Category Archives: Sullivan Chapter 5

Problem: Use the rational zeros theorem to find all real zeros of the polynomial function. Use the zeros to factor f over the real numbers.

f(x)=7x^3-x^2+7x-1

Since f is a polynomial function with integer coefficients use the rational zeros theorem to find the possible zeros.

The factors of the constant term, 1 are p.
p: pm 1

The factors of the leading coefficient, 7 are q.
q: pm 1, pm 7

The possible rational zeros can be found by working out all of the possible combinations of p/q.
p/q : {pm 1}/{pm 1}, {pm 1}/{pm 7}

Simplifying these combinations give p/q : 1, -1, 1/7, -{1/7}

To test if any of these potential zeros are actual zeros, evaluate the function at these values.

x	f(x)	f(x)
-1	7(-1)^3-(-1)^2+7(-1)-1	-16
-1/7	7(-1/7)^3-(-1/7)^2+7(-1/7)-1	-2.041
1/7	7(1/7)^3-(1/7)^2+7(1/7)-1	0
1	7(1)^3-(1)^2+7(1)-1	12

This can be completed quickly using the ask feature in your calculator.

Since f(1/7) is zero, 1/7 is a zero of the function. Since the function has a zero of x=1/7 then the function has a factor of x-1/7

Use long division or synthetic division to to reduce the polynomial.

Write the factor on the outside and the function on the inside of the long division symbol. Make sure both are written in descending order and to use place holders where needed.
Divide the first term of the factor into the first term of the function. Write that value on top of the long division symbol. For this example ${7x^3}/x=7x^2$
Multiply this value by all of the terms in the expression being divided by. Write the terms under the expression you are dividing into and be sure the line up the like terms.
Change the signs of all the terms you just multiplied.
Combine like terms and bring down the next set of terms.
Repeat the process over again. Divide the first term in the factor into the new first term of the function. Write that value on top of the long division symbol. For this example
Multiply this value by all of the terms in the expression being divided by. Write the terms under the expression you are dividing into and be sure the line up the like terms.
Change the signs of all the terms you just multiplied.
Combine like terms. The steps of this process (the division algorithm) are repeated until the degree of the remainder is less than the degree of expression you are dividing by.

Write the function in factored form using the results of the long division.

f(x) =7x^3-x^2+7x-1
f(x) =(x-1/7)(7x^2+7)

Factor completely.

f(x) =(x-1/7)(7x^2+7)
f(x) =7(x-1/7)(x^2+1)

5.2 The Real Zeros of a Polynomial Function

Intermediate Value Theorem

July 5, 2018 admin

Problem: Use the Intermediate Value Theorem to show that the following function has a zero in the given interval. Approximate the zero to two decimal places.

f(x)=9x^3+9x^2-9x+6; [-2,-1]

Solution:

To determine if there is a zero in the interval use the Intermediate Value theorem. To use the Intermediate Value Theorem, the function must be continuous on the interval [-2,-1] . The function f(x)=9x^3+9x^2-9x+6 is a polynomial function and polynomial functions are defined and continuous for all real numbers.

Evaluate the function at the endpoints and if there is a sign change. If there is a sign change, the Intermediate Value Theorem states there must be a zero on the interval. To evaluate the function at the endpoints, calculate f(-2) and f(-1) .

f(-2)=9(-2)^3+9(-2)^2-9(-2)+6
f(-2)=9(-8)+9(4)+18+6
f(-2)=-72+36+24
f(-2)=-72+60
f(-2)=-12

f(-1)=9(-1)^3+9(-1)^2-9(-1)+6
f(-1)=9(-1)+9(1)+9+6
f(-1)=-9+9+9+6
f(-1)=0+15
f(-1)=15

Since one endpoint gives a negative value and one endpoint gives a positive value, there must be a zero in the interval.

We can get a better approximation of the zero by trying to figure out the next decimal point. Write out all of the values to one decimal point between -2 and -1.

x	f(x)	f(x)
-2.0	9(-2)^3+9(-2)^2-9(-2)+6	-12
-1.9
-1.8
-1.7
-1.6
-1.5
-1.4
-1.3
-1.2
-1.1
-1.0	9(-1)^3+9(-1)^2-9(-1)+6	15

Fill the table. There are functions in your calculator that make this easier.

x	f(x)	f(x)
-2.0	9(-2)^3+9(-2)^2-9(-2)+6	-12
-1.9	9(-1.9)^3+9(-1.9)^2-9(-1.9)+6	-6.141
-1.8	9(-1.8)^3+9(-1.8)^2-9(-1.8)+6	-1.128
-1.7	9(-1.7)^3+9(-1.7)^2-9(-1.7)+6	3.093
-1.6	9(-1.6)^3+9(-1.6)^2-9(-1.6)+6	6.576
-1.5	9(-1.5)^3+9(-1.5)^2-9(-1.5)+6	9.375
-1.4	9(-1.4)^3+9(-1.4)^2-9(-1.4)+6	11.544
-1.3	9(-1.3)^3+9(-1.3)^2-9(-1.3)+6	13.137
-1.2	9(-1.2)^3+9(-1.2)^2-9(-1.2)+6	14.208
-1.1	9(-1.1)^3+9(-1.1)^2-9(-1.1)+6	14.811
-1.0	9(-1)^3+9(-1)^2-9(-1)+6	15

Use the Intermediate Value Theorem again. Look for a sign change. Looking down the table, there is a sign change between -1.8 and -1.7. With this information we now know the zero is between these two values.

Repeat this process again with two decimal places between -1.8 and -1.7.

x	f(x)	f(x)
-1.80	9(-1.8)^3+9(-1.8)^2-9(-1.8)+6	-1.128
-1.79	9(-1.79)^3+9(-1.79)^2-9(-1.79)+6	-.6712
-1.78	9(-1.78)^3+9(-1.78)^2-9(-1.78)+6	-.2222
-1.77	9(-1.77)^3+9(-1.77)^2-9(-1.77)+6	.219
-1.76	9(-1.76)^3+9(-1.76)^2-9(-1.76)+6	.65242
-1.75	9(-1.75)^3+9(-1.75)^2-9(-1.75)+6	1.0781
-1.74	9(-1.74)^3+9(-1.74)^2-9(-1.74)+6	1.4962
-1.73	9(-1.73)^3+9(-1.73)^2-9(-1.73)+6	1.9066
-1.72	9(-1.72)^3+9(-1.72)^2-9(-1.72)+6	2.3096
-1.71	9(-1.71)^3+9(-1.71)^2-9(-1.71)+6	2.705
-1.70	9(-1.7)^3+9(-1.7)^2-9(-1.7)+6	3.093

Use the Intermediate Value Theorem. Look for a sign change. Looking down the table, there is a sign change between -1.78 and -1.77. With this information we now know the zero is between these two values and the zero to two decimal places is -1.77 since all the numbers between -1.78 and -1.77 start with -1.77.

5.2 The Real Zeros of a Polynomial Function

Remainder Theorem and Factor Theorem

June 23, 2018 admin

Problem: Use the remainder theorem to find the remainder when f(x) is divided by x-1/5 . Then use the factor theorem to determine whether x-1/5 is a factor of f(x).

f(x)=5 x^4-x^3-5x-1

Solution:

The Remainder Theorem: Let f be a polynomial function. If f(x) is divided by x-c , then the remainder is f(c) .

The remainder theorem gives a way to find the remainder without performing the division. The remainder when dividing by x-c is f(c). Calculate f(c) to find the remainder where c is the zero of the factor x-c .

For the example above, f(x) is divided by x-1/5 . The zero of the factor can be found by setting the factor equal to zero and solve.

x-1/5=0
x-1/5+1/5=0+1/5
x=1/5

Use c=1/5 and calculate f(1/5) .

f(1/5)=5 (1/5)^4-(1/5)^3-5(1/5)-1
f(1/5)=5 *(1/625)-1/125-1-1
f(1/5)=5/625-1/125-2
f(1/5)=1/125-1/125-2
f(1/5)=-2

The remainder when f(x) is divided by x-1/5 is -2.

Factor Theorem: Let f be a polynomial function. Then x-c is a factor of f(x) if and only if f(c)=0 .

This theorem gives of the result that if f(x) is divided by x-c and the remainder is zero, then x-c is a factor of f.

For the example above, the remainder when f(x) is divided by x-1/5 is -2. Since the remainder is not zero x-1/5 is not a factor.

4.6 Rational Functions and Their Graphs, 5.4 Properties of Rational Functions, 5.5 The Graph of a Rational Function

Rational Function: Sketch the graph of the rational function

April 2, 2018 admin

Problem:

Sketch the graph of the rational function using algebra techniques.

$f(x)= 1/{x^2-9}$

Solution:

To start sketching the graph, gather the following information about the graph.

Domain, vertical asymptotes, holes/removable discontinuities, horizontal asymptotes, oblique/slant asymptotes, points where the graph crosses the horizontal asymptotes, x-intercepts, and y-intercepts.

Find the Domain of a Rational Function

Division by zero is undefined. Having a zero as the denominator is equivalent to division by zero thus is also undefined. The rational function is undefined for any value of the variable that gives a zero denominator. Find these values by creating an equation to solve. The equation is the expression in the denominator equal to zero.

Set the denominator equal to zero.
Solve the equation. This equation is a quadratic equation and can be solved by factoring, the square root method, completing the square or the quadratic formula.
Solve by the square root method. Start by isolating the square.
Cancel out the square by square rooting both sides.	$sqrt{x^2}=sqrt{9}$
The solutions to the equation are 3 and -3.	or

The values 3 and -3 give a zero value in the denominator and must be excluded from the domain.

In set builder notation, the domain is delim{lbrace}{ x delim{ |} x!=3 and x!=-3}{rbrace}

In interval notation, the domain is (- infty,-3) union (-3,3) union (3,infty)

Find Vertical Asymptotes and Holes/Removable Discontinuities

A vertical asymptote occurs when there is a non-zero in the numerator and a zero in the denominator. To find the location of any vertical asymptotes, find the values where the denominator is zero but only after reducing the fraction so that we can guarantee that there is a non-zero in the numerator.

Factor the rational function and write it in lowest terms by canceling any common factors.	$f(x)= 1/{x^2-9}$
The rational function has no common factors and is in lowest terms	$f(x)= 1/{x^2-9}$
Find the values where the denominator is zero and the numerator is non-zero.
Solve the equation. This equation is a quadratic equation and can be solved by factoring, the square root method, completing the square or the quadratic formula.
Solve by factoring. Factor the expression on one side.
Use the zero product property and set each factor equal to zero.	or
Solve each equation.	or or or

The rational function has vertical asymptotes at x=-3 and x=3. That means that y approaches infty or - infty as x approaches the values of 3 and -3. Knowing this will guide us when sketching the graph.

Determine if the graph has any holes/removable discontinuities. In this example, since the rational function was in lowest terms there is no hole/removable discontinuity.

Find Horizontal Asymptotes and Oblique/Slant Asymptotes

Find the horizontal asymptotes using a set of rules based on the degree of the numerator and the degree of the denominator.

Horizontal Asymptote and Oblique/Slant Cases

Case 1: The degree of the numerator is less than the degree of the denominator.	The horizontal asymptote is y=0. There is no oblique/slant asymptote.
Case 2: The degree of the numerator is equal to the degree of the denominator.	The horizontal asymptote is y= the ratio of the leading coefficients. There is no oblique/slant asymptote.
Case 3: The degree of the numerator is one more than the degree of the denominator.	There is no horizontal asymptote. There is an oblique/slant asymptote. Find it using long division.
Case 4: The degree of the numerator is two or more than the degree of the denominator.	There is no horizontal asymptote and there is no oblique/slant asymptote.

For this example, the numerator is 1 and the degree of the numerator is 0. The denominator is x^2-9 and the degree of the numerator is 2. Since the degree of the numerator is less than the degree of the denominator (case 1) the horizontal asymptote is y=0. That means that y approaches 0 as x approaches infty or - infty . Knowing this will guide us when sketching the graph.

Find any Points that Cross the Horizontal or Oblique Asymptote

The horizontal and oblique asymptotes describe the graph as x approaches infty or - infty , usually called the end behavior. The graph will approach the asymptote but not reach the asymptote at the ends of the graph. However, the graph can cross the horizontal or oblique asymptote in the middle section of the graph. To find were the graph crosses the horizontal or oblique asymptote, set the value of the asymptote equal to the function.

For this example, set the horizontal asymptote (y=0) equal to the function $f(x)=1/{x^2-9}$ .

Horizontal Asymptote = Function

$0=1 /{x^2-9}$

$0(x^2-9)=1 /{x^2-9}(x^2-0)$

0=1

Since there is no solution to this equation, there are no points where the graph crosses the horizontal asymptote.

Find the Intercepts of the Rational Function

To find a y-intecept let x=0.

$f(0)=1/{0^2-9}$

f(0)=1/{-9}

f(0)=-{1/9}

The y-intercept is (0,-{1/9})

To find the x-intercept let y=0 or f(x)=0. Exclude solutions that were found to be undefined for the function when the domain was found.

$0=1 /{x^2-9}$

$0(x^2-9)=1 /{x^2-9}(x^2-0)$

0=1

Since this equation has no solution. There is no x-intercept.

Sketch the Graph

Draw the asymptotes as dashed lines. These will be used as guides to draw the graph. For this example, the vertical asymptotes were at x=-3 and x=3 and the horizontal asymptote was at y=0.

Plot the intercepts, holes or places where the graph crosses an asymptote. For this example, the only value is the y-intercept.

More points are needed to sketch the graph. Choose points in every section of the graph, using the vertical asymptotes to divide the sections up.

For example, choose x=-4 to represent the left section. The middle section needs a few extra points since it is surrounded by asymptotes. Choose x=-2 and x=2 to represent the middle section. Choose x=4 to represent the right section.

x	y
-4	$1/{(-4)^2-9}=1/{16-9}=1/7$
-2	$1/{(-2)^2-9}=1/{4-9}=1/{-5}=-{1/5}$
2	$1/{2^2-9}=1/{4-9}=1/{-5}=-{1/5}$
4	$1/{4^2-9}=1/{16-9}=1/7$

Plot the points on the graph.

Use the definition of the asymptotes to draw the graph near each point.

In the left section, consider the graph as x approaches - infty ( x right - infty ).

Since the graph has a horizontal asymptote at y=0, y right 0 as x right - infty . As you draw the graph to the left of -4, approach the horizontal asymptote.

In the left section, consider the graph as x approaches -3 from the left ( x right-3^- .)

Since the graph has a vertical asymptote at x=-3, y right infty or y right - infty as x right -3^- . Since the point in the left section lies above the horizontal asymptote and it was determined that the graph does not cross the horizontal asymptote, . As you draw the graph to the right of -4 approach the vertical asymptote going towards positive infinity.

In the middle section, consider the graph as x approaches -3 from the right, ( x right-3^+ .)

Since the graph has a vertical asymptote at x=-3, y right infty or y right - infty as x right -3^+ . Since the point near -3 in the middle section lies below the horizontal asymptote and it was determined that the graph does not cross the horizontal asymptote, . As you draw the graph to the left of -2, approach the vertical asymptote going towards negative infinity.

In the middle section, the three points can be connected to each other since there are no undefined values between them.

In the middle section, consider the graph as x approaches 3 from the left, ( x right 3^- .)

Since the graph has a vertical asymptote at x=3, y right infty or y right - infty as x right 3^- . Since the point near 3 in the middle section lies below the horizontal asymptote and it was determined that the graph does not cross the horizontal asymptote, . As you draw the graph to the right of 2, approach the vertical asymptote going towards negative infinity.

In the right section, consider the graph as x approaches 3 from the right ( x right 3^+ .)

Since the graph has a vertical asymptote at x=3, y right infty or y right - infty as x right 3^+ . Since the point in the right section lies above the horizontal asymptote and it was determined that the graph does not cross the horizontal asymptote, . As you draw the graph to the right of 4 approach the vertical asymptote going towards positive infinity.

In the right section, consider the graph as x approaches infty ( x right infty ).

Since the graph has a horizontal asymptote at y=0, y right 0 as x right infty . As you draw the graph to the right of 4, approach the horizontal asymptote.

The sketch of the rational function $f(x)=1/{x^2-9}$ is below.

Here are some video examples that might help.

5.3 Complex Zeros; Fundamental Theorem of Algebra

Form a Polynomial given the Degree and Zeros

November 1, 2017 admin

Example: Form a polynomial f(x) with real coefficients having the given degree and zeros.

Degree 4; Zeros -2-3i; 5 multiplicity 2

Solution:

By the Fundamental Theorem of Algebra, since the degree of the polynomial is 4 the polynomial has 4 zeros if you count multiplicity.

There are three given zeros of -2-3i, 5, 5.

The remaining zero can be found using the Conjugate Pairs Theorem. f(x) is a polynomial with real coefficients. Since -2-3i is a complex zero of f(x) the conjugate pair of -2+3i is also a zero of f(x).

Now that all the zeros of f(x) are known the polynomial can be formed with the factors that are associated with each zero.

Since f(x) has a zero of 5, f(x) has a factor of x-5

Since f(x) has a second zero of 5, f(x) has a second factor of x-5

Since f(x) has a factor of -2-3i, f(x) has a factor of x-(-2-3i)

Since f(x) has a factor of -2+3i, f(x) has a factor of x-(-2+3i)

Form the polynomial using all of the factors. The leading coefficient will remain unknown.
Multiply the factors with complex numbers. Doing so will cancel the complex numbers from the expression Distribute the minus Multiply each term in one factor by each term in the other factor simplify combine like terms
Multiply the other pair of factors
Multiply the two trinomials by multiplying each term in the first trinomial by each term in the other trinomial and then combine like terms

The polynomial with degree 4 and zeros of -2-3i and 5 wiht multiplicity 2 is f(x)=a(x^4-6x^3-2x^2-30x+325)

4.6 Rational Functions and Their Graphs, 5.4 Properties of Rational Functions

Finding Horizontal and Oblique (Slant) Asymptotes

July 13, 2017 admin

4.6 Rational Functions and Their Graphs, 5.4 Properties of Rational Functions

Finding Vertical Asymptotes and Holes (Removable Discontinuity)

July 13, 2017 admin

http://https://www.youtube.com/watch?v=fWYmFpWzGTs

3.1 Functions, 3.1 Relations and Functions, 4.6 Rational Functions and Their Graphs, 5.4 Properties of Rational Functions

Finding Domain: Rational Function

July 13, 2017 admin

Example: Classify the function as a polynomial function, rational function, or root function, and then find the domain. Write the domain interval notation and set builder notation.

$h(x) = {x^2+8}/{x^2+x-20}$

Solution:

Classify the Function

Formal Definition	Practical Way of Identifying
Polynomial Function A polynomial function is a function of the form $f(x)=a_n x^n+a_{n-1} x^{n-1}+...+a_1 x+a_0$ where n is a non-negative integer {0, 1, 2, 3, 4, …} and the coefficients $a_n , a_{n-1,...,a_1,a_0$ are from the real numbers.	Look for the variables to be in the numerator. (If there is no fraction at all, they are in the numerator.) The variable should not be inside a radical or absolute value. The powers or exponents on the variables should be whole numbers. Whole numbers come from this list {0, 1, 2, 3, 4, …}
Rational Function A rational function is a function of the form where and are polynomial functions and is not equal to zero.	The numerator and denominator are polynomials. Most functions with variables in the denominator are considered Rational Functions but there are exceptions.
Root Function (even index) A root function is a function of the form where n is an even positive integer greater than or equal to 2.	The variable is inside or underneath a radical. The index of the radical is an even number. {2, 4, 6, 8, …} The square root is an even index although the index is not written.
Root Function (odd index) A root function is a function of the form where n is an odd positive integer greater than or equal to 2.	The variable is inside or underneath a radical. The index of the radical is an odd number. {3, 5, 7, 9, …} The cube root is an odd index.

Since the function $h(x) = {x^2+8}/{x^2+x-20}$ has a variable in the denominator and the numerator and denominator are polynomial functions this function is a rational function.

Find the Domain of a Rational Function

Set the denominator equal to zero.
Solve the equation. This equation is a quadratic equation and can be solved by factoring, completing the square or the quadratic formula.
Solve by factoring. Factor the expression on one side.
Use the zero product property and set each factor equal to zero.	or
Solve each equation.	or or or