2(125)2 = 31250
Maximum area of 31250 ft2 occurs when the
dimensions of the rectangle are 125 ft by 250 ft
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3.1 Additional Story Problem (like Hw 3.1G#5)
A piece of sheet metal is 2.9 times as long as it is wide. It is
to be made into a box with an open top by cutting 3-inch
squares from each corner and folding up the sides.
L = 2.9x
Restrictions for x:
If we are to cut away 3 inch squares then x > 6
x
Equation for the volume of the box:
V = 3(x – 6)(2.9x – 6)
The problem wants us to find the value(s) for x, in order for the box to have a
volume between 600 and 800 inches
In the calculator Y1 = 3(x – 6)(2.9x – 6) and Y2 = 600
calculate the intersection
In the calculator Y1 = 3(x – 6)(2.9x – 6) and Y2 = 800
calculate the intersection
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
x must be
between 12.6
& 13.8 inches
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3.4
Example 7 Approximating Real Zeros of a Polynomial
Function (page 349)
Approximate the real zeros of
The greatest degree term is
have end behavior similar to
, so the graph will
.
There are at most 3 real zeros.
f(0) = 10, so the y-intercept is 10.
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3.4
Example 7 Approximating Real Zeros of a Polynomial
Function (cont.)
The graph shows that there
are 3 real zeros.
Use the calculator to calculate
each zero.
The zeros are approximately –8.33594, –.9401088,
and 1.2760488.
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