Download Syllabus Objective: 2

2.1 Quadratic Functions
Date:
Precalculus Notes: Unit 2 – Polynomial Functions
Objective: The student will sketch the graph of a quadratic equation. The student will write the
equation of a quadratic function.
Quadratic Function: a polynomial function of degree 2
 Graph is u-shaped, called a parabola
 Parabolas are symmetric with respect to a line called the axis of symmetry
 Parabolas have a vertex, which is the point on the parabola where it intersects the axis of
symmetry, and is the maximum or minimum of the quadratic function

General Form: f  x   ax 2  bx  c
Axis of Symmetry: x  
o
x-intercepts can be found using the quadratic formula:
o
o
x-intercepts can be found by factoring.
x-intercepts can be found by completing the square.

Vertex Form (standard form): f  x   a  x  h   k

Vertex:
x
b
2a
b  b2  4ac
2a
2
o
 h, k 
Axis of Symmetry: x  h
In both forms, if a  0 , the parabola opens UP; if a  0 , the parabola opens DOWN
Graphing a Parabola
Ex1a: Write the equation in vertex form. Find the vertex and axis of symmetry. Then graph the
function. f  x   2 x 2  4 x  1
Ex. 1b. You Try: Write in vertex form. Find vertex & axis of symmetry:
f(x) = -6x + x2 – 3
Ex. 1c. Identify the x-intercepts of the quadratic function f  x    x2  6 x  8
Page 1 of 28
Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Precalculus Notes: Unit 2 – Polynomial Functions
Writing the Equation of a Parabola

Substitute the vertex for  h, k  in vertex form.

Substitute the given point for  x, y  and solve for a.
Ex2a: Write an equation for the parabola with vertex  3, 4 that passes through the point  4, 6 .
 h, k  
 x, y  
Ex.2b You try: Write the equation for a parabola with a vertex (1,2)
that passes through the point (0, 5).
Average Rate of Change (recall):
f b  f  a 
ba
Ex3: Find the average rate of change of f  x   x 2 from x  2 to x  4 .
Ave Rate of Change =
Vertical Free-Fall Motion: formula for the position of an object in free fall
1
s  t    gt 2  v0t  s0
2
s = position
g  32 ft/sec2  9.8 m/sec2 (on Earth)
g = acceleration due to gravity
v0 = initial velocity
s0 = initial position
Ex4: A flare is shot straight up from a ship’s bridge 75 feet above the water with an initial velocity of 76
ft/sec. How long does it take to reach its maximum height? How long does it take to reach 163 feet?
The flare reaches it max height at the vertex of the parabola.
It takes _____ sec for the flare to reach its maximum height.
Note: This is NOT the graph of the path of the flare. The graph relates the position of the
flare to time.
The flare reaches 163 feet when 163  16t 2  76t  75 .
It reaches 163 feet at ________________
Algebraic Method:
Page 2 of 28
Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Precalculus Notes: Unit 2 – Polynomial Functions
Ex. 5 A soda pop company’s sales average 26,000 cans per month when the cans sell for 50
cents each. For each nickel increase in the price, the sales per month drop by 1000 cans. Find the
maximum revenue.
You Try: Write the function in vertex form. Find the vertex and axis of symmetry. Then graph.
f  x   2 x  x 2  3
Reflection: Using a table of values, how can you determine whether it is a linear or quadratic function?
Page 3 of 28
Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Precalculus Notes: Unit 2 – Polynomial Functions
Date:
2.2 Polynomial Functions of Higher Degree
Syllabus Objective: 2.9 – The student will sketch the graph of a polynomial, radical, or rational function.
Polynomial
Function:
a function that can be written in the form
f  x   an x n  an1x n1  ...  a2 x 2  a1x  a0 , where n is a nonnegative integer and an  0
Note: This is called Standard Form – when the exponents of x descend
Degree: the largest exponent, n , of x
Leading Coefficient: the coefficient of the first term when written in standard form
Constant Term: the numerical term, a0
Classifying Polynomial Functions
I. Number of Terms
 1 term = monomial
 2 terms = binomial
 3 terms = trinomial
 4 or more terms = polynomial
II. Degree
The graph of a
polynomial
function is a
continuous
curve with
smooth round
turns.
Degree
Name
(Degree)
Standard Form
Example
Classification
of Example
0
Constant
f  x  a
y  3
Constant
Monomial
1
Linear
f  x   ax  b
y  2x
Linear
Monomial
2
Quadratic
f  x   ax 2  bx  c
y   x 2  5x  7
Quadratic
Trinomial
3
Cubic
f  x   ax3  bx 2  cx  d
y  5x 3  2
Cubic
Binomial
4
Quartic
f  x   ax 4  bx3  cx 2  dx  e
y  x4  x2  9 x  1
Quartic
Polynomial
5
Quintic
f  x   ax5  bx 4  cx3  dx 2  ex  f
y  x5
Quintic
Monomial
Page 4 of 28
Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Precalculus Notes: Unit 2 – Polynomial Functions
Ex1: Which of the following are polynomial functions? State the degree and the leading coefficient or
explain why it is not a function.
a) g  x   3x 8  17
b) h  x   13x 2  7 x5  11
Ex2 Describe how to transform the graph. Name the y-intercept.
f  x   2  x  4   7
3
a.)
b.) g(x) = 2(6 – 3x)4 – 1
The Leading Coefficient Test to Predict End Behavior of Polynomial Functions
Odd Degree:
& lim  .
If the leading coefficient is positive, then lim 
x 

x 
If the leading coefficient is negative, then lim 
x 
& lim  .
x 
Even Degree:

If the leading coefficient is positive, then lim 

LIf the leading coefficient is negative, then lim 
x 
x 
& lim  .
x 
& lim  .
x 
Ex2: Indicate if the degree of the polynomial function shown in the graph is odd or even and
indicate the sign of the leading coefficient. (see pg. 105 exploration)
a.)
b.)
Page 5 of 28
Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Precalculus Notes: Unit 2 – Polynomial Functions
Zeros (x-intercepts, Roots) of Polynomial Functions: nth-degree polynomials have at most n  1
extrema (relative minima or maxima) and n zeros.
Recall that a zero of a function f is a number x for which f(x)=0.
1. x=a is a zero of function f.
2. x=a is solution of the polynomial function f(x)=0
3. (x-a) is a factor of the polynomial f(x).
4. (a,0) is an x-intercept of the graph of f.
Zeros of Polynomial Functions
Multiplicity: “repeated” zeros; If a polynomial function f has a factor of  x  c  , and not  x  c 
m
m 1
,
then c is a zero of multiplicity m of f.

Odd Multiplicity: f crosses the x-axis at c; f  x  changes signs

Even Multiplicity: f “bounces” or is tangent to the x-axis at c; f  x  doesn’t change signs
Ex. 3 Sketch the graph of g  x   2  x  2 
3
 x  1
2
. Describe the multiplicity of the zeros.
End behavior (degree of _______, odd or even, _________________ leading coefficient):
lim 
; lim 
y
x
x













x


Ex4: Graph and analyze the function: f  x   x
Domain:
Continuity:
Symmetry:
Local Extrema:
Asymptotes:
End Behavior:
3

Range:
Increasing/Decreasing:
Boundedness:
y





lim f  x   ____, lim f  x   ____
x 
x 










Page 6 of 28
Precalculus – Graphical, Numerical, Algebraic: Chapter 2

x
x
−2
−1
0
1
2
1
2
y
Precalculus Notes: Unit 2 – Polynomial Functions
Ex5: Graph the function and list its characteristics. f  x   x 3  2 x 2
Domain:
Range:
Symmetry:
Boundedness: Extrema: Local Max:
End Behavior: lim f ( x )  ;
x 
Increasing:
Decreasing:
Local Min:
lim f ( x) 
x 
Zeros:
1
1
Ex6: Find the zeros and extrema of the function f  x   x4  5x 2  4 . Graph:
Zeros:
Extrema:
y













x



Newton’s Law of Cooling
Ex7: The rate at which an object cools varies as the difference between its temperature and the
temperature of the surrounding air. When a 270° C steel plate is placed in air that is at 20° C, it is cooling
at 50° C per minute. How fast is it cooling when its temperature is 100° C?
R = rate of cooling, T0 = initial temp, TS = surrounding temp:
R  k T0  TS 
Solve for k first:
Page 7 of 28
Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Precalculus Notes: Unit 2 – Polynomial Functions
Watch for hidden behavior:
Ex8: Graph the function in a Standard window and find the zeros.
f  x   x3  31x 2  28x  60
f  x  is degree 3, so the end behavior is lim   , so we know it must cross the x-axis again to the right.
x
New window:
Zeros are
Intermediate Value Theorem (Location Principle): If a and b are real numbers with a  b , and if f is
continuous on  a, b , then f takes on every value between f  a  and f  b  .
Therefore, if f  a  and f  b  have opposite signs, then f  c   0 for some number c in  a, b .
Ex9: Use the Intermediate Value Theorem to show f  x   12 x3  55x 2  18x  40 has zeros
in these intervals:
 1,0 and 1, 2
f changes sign (negative to positive) in the interval  1,0 , so f must have at least one zero in  1,0 .
f changes sign (positive to negative) in the interval 1, 2  , so f must have at least one zero in 1, 2  .
Application of Polynomial Functions
Ex10: You cut equal squares from the corners of a 22 by 30 inch sheet of cardboard to make a
box with no top. What size squares would need to be cut for the volume to be 300 cubic inches?
Define x as the length of the sides of the squares. Write a formula for the volume of the box.
V  x 
Solve the equation ____________________________________ by graphing.
Squares with lengths of approximately ____________________ inches should be cut.
Page 8 of 28
Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Precalculus Notes: Unit 2 – Polynomial Functions
Ex. 11
Find polynomial functions with the following zeros. (There are many correct solutions.)
a.)
1
 ,3,3
2
b.) 3, 2  11, 2  11
You Try: Sketch the graph of the polynomial function.
1
g(x) = - (x-2)3(x + 1)2
3
f(x) = .2(x +1) 2(x – 3)(x + 5)
Reflection: How many zeros can a function of degree n have? Explain your answer.
Page 9 of 28
Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Precalculus Notes: Unit 2 – Polynomial Functions
Date:
2.3 Real Zeros of Polynomial Functions
Syllabus Objectives: 2.2 – The student will calculate the intercepts of the graph of a given relation.
Review: Long Division
23024  23
23 2302
Long Division of Polynomials (same process!)
Ex1a: Find the quotient.
x
4
 8 x 3  11x  6    x  3
Note: Every term of the polynomial in the dividend must be represented. Since this polynomial is
missing an x 2 term, we must include the term 0x 2 .
x  3 x 4  8 x 3  0 x 2  11x  6
Solution:
Remainder Theorem: If f  x  is divided by x  k , then the remainder, r  f  k  .
Ex2a: Find the remainder without using division for x 4  8 x 3  11x  6 divided by x  3 .
f  x   x 4  8x3  11x  6
r  f  3 
Note: Compare your answer to the long division problem above.
Page 10 of 28
Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Precalculus Notes: Unit 2 – Polynomial Functions
Ex. 2b. You try: (2x3 – 3x2 – 5x + 6) ÷ (x2 – 3)
Factor Theorem:
x  k
is a factor of f  x  if and only if f  k   0 .
Ex3: Determine if x  3 is a factor of 2 x 3  3x 2  5 x  12 without dividing.
Show that r  f  3  0 :
so by the Factor Theorem, x  3 is a factor of 2 x 3  3x 2  5 x  12 .
Synthetic Substitution:
Ex4: Find f  2  if f  x   3x4  x3  5x2  6 x  1 using synthetic substitution.




Using the polynomial in standard form, write the coefficients in a row.
Put the x-value to the upper left.
Bring down the first coefficient, then multiply by the x-value.
Add straight down the columns, and repeat.
The number in the bottom right is the value of f  2  .
So : f (2) 
This also means
Ex 5: Find f  3 if f  x   x5  2 x3  7 x 2  11 using synthetic substitution.
that  3, 137  is
This polynomial function is in standard form, however it is missing two terms. We can
rewrite the function as f  x   x5  0 x 4  2 x 3  7 x 2  0 x  11 to fill in the missing
an ordered pair
that would be a
point on the graph.
And the remainder
when dividing
terms.
f  x  by x  3 is
−137.
Page 11 of 28
Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Precalculus Notes: Unit 2 – Polynomial Functions
Synthetic Division:
Ex.6
a) (2x3 – 3x2 – 5x – 12) ÷ (x – 3)
b) Divide x4 – 8x3 + 11x – 6 by x + 3.
Recall: f  x   Qxn  an1 x x1  ...  a2 x2  a1x  P
Possible Rational Zeros of a Polynomial Function =
factors of constant  P 
factors of leading coefficient  Q 
Ex7: Find the possible rational zeros of f ( x)  x 2  7 x  12
This also means
that 5 could not be
a zero, 7 could not
Step 1: The leading coefficient is _____. _____ is the only factor of ________.
Step 2: The constant is ____. All of the factors of _____ are
1
be a zero, could
2
Step 3: List the possible factors -
not be a zero, ...
*If we tested for actual zeros using synthetic substitution
we would find that 3 and  4 are zeros.
Descartes’ Rule of Signs: The number of real zeros equals the number of sign changes in f  x  or that
number less a multiple of 2. The number of negative real zeros equals the number of sign changes in
f   x  or that number less a multiple of 2.
Ex8: Use Descartes’ Rule of Signs to determine the possible number of positive and negative
real zeros of the function f  x   3x3  4 x 2  5x  2 .
Number of sign changes of f  x  : f  x   3x3  4 x 2  5x  2
1 sign change
Number of sign changes of f   x  : f   x   3x3  4 x 2  5x  2
2 sign changes
There is 1 possible positive real zero and 2 or 0 possible negative real zeros.
Page 12 of 28
Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Precalculus Notes: Unit 2 – Polynomial Functions
Upper and Lower Bound Rules: used to determine if there is no zero larger or smaller than a number c,
if the leading coefficient is POSITIVE  an  0  , using synthetic division for f  x    x  c 

Upper Bound: c is an upper bound for the real zeros of f if c  0 and every number in the last
line of f  x    x  c  using synthetic division has signs that are all nonnegative.

Lower Bound: c is a lower bound for the real zeros of f if c  0 and every number in the last line
of f  x    x  c  using synthetic division has signs that are alternately nonnegative and
nonpositive.
Ex9: Verify that all of the real zeros of the function f  x   3x3  4 x 2  5x  2 lie on the
interval  3,2 .
f  2 :
c  2  0 and signs are all nonnegative, so c  2 is an upper bound of the zeros
f  3 :
c  3  0 and signs alternate, so c  3 is an lower bound of the zeros
Conclusion: ________________________________________________
Finding the Real Zeros of a Polynomial:
Ex10: Find all real solutions of the polynomial equation. f  x   3x3  4 x 2  5x  2
Note: There is one sign change, so________________________________
Possible Rational Zeros:
Synthetic Substitution (try 1):
f 1  1 3 4  5  2
Note: Remember,
if 1 is a zero,
x – 1 is a factor


1 is a zero, so we can write f  x    x  1 3x 2  7 x  2  0 .
Factor the quadratic:
Solutions:
You Try: Find the real zeros of f  x   2 x 4  7 x3  8x 2  14 x  8 .
Reflect: How is the Factor Theorem connected to the Remainder Theorem?
Page 13 of 28
Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Precalculus Notes: Unit 2 – Polynomial Functions
Date:
2.4 Complex Numbers
Objective: Students will use imaginary unit i to write complex numbers. Add, subtract and multiply
complex numbers, use complex conjugates to write the quotient of two complex numbers in standard form,
plot complex numbers in the complex plane.
I.
Number Sets
C:
R:
Q:
Z
Complex: a + bi
Imaginary
Real
Rational
I: Irrational
: Integers
Transcendental
W: Whole
N:
Natural
Transcendental numbers are not the roots of any algebraic equations. All other complex numbers are
algebraic.
Powers of i:
Cont to i7
i0 = 1
*
i1  1  i
i2 = -1
i3 =
i4 
Page 14 of 28
Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Precalculus Notes: Unit 2 – Polynomial Functions
Mult.
complex #s
is
like FOIL,
but i is
not a
variable.
Ex.1
a) (3 + 7i) – (2 – 4i)
Ex.2
a)


3  2i  3 4i 
Complex Conjugate of z: (a + bi) is
The are many
examples of
conjugates, the
difference of two
squares for
example.
Ex.3
b) (3 + 7i)(2 – 4i)
b) (2 – i)3
z : (a – bi)
a) Multiply 3 – 4i by its conjugate.
b) Find x & y. (2 + 3i)(x – i) = 13 + yi
b) (3i)(5 2i)
Ex.4 a) (2i)(4 2i)
1 2i
4 i
Complex Solutions to Quadratic Equations
Discriminant:
D<0
Ex.5 a) 5x2 + 2x + 1 = 0
b) 3x2 + 4 = -5x
Reflect:
Page 15 of 28
Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Date:
2.5 The Fundamental Theorem of Algebra
Precalculus Notes: Unit 2 – Polynomial Functions
Syllabus Objectives: 1.7 – The student will solve problems using the Fundamental Theorem of
Algebra. 1.8 – The student will calculate the complex roots of a given function.
Fundamental Theorem of Algebra
 A polynomial function of degree n has n complex zeros (some zeros may repeat)
 Complex zeros come in conjugate pairs
Review: a  bi  conjugate  a  bi
 A polynomial function with odd degree and real coefficients has at least one real zero
Calculator Exercise: State how many complex zeros the function has. Use
a graphing calculator for b & c to determine how
many of the zeros are real.
a) f(x) = 2x2 – 12x + 19
b) f(x) = 5x3 + 4x2 + 3x – 4
c) f(x) = x5 + 2x4 – 22x2 + 2x3 – 18x + 55
Ex1: Find all the zeros of the function and write the polynomial as a product of linear factors.
f  x   x 4  6 x3  10 x 2  6 x  9
Step One: Check for any rational zeros.
Possible rational zeros:
factors of 9
1 3 9
  , ,
factors of 1
1 1 1
Use synthetic substitution:
So ________ is a zero, and _____________ is a factor.
Step Two: Rewrite the last line as a polynomial and find the zeros of the polynomial.
f  x 
x 3  3x 2  x  3  0 Factor by grouping:
Step Three: List the zeros. Check that you have the correct number according to the FTA.
The polynomial was degree 4, and there are 4 zeros (3 is repeated).
Note that i and –i are complex conjugates.
Page 16 of 28
Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Precalculus Notes: Unit 2 – Polynomial Functions
Step Four: Write the polynomial as a product of linear factors.
f ( x) 
Ex2: Find a polynomial function with real coefficients in standard form that has the zeros −2 and
1  2i .
Write the polynomial as a product of linear factors using the zeros.
Note: If 1  2i is a zero, then its complex conjugate, ______________must also be a zero.
Write the polynomial in standard form.
Irreducible Quadratic Factor: A quadratic factor of a polynomial function is irreducible over the reals
if it has real coefficients but no real zeros.
Ex3: Write the polynomial as the product of factors that are irreducible over the reals. Then
write the polynomial in completely factored form. f  x   3x5  x 4  3x3  x 2  60 x  20
Step One: Check for any rational zeros.
Possible rational zeros:
Step Two: Use synthetic substitution:
Step Three: Rewrite the last line as a polynomial and find the zeros of the polynomial.
Product of factors irreducible over the reals:
Step Four: Completely factored form:
Page 17 of 28
Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Precalculus Notes: Unit 2 – Polynomial Functions
Ex4: Use the given zero to find ALL the zeros of the function. Zero = 1  i ;
f  x   x 4  2 x3  x 2  6 x  6
Note: By the FTA, we know that f  x  must have ___________ zeros.
Since 1  i is a zero, we know that ______ (its conjugate) must also be a zero.
Two factors of f  x  are: ____________and _________________
Multiply:
Use long division:
Zeros of
Zeros of f  x  :
You Try: Find a polynomial function with real coefficients in standard form that has the zeros 3, and −1
with multiplicity 3.
Reflection: Is the polynomial function found in the “You Try” unique? Explain your answer.
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Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Date:
2.6 Rational Functions and Asymptotes
Precalculus Notes: Unit 2 – Polynomial Functions
Syllabus Objectives: 1.9 – The student will solve rational equations in one variable.
Rational Function: the ratio of two polynomial functions
r  x 
f  x
, g  x  0
g  x
Domain of a Rational Function: the set of all real numbers except the zeros if its denominator
Recall: Reciprocal Function f  x  
1
x
Rational Functions that are Transformations of the Reciprocal Function
Ex1: Describe the transformations on f  x  
3
2
x 1
3
2
3. g  x  
x 1
1. g  x  
_______________
_______________
1
3
2
. State the domain. g  x  
x
x 1
3
 2 _______________
2. g  x  
x 1
3
 2 _______________
4. g  x  
x 1
Domain: All reals, x  1
Using Long Division (used when variable is in the numerator and denominator)
Ex2: Describe the transformations on f  x  
3
Long division: x  5 3x  4
3x  15
1
4  3x
. State the domain. g  x  
x
x 5
g  x   3 
11
x 5
 11
11
_________________
x 5
11
3. g  x   3 
_________________
x 5
1. g  x   3 
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11
______________
x 5
11
4. g  x   3 
_______________
x 5
2. g  x   3 
Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Precalculus Notes: Unit 2 – Polynomial Functions
Recall:
Limit: the value that f  x  approaches as x approaches a given value
Notation: lim f  x  “The limit as x approaches a of f  x  .” (From both sides.)
xa
lim f  x  “The limit of f  x  as x approaches a from the left.”
x a 
lim f  x  “The limit of f  x  as x approaches a from the right.”
x a 
Ex3: Evaluate the limits based on the graph.
1)
lim f  x  
2)
lim f  x  
3)
x1
x1
lim f  x  
x 
4) lim f  x  
x 
End Behavior Asymptotes of f  x  
an x n  ...
bm x m  ...

If n  m : horizontal asymptote y  0

If n  m : horizontal asymptote y 

If n  m : No horizontal asymptote. End behavior is the function that is the result after
an
(ratio of the leading coefficients)
bn
performing the division on f  x  .
Note: If the n is one more than m, then the function describing end behavior is linear. This is called a
slant asymptote.
Vertical Asymptote: Set the denominator equal to zero.
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Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Precalculus Notes: Unit 2 – Polynomial Functions
Ex4: Describe the end behavior of the following functions.
y
a.)
x 3  3x 2  x  1
x 1
degree of the numerator > degree of the denominator
___________________________________________
Long Division: y  x 2  4 x  5 
6
x 1
End Behavior: Approaches the graph of the function y  x 2  4 x  5
y
b.)
6x2
3  2 x  1
2
degree of the numerator = degree of the denominator
Lead coefficient of numerator = _______


Lead coefficient of denominator =______: 3  2 x  1  3 4 x 2  4 x  1  12 x 2  12 x  3
2
End Behavior: horizontal asymptote of y 
y
c.)
d.) y 
x2
x3  2 x 2
6 x2
3x  2
Slant:

Ex. 5 Find the horizontal and vertical asymptotes:
y
6 x2
x2  4
Reflection:
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Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Precalculus Notes: Unit 2 – Polynomial Functions
Date:
2.7 Graphs of Rational Functions
Syllabus Objectives 2.9 – The student will sketch the graph of a polynomial, radical, or rational
function.
Graphing Rational Functions: To sketch the graph of a rational function, find the domain, asymptotes,
and intercepts. The plot points in each region created by the asymptotes to locate the curve and use the
asymptotes to guide the sketching of the curve.
Ex1: Graph g  x  
x2  x
.
x 1
Domain:
Vertical Asymptote:
x-intercepts (zeros of numerator):
y-intercepts ( g  0 ): g  0 
End Behavior: asymptote___________________; long division results in g  x   x  2 
Slant Asymptote: ____________
2
x 1
2
0
x  x  1
Note: lim
Plot a few points to graph the lower part of the curve.
Ex2: Graph g  x  
2 x2  8x  6
.
x2  4
Domain: _____________________
Vertical Asymptotes: ______________________
x-intercepts (zeros of numerator): ______________________________________________
y-intercepts ( g  0 ): g  0 
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End Behavior: horizontal asymptote y  ___________
Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Precalculus Notes: Unit 2 – Polynomial Functions
Removable Discontinuity: a rational function has a “hole” in the graph if one or more of the linear
factors in the denominator can be cancelled with one of the factors in the numerator.
2 x 2  18
Ex3: Sketch the graph of g  x   2
x  3x
Domain:
Vertical Asymptotes:
Note: x  _________ is NOT a vertical asymptote. It is a hole.
x-intercepts (zeros of numerator):
y-intercepts ( g  0 ):
End Behavior: horizontal asymptote y  _________
You Try: Sketch the graph of y 
x2  x  2
. Describe all relevant features.
x 5
QOD: Can a function cross a vertical asymptote? Explain. Can a function cross a horizontal asymptote?
Explain.
Reflection:
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Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Precalculus Notes: Unit 2 – Polynomial Functions
Date:
Solve Rational Functions
Syllabus Objective: 1.9 – The student will solve rational equations in one variable.
Solving a Rational Equation:
1. Multiply every term by the LCD to wipe out the fractions.
2. Solve the resulting equation.
3. Check for extraneous solutions in the original equation, which can result when multiplying
by a variable expression.
Ex1: Solve the equation.
1
1
4

 2
y2 y2 y 4
Factor to find LCD:
Multiply by LCD:
Solve the resulting equation:
Check for extraneous solutions:
1
1
4

 2
22 22 2 4
y  2 makes one or more of the denominators in the original equation equal zero, so it is extraneous.
Therefore, there is___________________________.
Ex2: Solve the equation:
2x
3

5 x  4 3x  8
This equation is a proportion, so we will cross multiply.
Solve the equation by factoring:
Check for extraneous solutions: neither solution results in a denominator of zero in the original equation
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Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Precalculus Notes: Unit 2 – Polynomial Functions
Ex3: Find the solution set of the equation
3
x4
2
.
x  7 x  10
x 5
2
Factor to find LCD:
Multiply by LCD:
Solve the resulting equation:
Check for extraneous solutions:
x  _____ does not make one or more of the denominators in the original equation equal zero, so it is a
solution. x  ______ does make one or more of the denominators in the original equation equal zero,
so it is an extraneous solution. Solution: x  ____
You Try: Solve the equation
x
3
4

 2
.
x2 x3 x  x6
Reflection: Explain two ways you can check your solution graphically when solving a rational equation.
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Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Precalculus Notes: Unit 2 – Polynomial Functions
Sample SAT Question(s): Taken from College Board online practice problems.
1. The projected sales volume of a video game cartridge is given by the function s  p  
3000
,
2p  a
where s is the number of cartridges sold, in thousands; p is the price per cartridge, in dollars; and
a is a constant. If according to the projections, 100,000 cartridges are sold at $10 per cartridge,
how many cartridges will be sold at $20 per cartridge?
(A) 20,000
(B) 50,000
(C) 60,000
(D) 150,000
(E) 200,000
2.
S
a c 1
  . If 0  a  b  c  d  e in the equation given, then the greatest increase in S
b d e
would result from adding 1 to the value of which variable?
(A) a
(B) b
(C) c
(D) d
(E) e
3. If ☼ is defined for all positive numbers a and b by a☼b =
ab
, then 10☼2 =
ab
5
3
5
(B)
2
(A)
(C) 5
(D)
20
3
(E) 20
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Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Bonus: Solve Inequalities
Precalculus Notes: Unit 2 – Polynomial Functions
Date:
Syllabus Objective: 2.10 – The student will solve and graph inequalities in one variable.
Solving an Inequality:
1. Find the zeros of the corresponding equation and the zeros of the denominators of any rational
expressions (called critical values).
2. Make a sign chart (choose a test value) to find the correct interval solutions.
Ex1: Solve the inequality  x  5 x  2 
4
 x  8  0
Critical Values:
−
+
Sign Chart:
−
+
Solution:
Note: If the inequality was  x  5 x  2 
4
 x  8  0 , the solution would be  5,8 .
3x  2
0
x3
Ex2: Solve the inequality:
Critical Values:
+
−
+
Sign Chart:
Solution:
Note: x  3 is not in the domain of the original function.
Ex3: Solve the inequality.
1
1

x 1
x2
Rewrite the inequality:
Critical Values:
−
+
−
+
Sign Chart:
Solution:
Page 27 of 28
Precalculus – Graphical, Numerical, Algebraic: Chapter 2
Precalculus Notes: Unit 2 – Polynomial Functions
Solving an Inequality Graphically
Ex4: Solve graphically: 2 x 3  3x 2  4 x  1  0
Find the zeros (watch for hidden zeros!):
Note: The 2nd and 3rd graphs were created by zooming in to the local minimum 3 times!
Solution:
Application Problem
Ex5: A cylindrical can must have a volume of 58 cubic inches. What dimensions can you use if you
must keep the surface area below 100 square inches?
Volume: V   r 2h  58
Surface Area: SA  2 r 2  2 rh  100
Solve for h in the volume equation and substitute into the surface area inequality.
h
58
 r2
2 r 2  2 r
58
116
 100  2 r 2 
 100
2
r
r
Solve by graphing:
Solution: Radius must be between _____and ______ inches. Height must be between 1.827 and 10.975
inches.
You Try: Solve the inequality
2 x2  x  1
0.
x2  4 x  4
QOD: Explain why you need only choose one test point within the intervals created by the critical value.
Reflection:
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Precalculus – Graphical, Numerical, Algebraic: Chapter 2