copyright © 2014, 2010 pearson education, inc. chapter 2 polynomials and rational functions...

Copyright © 2014, 2010 Pearson Education, Inc.

Chapter 2

Polynomials and Rational

Functions



Section 2.1 Quadratic Functions

1. Graph a quadratic function in standard form.2. Graph any quadratic function.3. Solve problems modeled by quadratic functions.

SECTION 1.1

Copyright © 2014, 2010 Pearson Education, Inc. 3

QUADRATIC FUNCTION

A function of the form

where a, b, and c, are real numbers with a ≠ 0, is called a quadratic function.

f x ax2 bx c ,


THE STANDARD FORM OFA QUADRATIC FUNCTION

The quadratic function

is in standard form. The graph of f is a parabola with vertex (h, k). The parabola is symmetric with respect to the line x = h, called the axis of the parabola. If a > 0, the parabola opens up, and if a < 0, the parabola opens down.

f x a x h 2 k , a 0


Find the standard form of the quadratic function whose graph has vertex (–3, 4) and passes through the point (–4, 7).

Let y = f (x) be the quadratic function.

2

2

2

3 4

3 4

y a x

y a x

a x

kh

y

2

2

7 3 4

3

Hence,

3 3 4

–4a

a

y x

Example: Writing the Equation of a Quadratic Function


Example: Graphing a Quadratic Function in Standard Form

Sketch the graph of f (x) = a(x – h)2 + k

Step 1 The graph is a parabola because it has the form f(x) = a(x – h)2 + k Identify a, h, and k.

Sketch the graph of

1. The graph of f(x) = –3(x + 2)2 + 12

is the parabola: a = –3, h= – 2, and k = 12.

23 2 12.f x x



Step 2 Determine how the parabola opens. If a > 0, the parabola opens up. If a < 0, the parabola opens down.

Step 3 Find the vertex. The vertex is (h, k). If a > 0 (or a < 0), the function f has a minimum (or a maximum) value k at x = h.

2. Since a = –3, a < 0, the parabola opens down.

3. The vertex (h,k) = (–2, 12). Since the parabola opens down, the function f has a maximum value of 12 at x = –2.



Step 4 Find the x-intercepts (if any). Set f (x) = 0 and solving the equation a(x – h)2 + k = 0 for x. If the solutions are real numbers, they are the x-intercepts. If not, the parabola lies above the x-axis (when a > 0) or below the x-axis (when a < 0).

4.

2

2

2

0 3 2 12

12 3 2

4 2

x

x

x

2 2

0 or 4

-intercepts: 0 and 4

x

x x

x



Step 5

Find the y-intercept. Replace x with 0. Thenf (0) = ah2 + k is the y-intercept.

5.

20 3 0 2 12

3 4 12 0

-intercept is 0 .

f

y



Step 6

Sketch the graph. Plot the points found in Steps 3–5 and join them by a parabola. Show the axis x = h of the parabola by drawing a dashed line.

6. axis: x = –2


Example: Graphing a Quadratic Function

Graph f(x) = ax2 + bx + c, a ≠ 0.

Step 1 Identify a, b, and c.

Sketch the graph of

1. In the equation y = f(x) = 2x2 + 8x – 10,

a = 2, b = 8, and c = – 10.

22 8 10.f x x x



Step 2 Determine how the parabola opens. If a > 0, the parabola opens up. If a < 0, the parabola opens down.

Step 3 Find the vertex.

2. Since a = 2 > 0, the parabola opens up.

3.

2

2

bh

ab

k fa

2

82

2 2 2

2

2 2 8 2 10 18

, 2, 18

bh

a

k f

h k



Step 4 Find the x-intercepts (if any). Set f (x) = 0 and solving the equation a(x – h)2 + k = 0 for x. If the solutions are real numbers, they are the x-intercepts. If not, the parabola lies above the x-axis (when a > 0) or below the x-axis (when a < 0).

4. 22 8 10 0x x

22 4 5 0

2 5 1 0

x x

x x

5 0 or 1 0

5 1

x x

x x



Step 5

Find the y-intercept. Let x = 0. The result,f (0) = c, is the y-intercept.

5. Let x = 0.

20 2 0 8 0 10

-intercept is 10 .

f

y



Step 6

The parabola is symmetric with respect to its axis,

Use this symmetry to find additional points.

6.

Axis of symmetry is x = –2. The symmetric image of (0, –10) with respect to the axis x = –2 is (–4, –10).

.2

bx

a



Step 7

Draw a parabola through the points found in Steps 3–6.

7. Sketch the parabola passing through the points found in Steps 3–6.

.2

bx

a


The graph of f (x) = –2x2 +8x – 5 is shown.a. Find the domain and range of f.b. Solve the inequality –2x2 +8x – 5 > 0.

b. The graph is above the x-axis in the interval

4 6 4 6, .

2 2

a. The domain is (–∞, ∞). The range is (–∞, 3].

Example: Identifying the Characteristics of a Quadratic Function from Its Graph


Section 2.2 Polynomial Functions

1. Learn properties of the graphs of polynomial functions.2. Determine the end behavior of polynomial functions.3. Find the zeros of a polynomial function by factoring.4. Identify the relationship between degrees, real zeros, and

turning points.5. Graph polynomial functions.

SECTION 1.1


Definitions

A polynomial function of degree n is a function of the form

where n is a nonnegative integer and the coefficients an, an–1, …, a2, a1, a0 are real numbers with an ≠ 0.

f x an xn an 1xn 1 ... a2 x2 a1x a0 ,


Definitions

The term anxn is called the leading term.

The number an is called the leading coefficient, and a0 is the constant term.

A constant function f (x) = a, (a ≠ 0) which may be written as f (x) = ax0, is a polynomial of degree 0.


Definitions

The zero function f (x) = 0 has no degree assigned to it.

Polynomials of degree 3, 4, and 5 are called cubic, quartic, and quintic polynomials.


Definitions

an xn an 1xn 1 ... a2 x2 a1x a0

The expression

is a polynomial, the function f defined by

is a polynomial function, and the equation

is a polynomial equation.

f x an xn an 1xn 1 ... a2 x2 a1x a0

an xn an 1xn 1 ... a2 x2 a1x a0 0


COMMON PROPERTIES OFPOLYNOMIAL FUNCTIONS

1. The domain of a polynomial function is the set of all real numbers.


2. The graph of a polynomial function is a continuous curve.


3. The graph of a polynomial function is a smooth curve.


State which functions are polynomial functions. For each polynomial function, find its degree, the leading term, and the leading coefficient.a. f (x) = 5x4 – 2x + 7b. g(x) = 7x2 – x + 1, 1 x 5

a. f (x) is a polynomial function. Its degree is 4, the leading term is 5x4, and the leading coefficient is 5.b. g(x) is not a polynomial function because its domain is not (–, ).

Example: Polynomial Functions


POWER FUNCTION

A function of the form

is called a power function of degree n, where a is a nonzero real number and n is a positive integer.

f x axn


POWER FUNCTIONS OF EVEN DEGREE

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

The graph of y = xn (n is even) is similar to the graph of y = x2.

Let f x axn . If n is even, then x n xn .

Then f x a x n axn f x .


POWER FUNCTIONS OF ODD DEGREE

The graph is symmetric with respect to the origin.

The graph of y = xn (n is odd) is similar to the graph of y = x3.

Let f x axn . If n is odd, then x n xn .

Then f x a x n axn f x .


END BEHAVIOR OF POLYNOMIAL FUNCTIONS

Case 1a > 0

The graph rises to the left and right, similar to y = x2.



Case 2a < 0

The graph falls to the left and right, similar to y = –x2.



Case 3a > 0

The graph rises to the right and falls to the left, similar to y = x3.



Case 4a < 0

The graph rises to the left and falls to the right, similar to y = –x3.


Let be a polynomialfunction of degree 3. Show thatwhen |x| is very large.

P x 2x3 5x2 7x 11P x 2x3

P x x3 2 5

x

7

x2 11

x3

When |x| is very large are close to 0.5

x,

7

x2 and 11

x3

P x x3 2 0 0 0 2x3.Therefore,

Example: Understanding the End Behavior of a Polynomial Function


THE LEADING–TERM TEST

Its leading term is anxn.

The behavior of the graph of f as x → ∞ or as x → –∞ is similar to one of the following four graphs and is described as shown in each case.

The middle portion of each graph, indicated by the dashed lines, is not determined by this test.

Let 11 1 0... 0n n

nn nf x a x ax ax aa

be a polynomial function.


Case 1

an > 0



Case 2

an < 0



Case 3

an > 0



Case 4

an < 0



Use the leading-term test to determine the end behavior of the graph of

y f x 2x3 3x2 4.

Here n = 3 (odd) and an = –2 < 0. Thus, Case 4 applies. The graph of f (x) rises to the left and falls to the right. This behavior is described as y → ∞ as x → –∞ and y → –∞ as x → ∞.

Example: Using the Leading-Term Test


REAL ZEROS OF POLYNOMIAL FUNCTIONS

1. c is a zero of f .2. c is a solution (or root) of the equation f (x) = 0.3. c is an x-intercept of the graph of f . The point

(c, 0) is on the graph of f .

If f is a polynomial function and c is a real number, then the following statements are equivalent.


Find all zeros of each polynomial function.

4 3 2

3 2

a. 2 3

b. 2 2

f x x x x

g x x x x

Factor f (x) and then solve f (x) = 0.

4 3 2

2

2

2

2

a. 2 3

3

0, 1 0, 3 0

0,

2 3

1

1, 3

f x x x x

x

x x

x x

x

x

x

x

x

x

x

Example: Finding the Zeros of a Polynomial Function


3 2

2

2

2

2

b. 2 2

1

1

0 2 1

2 0 or

2 2

2

1 0

2

f x x x x

x

x

x x

x

x

x

x

x

x

The only zero of g(x) is 2, since x2 + 1 > 0 for all real numbers x.

Example: Finding the Zeros of a Polynomial Function


REAL ZEROS OF POLYNOMIAL FUNCTIONS

A polynomial function of degree n with real coefficients has at most n real zeros.


Find the number of distinct real zeros of the following polynomial functions of degree 3.

a. Solve f (x) = 0.

22

a. 1 2 3

b. 1 1 c. 3 1

f x x x x

g x x x h x x x

1 2 3 0

1 0 or 2 0 or 3 0

1 or 2 or 3

f x x x x

x x x

x x x

f (x) has three real zeros: –2, 1, and 3.

Example: Finding the Number of Real Zeros


2

2

b. 1 1 0

1 0 or 1 0

1 No Real Solutions

g x x x

x x

x

h(x) has two distinct real zeros: –1 and 3.

2c. 3 1

3 0 or 1 0

3 or 1

h x x x

x x

x x

g(x) has only one real zero: –1.

Example: Finding the Number of Real Zeros


MULTIPLICITY OF A ZERO

If c is a zero of a polynomial function f (x) and the corresponding factor (x – c) occurs exactly m times when f (x) is factored, then c is called a zero of multiplicity m.

1.If m is odd, the graph of f crosses the x-axis at x = c.

2. If m is even, the graph of f touches but does not cross the x-axis at x = c.


Find the zeros of the polynomial function f (x) = x2(x + 1)(x – 2), and give the multiplicity of each zero.

f (x) is already in factored form.

f (x) = x2(x + 1)(x – 2) = 0 x2 = 0, or x + 1 = 0, or x – 2 = 0 x = 0 or x = –1 or x = 2

The zero x = 0 has multiplicity 2, while each of the zeros –1 and 2 have multiplicity 1.

Example: Finding the Zeros and Their Multiplicities


TURNING POINTS

A local (or relative) maximum value of f is higher than any nearby point on the graph.

A local (or relative) minimum value of f is lower than any nearby point on the graph.

The graph points corresponding to the local maximum and local minimum values are called turning points. At each turning point the graph changes from increasing to decreasing or vice versa.


TURNING POINTS

The graph of f has turning points at (–1, 12) and at (2, –15).

f x 2x3 3x2 12x 5


NUMBER OF TURNING POINTS

If f (x) is a polynomial of degree n, then the graph of f has at most (n – 1) turning points.


Use a graphing calculator and the window –10 x 10; –30 y 30 to find the number of turning points of the graph of each polynomial.

4 2

3 2

3 2

a. 7 18

b. 12

c. 3 3 1

f x x x

g x x x x

h x x x x

Example: Finding the Number of Turning Points


f has three total turning points; two local minimum and one local maximum.

a. f x x4 7x2 18



g has two total turning points; one local maximum and one local minimum.

b. g x x3 x2 12x



h has no turning points, it is increasing on the interval (–∞, ∞).

c. h x x3 3x2 3x 1



Example: Graphing a Polynomial Function

Sketch the graph of a polynomial function

Step 1 Determine the end behavior. Apply the leading-term test.

Sketch the graph of

1. Degree = 3 Leading coefficient = –1End behavior shown.

3 24 4 16.f x x x x



Step 2 Find the zeros of the polynomial function.

Set f (x) = 0 and solve. The zeros give the x-intercepts.

2.

Each zero has multiplicity 1, the graph crosses the x-axis at each zero.

3 24 4 16 0

4 2 2 0

4, 2, or 2

x x x

x x x

x x x



Step 3 Find the y-intercept by computing f (0).

Step 4 Use symmetry to check whether the function is odd, even, or neither.

3. The y-intercept is f(0) = 16. the graph passes through the point (0,16).

4.

There is no symmetry in the y-axis nor with respect to the origin.

3

3 2

4 16

4 4 16

f x x x

x x x



Step 5 Determine the sign of f(x) by using arbitrarily chosen “test numbers” in the intervals defined by the x-intercepts. Find the intervals on which the graph lies above or below the x-axis.

5.



Step 6 Draw the graph. Use the fact that the number of turning points is less than the degree of the polynomial to check whether the graph is drawn correctly.

6. Draw the graph.

The number of turning points is 2, which is less than 3, the degree of f.


Section 2.3 Dividing Polynomials and the Rational Zeros Test

1. Learn the Division Algorithm.2. Use the Remainder and Factor Theorems.3. Use the Rational Zeros Test.

SECTION 1.1


POLYNOMIAL FACTOR

A polynomial D(x) is a factor of a polynomial F(x) if

there is a polynomial Q(x) such that F(x) = D(x) ∙

Q(x).


THE DIVISION ALGORITHM

If a polynomial F(x) is divided by a polynomial D(x), with

D(x) ≠ 0, there are unique polynomials Q(x) and R(x)

such that

Either R(x) is the zero polynomial, or the degree of

R(x) is less than the degree of D(x).


Use long division and synthetic division to find the quotient and remainder when

2x4 + x3 − 16x2 + 18 is divided by x + 2.

We will start by performing long division.

Example: Using Long Division and Synthetic Division


quotient: 2x3 – 3x2 – 10x + 20 remainder: –22

The result is



THE REMAINDER THEOREM

If a polynomial F(x) is divided by x – a, then the remainder R is given by

.R F a


Find the remainder when the polynomialF x 2x5 4x3 5x2 7x 2

is divided by x 1.

5 3 21 1 1 1 12 4 5 7 2

2 4 5 7 2 2

F

By the Remainder Theorem, F(1) is the remainder.

The remainder is –2.

Example: Using the Remainder Theorem


Let 4 3 23 5 8 75.f x x x x x Find f 3 .

One way is to evaluate f (x) when x = –3.

Another way is to use synthetic division.

4 3 23 53 3 3 3 638 75f

Either method yields a value of 6.

1 3 5 8 75

3 0 15 69

1 0 5 23

3

6

Example: Using the Remainder Theorem


THE FACTOR THEOREM

A polynomial F(x) has (x – a) as a factor if and only if F(a) = 0.


Given that 2 is a zero of the function

solve the polynomial equation

Since 2 is a zero of f (x), f (2) = 0 and (x – 2) is a factor of f (x). Perform synthetic division by 2.

f x 3x2 2x2 19x 6,

3x2 2x2 19x 6 0.

2 3 2 19 6

6 16 6

3 8 3 0

Example: Using the Factor Theorem


2 2 23 2 19 6 2 3 8 3f x x x x x x x

Since the remainder is 0,

To find other zeros, solve the depressed equation.

23 8 3 0

3 1 3 0

3 1 0 or 3 0

1 or 3

3

x x

x x

x x

x x



Including the original zero of 2, the solution set is

3,1

3, 2

.



THE RATIONAL ZEROS TEST

1. p is a factor of the constant term a0;

2. q is a factor of the leading coefficient an.

is a polynomial function with integer coefficients (an ≠ 0, a0 ≠ 0) and

is a rational number in lowest terms that is a zero of F(x), then

1 21 2 1 0If ...n n

n nF x a x a x a x a x a

p

q


Find all the rational zeros of

3 25 4 .2 3F x x x x List all possible zeros

Factors of the constant term,

Factors of the leading coefficien

3

t, 2

p

q

Factors of : 1, 3

Factors 2 1

3

of : , 2

Possible rational zeros are: 1, 1

2,

3

2, 3.

Example: Using the Rational Zeros Test


Begin testing with 1. if it is not a rational zero, then try another possible zero.

1 2 5 4 3

2 7 3

2 7 3 0

The remainder of 0 tells us that (x – 1) is a factor of F(x). The other factor is 2x2 + 7x + 3. To find the other zeros, solve 2x2 + 7x + 3 = 0.



22 7 3 0

2 1 3 0

2 1 0 or 3 0

1 or 3

2

x x

x x

x x

x x

The solution set is 1, 1

2, 3

.

The rational zeros of F are 3, 1

2, and 1.



Section 2.4 Zeros of a Polynomial Function

1. Learn basic facts about the complex zeros of polynomials.2. Use the Conjugate Pairs Theorem to find zeros of

polynomials.

SECTION 1.1


If we extend our number system to allow the coefficients of polynomials and variables to represent complex numbers, we call the polynomial a complex polynomial. If P(z) = 0 for a complex number z we say that z is a zero or a complex zero of P(x).

In the complex number system, every nth-degree polynomial equation has exactly n roots and every nth-degree polynomial can be factored into exactly n linear factors.

Definitions


FUNDAMENTAL THEOREM OF ALGEBRA

Every polynomial

with complex coefficients an, an – 1, …, a1, a0 has at least one complex zero.

11 1 0... 1, 0n n

n n nP x a x a x a x a n a


FACTORIZATION THEOREM FOR POLYNOMIALS

If P(x) is a complex polynomial of degree n ≥ 1, it can be factored into n (not necessarily distinct) linear factors of the form

where a, r1, r2, … , rn are complex numbers.

1 2 ... ,nP x a x r x r x r


Find a polynomial P(x) of degree 4 with a leading coefficient of 2 and zeros –1, 3, i, and –i. Write P(x)

a. Since P(x) has degree 4, we write

1 2 3 4

1

2 1 3

32

P x x x x x

x x x x

x x x i x i

r r r

i i

a r

a. in completely factored form;b. by expanding the product found in part a.

Example: Constructing a Polynomial Whose Zeros Are Given


b. Expand the product found in part a.

P x 2 x 1 x 3 x i x i 2 x 1 x 3 x2 1 2 x 1 x3 3x2 x 3 2 x4 2x3 2x2 2x 3 2x4 4x3 4x2 4x 6

Example: Constructing a Polynomial Whose Zeros Are Given


CONJUGATE PAIRS THEOREM

If P(x) is a polynomial function whose coefficients are real numbers and if z = a + bi is a zero of P, then its conjugate, is also a zero of P.,z a bi


ODD–DEGREE POLYNOMIALSWITH REAL ZEROS

Any polynomial P(x) of odd degree with real coefficients must have at least one real zero.


A polynomial P(x) of degree 9 with real coefficients has the following zeros: 2, of multiplicity 3; 4 + 5i, of multiplicity 2; and 3 – 7i. Write all nine zeros of P(x).

Since complex zeros occur in conjugate pairs, the conjugate 4 – 5i of 4 + 5i is a zero of multiplicity 2, and the conjugate 3 + 7i of 3 – 7i is a zero of P(x). The nine zeros of P(x) are:

2, 2, 2, 4 + 5i, 4 – 5i, 4 + 5i, 4 – 5i, 3 + 7i, 3 – 7i

Example: Using the Conjugate Pairs Theorem


FACTORIZATION THEOREM FOR A POLYNOMIAL WITH REAL COEFFICIENTS

Every polynomial with real coefficients can be uniquely factored over the real numbers as a product of linear factors and/or irreducible quadratic factors.


Given that 2 – i is a zero of

find the remaining zeros.

The conjugate of 2 – i, 2 + i is also a zero.

P x x4 6x3 14x2 14x 5,

x 2 i x 2 i x 2 i x 2 i

x 2 2 i2 x2 4x 5

So P(x) has linear factors: x 2 i x 2 i

Example: Finding the Complex Zeros of a Polynomial


Divide P(x) by x2 – 4x + 5.

x2 4x 5 x4 6x3 14x2 14x 5

x4 4x3 5x2

2x3 9x2 14x

2x3 8x2 10x

x2 4x 5

x2 4x 5

0

x2 2x 1



Therefore

The zeros of P(x) are 1 (of multiplicity 2), 2 – i, and 2 + i.

2 2

2

2 1 4 5

1 1 4 5

1 1 2 2

P x x x x x

x x x x

x x x i x i



Find all zeros of the polynomial P(x) = x4 – x3 + 7x2 – 9x – 18.

Possible zeros are: ±1, ±2, ±3, ±6, ±9, ±18Use synthetic division to find that 2 is a zero.

2 1 1 7 9 18

2 2 18 18

1 1 9 9 0

(x – 2) is a factor of P(x). Solve x3 x2 9x 9 0

Example: Finding the Zeros of a Polynomial


3 2

2

2

2

2

9 9 0

9 0

9 0

1 0 or 9 0

1 or 9

1 or

1

3

1

1

x x x

x

x

x

x x

x

x

x x

x x i

The four zeros of P(x) are –1, 2, –3i, and 3i.

Example: Finding the Zeros of a Polynomial


Section 2.5 Rational Functions

1. Define a rational function.2. Define vertical and horizontal asymptotes.3. Graph translations of 4. Find vertical and horizontal asymptotes (if any).5. Graph rational functions.6. Graph rational functions with oblique asymptotes.7. Graph a revenue curve.

SECTION 1.1

1( ) .f x

x


RATIONAL FUNCTION

A function f that can be expressed in the form

where the numerator N(x) and the denominator D(x) are polynomials and D(x) is not the zero polynomial, is called a rational function. The domain of f consists of all real numbers for which D(x) ≠ 0.

f x N x D x ,


Find the domain of each rational function.

a. f x 3x2 12

x 1

a. The domain of f (x) is the set of all real numbers for which x – 1 ≠ 0; that is, x ≠ 1 .In interval notation:

b. g x x

x2 6x 8

c. h x x2 4

x 2

,1 U 1,

Example: Finding the Domain of a Rational Function


b. Find the values of x for which the denominator x2 – 6x + 8 = 0, then exclude those values from the domain.

The domain of g (x) is the set of all real numbers such that x ≠ 2 and x ≠ 4 .

In interval notation: x 2 x 4 0

x 2 0 or x 4 0

x 2 or x 4

,2 U 2, 4 U 4,



,2 U 2,


c. The domain of

is the set of all real numbers for which x – 2 ≠ 0; that is, x ≠ 2 .

The domain of g (x) is the set of all real numbers such that x ≠ 2.

In interval notation:

2 4

2

xh x

x


VERTICAL ASYMPTOTES

The line with equation x = a is called a vertical asymptote of the graph of a function f if

as or as

or if as or as .

f x x a x a

f x x a x a


VERTICAL ASYMPTOTES


LOCATING VERTICAL ASYMPTOTES OF RATIONAL FUNCTIONS

If is a rational function,

where the N(x) and D(x) do not have a common factor and a is a real zero of D(x), then the line with equation x = a is a vertical asymptote of the graph of f.

f x N x D x


Find all vertical asymptotes of the graph of each rational function.

a. No common factors, zero of the denominator is x = 1. The line with equation x = 1 is a vertical asymptote of f (x).

a. f x 1

x 1

b. g x 1

x2 9

c. h x 1

x2 1

Example: Finding Vertical Asymptotes


b. No common factors. Factoring x2 – 9 = (x + 3)(x – 3), we see the zeros of the denominator are –3 and 3. The lines with equations x = – 3 and x = 3 are the two vertical asymptotes of g (x).

c. The denominator x2 + 1 has no real zeros. Hence, the graph of the rational function h (x) has no vertical asymptotes.

Example: Finding Vertical Asymptotes


Find all vertical asymptotes of the graph of each rational function.

The graph is the line with equation y = x + 3, with a gap (hole) corresponding to x = 3.

2 9

a. 3

xh x

x

b. g x x 2

x2 4

2 3 39a.

3 33, 3

x xxh x

x xx x

Example: Rational Function Whose Graph Has a Hole


The graph has a hole at x = –2. However, the graph of g(x) also has a vertical asymptote at x = 2.

b. g x x 2

x2 4

x 2

x 2 x 2

1

x 2, x 2



HORIZONTAL ASYMPTOTES

The line with equation y = k is called a horizontal asymptote of the graph of a function f if

f x k as x or x .


RULES FOR LOCATINGHORIZONTAL ASYMPTOTES

Let f be a rational function given by

To find whether the graph of f has one horizontal asymptote or no horizontal asymptote, we compare the degree of the numerator, n, with that of the denominator, m:

an xn an 1x

n 1 ... a2 x2 a1x a0

bm xm bm 1xm 1 ... b2 x2 b1x b0

,an 0,bn 0

f x N x D x


1. If n < m, then the x-axis (y = 0) is the horizontal asymptote.

2. If n = m, then the line with equationis the horizontal asymptote.

3. If n > m, then the graph of f has no horizontal asymptote.

y an

bm

RULES FOR LOCATINGHORIZONTAL ASYMPTOTES


Find the horizontal asymptote (if any) of the graph of each rational function.

a. f x 5x 2

1 3x

a. Numerator and denominator have degree 1.

is the horizontal asymptote.

b. g x 2x

x2 1

c. h x 3x2 1

x 2

5

3

5

3y

Example: Finding the Horizontal Asymptote


degree of denominator > degree of numerator y = 0 (the x-axis) is the horizontal asymptote.

b. g x 2x

x2 1

c. h x 3x2 1

x 2

degree of numerator > degree of denominator The graph has no horizontal asymptote.

Example: Finding the Horizontal Asymptote


Example: Graphing a Rational Function

Graph where f(x) is in lowest terms.

Step 1 Find the intercepts. Since f is in lowest terms, the x-intercepts are found by solving the equation N(x) = 0. The y-intercept is, if there is one, is f (0).

Sketch the graph of

1.

( )

,( )

N xf x

D x

2

2

2 2.

9

xf x

x

22 2 0

2 1 1 0

1 0 or 1 0

1 or 1

x

x x

x x

x x

The x-intercepts are 1 and 1, so the graph passes through (1, 0) and (1, 0).



Step 1 Find the intercepts. Since f is in lowest terms, the x-intercepts are found by solving the equation N(x) = 0. The y-intercept is, if there is one, is f (0).

1.

The y-intercept is , so the graph of f passes through the point (0, ).

2

2

2 0 2 20

90 9f

2

9

2

9



Step 2 Find the vertical asymptotes (if any). Solve D(x) = 0 to find the vertical asymptotes of the graph. Sketch the vertical asymptotes.

2.

The vertical asymptotes are x = 3 and x = –3.

2Solve 9 0, 3x x



Step 3 Find the horizontal asymptotes (if any). Use the rules from the previous slide.

3. Since n = m = 2 the horizontal asymptote is

Leading coefficient of 22.

Leasing coefficient of ( ) 1

N xy

D x



Step 4 Test for symmetry. If f (–x) = f (x), then f is symmetric with respect to the y-axis. If f (–x) = –f (x), then f is symmetric with respect to the origin.

4.

The graph of f is symmetric in the y-axis. This is only one symmetry.

2

2

2

2

2 2

9

2 2.

9

xxf

xf x

x

x



Step 5 Locate the graph relative to the horizontal asymptote (if any). Use the sign graphs and test numbers associated with the zeros of N(x) and D(x), to determine where the graph of f is above the x-axis and where it is below the x-axis.

5.

R(x) = 16 has no zeros and D(x) has zeros –3 and 3. These zeros divide the x-axis into three intervals. We choose test points –4, 0, and 4 to test the sign of

2

2 2

2 2 162

9 9

xf

x xx

2

16.

9x



Step 6 Sketch the graph. Plot some points and graph the asymptotes found in Steps 1-5; use symmetry to sketch the graph of f.

6.


Sketch the graph of

Step 1 Since x2 + 2 > 0, no x-intercepts.

; y-intercept is –1.

Step 2 Solve (x + 2)(x – 1) = 0; x = –2, x = 1Vertical asymptotes are x = –2 and x = 1.

f x x2 2

x 2 x 1 .

20

00

21

2 10f



Step 3 degree of den = degree of numy = 1 is the horizontal asymptote

Step 4 Symmetry. None

Step 5 The zeros of the denominator –2 and 1 yield the following figure:



Step 6 Sketch the graph.



Sketch a graph of f x x2

x2 1.

Step 1 Since f (0) = 0 and setting f (x) = 0 yields 0, x-intercept and y-intercept are 0.

Step 2 Because x2 +1 > 0 for all x, the domain is the set of all real numbers. Since there are no zeros for the denominator, there are no vertical asymptotes.



Step 3 degree of den = degree of numy = 1 is the horizontal asymptote.

Step 4 Symmetry.

Symmetric with respect to the y-axis.

Step 5 The graph is always above the x-axis, except at x = 0.

2 2

2 2 11

xx

xfx

xf

x




f x x2

x2 1



OBLIQUE ASUMPTOTES

Suppose f x N x D x ,

is greater than the degree of D(x). Then

and the degree of N(x)

f x N x D x Q x R x

D x .

Thus, as x , f x Q x 0 Q x .

That is, the graph of f approaches the graph of the oblique asymptote defined by Q(x).


Sketch the graph of f x x2 4

x 1.

Step 2 Solve x + 1 = 0; x = –1; domain is set of all real numbers except –1.

Vertical asymptote is x = –1.

f 0 0 4

0 1 4 ; y-intercept: –4.

Step 1 Solve x2 – 4 = 0; x-intercepts: –2, 2

Example: Graphing a Rational Function with an Oblique Asymptote


y = x – 1 is an oblique asymptote.

Step 3 degree of numerator > degree of denominator

Step 4 Symmetry. None

Step 5 Use the intervals determined by the zeros of the denominator.

f x x2 4

x 1x 1

3

x 1




2 4

1

xf x

x


copyright © 2014, 2010 pearson education, inc. chapter 2 polynomials and rational functions...

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