College Notes

College Notes!

 

 

Table of Contents

 

Preliminaries – Exponents & Radicals Review 3

Exponents 3

Radicals 4

Factoring Review 6

Rational Expressions 10

Complex Numbers 13

Solving Equations and Inequalities 15

Linear Equations 15

Absolute Value Equations 18

Quadratic Equations 20

Other Algebraic Equations 25

Equations Containing Radicals 27

Solving Inequalities 31

Absolute Value Inequalities 36

Graphing and Functions 38

Graphing 38

Graphing Lines 41

Graphing Circles 45

Functions 46

Function Notation 47

Domain and Range 48

Polynomial Functions 49

Dividing Polynomials 49

Graphing Polynomials 52

Leading Term Test For Polynomials 54

Graphing Polynomials With Known Zeros 55

Multiplicity Rules For Polynomial Zeros 57

Exponential and Logarithm Functions 60

Exponential Functions 60

Logarithm Functions 62

Properties of Logarithms 65

Exponential and Logarithm Equations 67

Exponential Equations 67

Logarithm Equations 69

Applications for Exponential and Logarithm Equations 71

Exponential Growth and Decay 71

Earthquake Intensity 72

 

Preliminaries – Exponents & Radicals Review

 

Exponents

Let’s first recall the definition of exponentiation with positive integer exponents. 

 

So, for example,

35=33333 = 243

 

Below are the rest of the properties for exponents.

 

Exponent Properties

xmxn = xm+n

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Radicals

 

We’ll open this section with the definition of the radical. If n is a positive integer that is greater than 1 and a is a real number then,

where n is called the indexaa is called the radicand, and the symbol  is called the radical. The left side of this equation is often called the radical form and the right side is often called the exponent form.

 

Note as well that the index is required in these to make sure that we correctly evaluate the radical. There is one exception to this rule and that is square root. For square roots we have,

In other words, for square roots we typically drop the index.

 

Example

These are together to make a point about the importance of the index in this notation. Let’s take a look at both of these.

So, the index is important. Different indexes will give different evaluations so make sure that you don’t drop the index unless it is a 2 (and hence we’re using square roots).

 

Below are the properties for Radicals.

 

Radical Properties

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Factoring Review

 

Let’s start out by talking a little bit about just what factoring is. Factoring is the process by which we go about determining what we multiplied to get the given quantity. We do this all the time with numbers. For instance, here are a variety of ways to factor 12.

 

There are many more possible ways to factor 12, but these are representative of many of them.

A common method of factoring numbers is to completely factor the number into positive prime factors. A prime number is a number whose only positive factors are 1 and itself. For example, 2, 3, 5, and 7 are all examples of prime numbers. Examples of numbers that aren’t prime are 4, 6, and 12 to pick a few.

If we completely factor a number into positive prime factors there will only be one way of doing it. That is the reason for factoring things in this way. For our example above with 12 the complete factorization is,

 

Factoring polynomials is done in pretty much the same manner. We determine all the terms that were multiplied together to get the given polynomial. We then try to factor each of the terms we found in the first step. This continues until we simply can’t factor anymore. When we can’t do any more factoring we will say that the polynomial is completely factored.

Here is an example.

This is completely factored since neither of the two factors on the right can be further factored.

 

Greatest Common Factor

 

The first method for factoring polynomials will be factoring out the greatest common factor. When factoring in general this will also be the first thing that we should try as it will often simplify the problem.

To use this method all that we do is look at all the terms and determine if there is a factor that is in common to all the terms.

 

We notice that each term has an a in it and so we “factor” it out.

 

Below are the Factoring properties

 

Factoring Properties

 

a2 – b2 = (a + b)(a – b)

a4 – b4 = (a2 + b2)(a2 – b2) = (a2 + b2)(a + b)(a – b)

x2 + bx + c = (x + k1)(x + k2)  where k1 + k2 = b and k1● k2 = c.

ax2 + bx + c = (a1x + k1)(a2x + k2)  where a1● a2=a, k1● k2 = c,  and  a2k1 + a1k2 = b 

 

Factoring By Grouping

AC + BC + DA + DB = C(A + B) + D(A  + B) = (C + D)(A + B)

Example:  

2x2 + 9x + 10 =  

2x2 + 4x + 5x + 10  =  

2x(x + 2) + 5(x + 2) = 

(x + 2)(2x + 5) 

The key here is to find two numbers that add to 9 but multiply out to 2×10 = 20. Those numbers are 4 and 5 so you rewrite 9x as 4x+5x.

There are nearly countless other forms of factorization, but these are the ones most commonly used. 

Examples:

 

Factor 9x2 – 49y4

 

9x2 – 49y4 =

 

(3x)2 – (7y2)2 =

 

(3x + 7y2)(3x – 7y2

 

Factor x2 – 2x – 120

 

We need to find two numbers that add to –2 but multiply to –120. These two numbers will be 10 and –12.  The factorization is then

 

(x + 10)(x + -12)  = 

 

(x + 10)(x  – 12)

 

Factor 2x2 + 3x – 14  by listing & checking possibilities resulting in 2x2 and –14.

 

2x+ 3x – 14 =  the following possibilities that result in 2×2 and –14 when multiplied out. But when we check the multiplication, only (2x + 7)(x – 2) results in 

2x+ 3x – 14.

 

(2x – 2)(x + 7)  = 2x2 + 14x – 2x – 14 =  2x2 + 12x – 14   NO!

 

(2x – 7)(x + 2)  = 2x2 + 4x – 7x – 14 =  2x2 – 3x – 14     NO!

 

(2x + 2)(x – 7)  = 2x2 – 14x + 2x – 14 =  2x2 – 12x – 14     NO!

 

(2x + 7)(x – 2)  = 2x2 – 4x + 7x – 14 =  2x2 + 3x – 14     YES!

 

Note that there were 4 more possible combinations that did not work involving 14 and 1 as second terms like (2x – 1)(x + 14). Often you will find the correct combination by only trying a small fraction of the total possible combinations. 

 

Factor 2x2 – 2x – 84

 

First, factor out a greatest common factor of 2 to rewrite 2x2 – 2x – 84 as 

 

2(x2 – x – 42) = 

 

2(x – 7)(x + 6)

 

Factor 2x2 –  3y – 6x + xy   by Grouping

 

First  form 2 groups, each containing a common factor. Also, change all subtractions into the addition of negatives.  We get

 

2x2  + xy  + -3y + -6x  

 

Now, factor out common factors. 

 

x(2x + y) + -3(y + 2x)  which is the same as x(2x + y) – 3(2x + y)  and now factor out (2x + y) to get

 

(2x + y)(x – 3)

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Rational Expressions

 

We now need to look at rational expressions. A rational expression is nothing more than a fraction in which the numerator and/or the denominator are polynomials. Here are some examples of rational expressions.

 

The first topic that we need to discuss here is reducing a rational expression to lowest terms. A rational expression has been reduced to lowest terms if all common factors from the numerator and denominator have been canceled. We already know how to do this with number fractions so let’s take a quick look at an example.

 

With rational expression it works exactly the same way.

 

Example: 

 

We now need to move into adding, subtracting, multiplying and dividing rational expressions.

Let’s start with multiplying and dividing rational expressions. The general formulas are as follows,

 

Example:

 

 

Okay, it’s time to move on to addition and subtraction of rational expressions. Here are the general formulas.

 

Example:

 

 

Complex Numbers

 

If we try to solve x2 = -1, what happens?

 

We extract square roots to get x = +/- -1.  But if we try to evaluate the square root of –1 on a scientific calculator, we get ERROR!  But still, we need a way to define solutions like this so it is defined that

 

i2 = -1 and thus i = (-1).  

 

This means that the solutions of x2 = -1 are x = i and x = -i

 

We refer to such solutions as Complex Solutions

 

Furthermore, we refer to a number containing the quantity “i”, where i = -1, as an imaginary number.  This choice of words “imaginary” is actually not appropriate, since we use the number “i” in many real-world engineering applications!

 

Using Complex Numbers To Evaluate Square Roots

 

Given that i2 = -1 and thus i = (-1), we can use this fact to evaluate any square root. 

For example, (-13) = (-1 13) = (-1) 13  and we can replace (-1)  with i to get   

(-13) = i ∙√13.

 

In general, we can say that 

 

Standard Form of a Complex Number

Complex Numbers consist of the set of all numbers of the form a + bi where,  “a” is the Real Part and “bi” is the Imaginary Part.

 

It turns out that all numbers may be written in this form.  For numbers that are regular old Real numbers, there is no i-part so b=0.  For example, we may write 6 as 6 + 0i. Some numbers, like 3 + 2i, have both a real and imaginary part, with a = 3 and b = 2.   And some numbers, like 4i have no Real part and may be written as 0 + 4i.  We sometimes call numbers like 4i, that have no Real part, as purely imaginary.

 

Operations With Complex Numbers

All the standard properties that apply to Real Numbers, like the Distributive, Commutative, and Associative Properties, also apply to Complex Numbers.  This means that you can use the following simple rule:

 

EASY RULE OF THUMB

 

When performing operations with complex numbers, treat the “i” like any other variable, except replace any occurrence of i2 with –1. 

Example: Multiply out (3 + 2i)(5 – 6i) and write the answer in Standard Form.

 

We simply multiply this out using the Distributive Property (FOIL Method) as if we were multiplying out (3 + 2x)(5 – 6x).

 

(3 + 2i)(5 – 6i) 

=  3 5  –  3 6i  +  2i 5  –  2i 6i  by the Distributive Property

= 15 – 18i + 10i –12i2 after multiplying terms 

= 15 – 8i – 12i2  after combining like terms

= 15 – 8i – 12(-1) after replacing i2 with –1

= 27 – 8i  or  27 + (-8i)  after combining like terms

 

The Complex Conjugate

 

For any complex number a + bi, the complex conjugate is defined as a – bi. 

If we multiply (a+bi)(a – bi), we get

 

(a+bi)(a – bi) 

= a2 – abi + abi – b2i2 

= a2 – b2i2

= a2 – b2(-1)

= a2 + b2

The i-part always disappears!

 

Example: Multiply out (3 + 2i)(3 – 2i)

(3 + 2i)(3 – 2i) 

=  3 3  –  3 2i  +  2i 3  –  2i 2i  by the Distributive Property

=  9 – 6i + 6i – 4i2 after multiplying terms 

=  9 – 4i2  after combining like terms

=  9 – 4(-1) after replacing i2 with –1

=  13  after combining like terms

 

Or, if you used the generalization, you end up with a2 + b2 = 32 + 22 = 13.

 

Dividing Complex Numbers

 

To divide by a complex number, we can always get an answer in complex standard form by using the following rule:

 

Do not panic!  You don’t need to memorize this formula.  Simply remember to multiply top and bottom of the fraction by the conjugate of the denominator. Simplify your answer and it will all work out.  See the next page for an example.

 

Example: Divide (3 + 2i) ÷ (4 –3i) and write the answer in standard form. 

 

First, write this as a fraction.

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Solving Equations and Inequalities 

 

Linear Equations

 

linear equation is any equation that can be written in the form

 

where a and b are real numbers and x is a variable. This form is sometimes called the standard form of a linear equation. Note that most linear equations will not start off in this form. Also, the variable may or may not be an x so don’t get too locked into always seeing an x there.

 

To solve linear equations we will make heavy use of the following facts.

  1. If a=b then a+c = b+c for any c. All this is saying is that we can add a number, c, to both sides of the equation and not change the equation.
  2. If a=b then a−c = b−c for any c. As with the last property we can subtract a number, c, from both sides of an equation.
  3. If a=b then ac = bc for any c. Like addition and subtraction, we can multiply both sides of an equation by a number, c, without changing the equation.
  4. If a=b then ac=bc for any non-zero c. We can divide both sides of an equation by a non-zero number, c, without changing the equation.

Method To Solve Linear Equations:

 

To solve linear equations, remember to do the following:

 

  1. If the equation contains any fractions use the least common denominator to clear the fractions. We will do this by multiplying both sides of the equation by the LCD. 

    Also, if there are variables in the denominators of the fractions identify values of the variable which will give division by zero as we will need to avoid these values in our solution.
  2. Simplify both sides of the equation. This means clearing out any parenthesis and combining like terms.
  3. Use the first two facts above to get all terms with the variable in them on one side of the equations (combining into a single term of course) and all constants on the other side.
  4. If the coefficient of the variable is not a one use the third or fourth fact above (this will depend on just what the number is) to make the coefficient a one. 

    Note that we usually just divide both sides of the equation by the coefficient if it is an integer or multiply both sides of the equation by the reciprocal of the coefficient if it is a fraction.
  5. VERIFY YOUR ANSWER! This is the final step and the most often skipped step, yet it is probably the most important step in the process. With this step you can know whether or not you got the correct answer long before your instructor ever looks at it. We verify the answer by plugging the results from the previous steps into the original equation. It is very important to plug into the original equation since you may have made a mistake in the very first step that led you to an incorrect answer.

     

 

Example:

 

For this problem there are no fractions so we don’t need to worry about the first step in the process. The next step tells to simplify both sides. So, we will clear out any parenthesis by multiplying the numbers through and then combine like terms.

 

 

The next step is to get all the x’s on one side and all the numbers on the other side. So, for our case this will mean adding 4x to both sides and subtracting 15 from both sides. 

 

The next step says to get a coefficient of 1 in front of the x. In this case we can do this by dividing both sides by a 7.

 

The last and final step is to then check the solution.

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Absolute Value Equations

 

Equations with a variable or variables within absolute value bars are known as Absolute Value Equations

 

Examples:

 

x – 3 =  5 

 

2x – 3 + x =  2 

 

Method To Solve Absolute Value Equations:

To solve absolute value equations, remember to do the following:

 

  • Isolate the absolute value expression on one side of =. So for example, use the Addition Property of Equality to subtract x from both sides of 2x – 3 + x =  2  to result in 2x – 3 =  2 – x   .

 

  • Use the Absolute Value Equation Property to solve two cases without the absolute values, one positive and one negative.  In the above example, you would solve                                                     2x – 3 =  2 – x    and  2x – 3 =  – (2 – x).

  

  • After solving, check all answers.  You may get extraneous solutions!  In the example above, we would get answers of x=5/3 and x=1.  It turns out that both work. 

 

Example:  Solve 2x – 3 =  2   

 

First, apply the Absolute Value Equation Property to rewrite as two problems.

2x – 3 =  2    and   2x – 3 =  -2    

 

Next, use the Addition Property of Equality to move all terms to one side in both equations, resulting in 

2x – 3 + 3 =  2 + 3    and   2x – 3 + 3 =  -2 + 3

2x  =  5     and   2x  =  1

 

Use the Division Property of Equality  to solve both to get

x = 5/2  and  x = 1/2.  

 

Check both answers in 2x – 3 =  2   

2(5/2) – 3 =  2   and   2(1/2) – 3 =  2  are both true!

 

Example:  Could you solve x2 – 1 =  x  using this same method? Explain how. 

 

Answer: Yes

  

First, apply the Absolute Value Equation Property to rewrite as two problems.

x2 – 1 =  x   and   x2 – 1 =  – x      

 

Next, use the Addition Property of Equality to move all terms to one side in both equations, resulting in 

x2 – 1 + 1 =  x + 1    and   x2 – 1  + 1 =  -x + 1

x2  =  x + 1     and   x2  =  1 – x

 

You now have two quadratic equations that can be rewritten as 

x2 – x – 1 = 0     and   x2 +x – 1 =  0 by applying the Addition Property of Equality.

 

You could then solve these using the Quadratic Formula as shown below. You would then have to check all four solutions!

 

The easiest way to check these solutions is to use the decimal forms of the answers and plug them into x2 – 1 =  x  .

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Quadratic Equations

First, the standard form of a quadratic equation is

The only requirement here is that we have an x2 in the equation. We guarantee that this term will be present in the equation by requiring a≠0. Note however, that it is okay if b and/or c are zero.

There are 4 main ways to solve quadratic equations. 

 

  • Method 1: Solve by Factoring

 

As the heading suggests we will be solving quadratic equations here by factoring them. To do this we will need the following fact.

This fact is called the zero factor property or zero factor principle.

 

To solve a quadratic equation by factoring we first must move all the terms over to one side of the equation. Doing this serves two purposes. First, it puts the quadratics into a form that can be factored. Secondly, and probably more importantly, in order to use the zero factor property we MUST have a zero on one side of the equation. 

 

Example: 

First, get everything on side of the equation and then factor.

 

Now at this point we’ve got a product of two terms that is equal to zero. This means that at least one of the following must be true.

 

Note that each of these is a linear equation that is easy enough to solve. What this tell us is that we have two solutions to the equation, x=4 and x=−3. As with linear equations we can always check our solutions by plugging the solution back into the equation. We will check x=−3 and leave the other to you to check.

 

  • Method 2: Square Root Property

 

The second method of solving quadratics we’ll be looking at uses the square root property,

 

This is a fairly simple property to use, however it can only be used on a small portion of the equations that we’re ever likely to encounter. 

 

Example:

 

This is a fairly simple problem so here is the work for this equation.

 

 

So, there are two solutions to this equation, x=±10. Remember this means that there are really two solutions here, x=−10 and x=10.

 

  • Method 3: Solving the Square

 

We should first define just what completing the square is.

Let’s start with:

and notice that the x2 has a coefficient of one. That is required in order to do this. Now, to this lets add (b2)2. Doing this gives the following factorable quadratic equation.

 

This process is called completing the square and if we do all the arithmetic correctly we can guarantee that the quadratic will factor as a perfect square.

 

Example:

 

Step 1 : Divide the equation by the coefficient of the x2 term. Recall that completing the square required a coefficient of one on this term and this will guarantee that we will get that. We don’t need to do that for this equation however.

Step 2 : Set the equation up so that the x’s are on the left side and the constant is on the right side.

 

Step 3 : Complete the square on the left side. However, this time we will need to add the number to both sides of the equal sign instead of just the left side. This is because we have to remember the rule that what we do to one side of an equation we need to do to the other side of the equation.

First, here is the number we add to both sides.

Now, complete the square.

Step 4 : Now, at this point notice that we can use the square root property on this equation. That was the purpose of the first three steps. Doing this will give us the solution to the equation.

 

  • Method 4: The Quadratic Formula

 

This is the final method for solving quadratic equations and will always work. Not only that, but if you can remember the formula it’s a fairly simple process as well.

 

Provided that we start off in standard form,

 

and that’s very important, then the solution to any quadratic equation is,

 

Example:

 

So, the first thing that we need to do here is to put the equation in standard form.

 

At this point we can identify the values for use in the quadratic formula. For this equation we have.

 


At this point there really isn’t anything more to do other than plug into the formula.

 

We can further simplify this to:

 

Other Algebraic Equations

 

Including Factorable Higher Degree Polynomials, Rational Equations, Radical Equations, Equations With Rational Exponents, and Quadratic Form Equations

 

Factorable Higher Degree Polynomials

 

A factorable higher degree polynomial is defined here as one that has a greatest common monomial factor that factors out in such a way that the resulting factors consist of the monomial and a polynomial of degree 2 or less, allowing easy solution. 

 

Here is an example:  3x4 – 2x3 + x2 = 0 may be factored as x2(3x2 – 2x + 1) = 0 by applying the Distributive Property.

 

Method For Solving These Factorable Higher Degree Polynomials

 

  • Move all terms to one side of = leaving 0 on one side using the Addition Property of Equality.
  • Factor out the greatest common factor using the Distributive Property.
  • Let each factor = 0 and solve each equation using an appropriate method.

 

 

Example: Solve 2x4 – 3x3 = –x2

 

2x4 – 3x3 =  –x2              Given

 

2x4 – 3x3 + x2 = 0      Add x2 to both sides using Addition Property of Equality 

 

x2(2x2 – 3x + 1) = 0   Factor out x2 using the Distributive Property

 

x2 = 0  and                Apply the Zero Product Law

2x2 – 3x + 1 = 0

 

Solve x2 = 0

x = ±√0   x= 0        Extract Square Roots    

 

Solve 2x2 – 3x + 1 = 0

(2x – 1)(x – 1) = 0       Factor using the Distributive Property

x –1 = 0                     Apply the Zero Product Law

2x – 1 = 0

x = 1  and x = ½        Solve using Addition and Division Properties of Equality 

 

The three solutions are x = 0, x = 1, and  x = ½      

 

Remember to find all three solutions – many students forget the zero solution!

 

Equations Containing Rational Expressions

 

Equations containing rational expressions are here defined as those containing fractions with variables.  For example, 1/x + 1/3 = 1/(x+1) would be such an equation.  

 

Method For Solving Equations Containing Rational Expressions

  • Identify the Least Common Denominator (LCD) of all the fractions. In some instances you may need to factor the denominators first.
  • Multiply both sides (all terms) by the LCD using the Distributive Property. This will allow you to cancel all denominators and leave you with a fraction-less equation.
  • Solve this equation using the appropriate method.
  • Check your solution.  Sometimes you will get a false solution even though you did all the work correctly!

 

 

2x –x + 1 = 3      Apply the Distributive Property and multiply out by –1

x + 1 = 3             Combine Like Terms  

x = 2                   Add –1 to both sides using the Addition Property of Equality

Does x=2 check out in the original equation? If we plug it in, we get 2 – ½ = 3/2 which is true.  Thus, x=2 is the solution.

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Equations Containing Radicals

 

Equations containing radicals are defined here as those containing the variable that we are solving for within a radical sign.  For example, the equation   

 

Method For Solving Equations Containing Radicals

  • Isolate the radical on one side of equals using Properties of Equality.
  • Raise both sides of equation to a power that eliminates the radical. if this is a square root equation, square both sides. If it is an nth root equation, raise both sides to the nth power.
  • Solve this equation using the appropriate method.
  • Check your solution.  Many times you will get one or more false solutions, called extraneous solutions, even though you did all the work correctly!

 

 

    x2 – 1  =  (x – 3)(x – 3)

    x2 – 1 = x2 –6x + 9      Apply Distributive Property and combine like terms

   –10   = -6x                   Add –x2 and –9 to both sides using Addition Property of Equality

   10/6 = 5/3 = x                  Divide both sides by –6 using the Division Property of Equality 

 

But, we must check this solution!    

The answer is No Solution

 

Equations Containing Rational Exponents

Equations containing rational exponents are defined here as those with quantities raised to fractional powers on one side of equal and real numbers on the other such as the equation (x + 4)2/3 = 7.  This type of equation is solved as follows:

 

Method For Solving Equations Containing Quantities Raised To The M/N Power

  1. Isolate the quantity raised to the M/N power on one side of equals using Properties of Equality.
  2. Raise both sides to the N/M power. If M of M/N is even, include a  ± sign on the numerical side.  If M of  M/N is odd, do not include a ± sign.  The Multiplication Property of Exponents will cause the powers on your quantity to multiply to 1 and cancel. 
  3. Solve using the appropriate method.
  4. Check answers.  As with radical equations, it is possible to get extraneous solutions.

 

Example: Solve (3 + 2x)2/5 + 3 = 7

(3 + 2x)2/5 + 3 = 7           Given

(3 + 2x)2/5 = 4                 Add –3 to both sides using Addition Property of Equality

[(3 + 2x)2/5]5/2 = ±45/2         Raise both sides to 5/2 power.  Include ± sign.

3 + 2x = ±45/2                 Exponent Rule results in exponents canceling  

2x = -3 ± 45/2                  Add –3 to both sides using Addition Property of Equality

x = (-3  ± 45/2) / 2            Divide both sides by 2 using Division Property of Equality

x = (-3 ± 32)/2                Simplify the power of 4

x = 29/2  or x = -35/2     

 

Now, you must check these answers in (3 + 2x)2/5 + 3 = 7!

If x = 29/2, (3 + 2(29/2))2/5 + 3 =  (32)2/5 + 3 = 22 + 3 = 4 + 3 = 7

If x = -35/2, (3 + 2(-35/2))2/5 + 3 =  (-32)2/5 + 3 = (-2)2 + 3 = 4 + 3 = 7

Both answers check, so x = 29/2  or x = -35/2 are the solutions    

 

Why the ± sign in some cases and not in others?

 

When solving an equation with rational power with even numerator power, 

like (x + 3)2/3 = 9, you are really solving the same equation as 

To solve this equation, we extract square roots and then cube both sides. Extracting any even roots requires the use of a ± sign to result in two possible answers.

 

Conversely, when solving an equation like (x + 1)3/2 = 8, we are solving the same equation as 

To solve this equation, we take the cube root of  both sides and then square both sides to eliminate the radical.  Extracting cube roots does not require a ± sign. There will only one possible solution to this equation. 

 

Quadratic-Form Equations

Equations that may be written in the form ax2N + bxN + c = 0 are known as quadratic form equations. For example, the equation 2x8 + 4x4 + 2 = 0 is a quadratic-form equation since the power of the first term, 8, is twice the power of the second term 4. The third term is a real number, 2.   There are two methods for solving quadratic-form equations given below.

 

Method One For Solving Quadratic Form Equations

 

  • Move all terms to one side of = by using the Addition Property of Equality.
  • Factor the equation into 2 binomials, if possible using the Distributive Property.  If this is not possible, use Method Two.
  • Let each factor = 0 as justified by the Zero Product Law.
  • Solve each resulting equation using the appropriate method.
  • Check your solution if radical equations or rational exponents were involved, you may get extraneous solutions.

 

 

Example:  Solve x4 – x2  =  12 

x4 – x2  =  12 Given

x4 – x2 – 12 = 0 Add –12 to both sides using Addition Property of Equality

(x2 – 4)(x2 + 3) = 0 Factor using Distributive Property

x2 – 4 = 0 Let each factor = 0 by Zero Product Law

x2 + 3 = 0 

x2 =  4 Add 4 to both sides using Addition Property of Equality

x2 = – 3 Add –3 to both sides using Addition Property of Equality

x = ±√4 Extract Square Roots

x = ±√(-3) Extract Square Roots

x = 2, x = -2 Simplify radicals

x = (3)i, x = -(3)i    Simplify using definition of “i”

Since no radicals nor rational exponents were involved, there are no extraneous solutions to check for.

 

Method Two For Solving Quadratic Form Equations

 

  • Move all terms to one side of = by using the Addition Property of Equality.
  • For the equation ax2N + bxN + c = 0, let u = xN.  This results in u2 = x2N.  Make these substitutions into your equation.
  • Solve your equation au2 + bu + c = 0  for u using an appropriate method.
  • Let each solution for u be equal to xN. Solve each of these equations for x.
  • Check your solutions if radical equations or rational exponents were involved, you may get extraneous solutions.

 

 

Example:  Solve 2x4  = 5 + 3x2

2x4  = 5 + 3x2 Given

2x4 – 3x2–  5 = 0 Move Terms to one side using Addition Property of Equality

Let u = x2, u2 = x4 Make substitutions

2u2 – 3u – 5 = 0

 

Solve for u using quadratic formula

 

u = 10/4  or u = -1

 

x2 = 10/4 and x2 = -1 Let x2 equal each of the solutions for u

 

x = ±√(10/4) Extract square roots.  Then simplify radicals.

x = ±√(-1) Apply definition of “i”.

 

x = (10)/2   x = -(10)/2   x = i   x = -i

 

Note: We could have solved the quadratic by factoring 

 

Remember to let each solution for u equal your power of x and solve for x – many students forget to do this!

 

Solving Inequalities

 

An inequality is the result of replacing the = sign in an equation with <, >, , or .  For example, 3x – 2 < 7 is a linear inequality.  We call it “linear” because if the < were replaced with an = sign, it would be a linear equation.  Inequalities involving polynomials of degree 2 or more, like 2x3 – x > 0,  are referred to as polynomial inequalities, or quadratic inequalities if the degree is exactly 2.  Inequalities involving rational expressions are called rational inequalities.  Some often used inequalities also involve absolute value expressions. 

 

Solving Inequalities: A Summary

In a nutshell, solving inequalities is about one thing: sign changes. Find all the points at which there are sign changes – we call these points critical values.  Then determine which, if any, of the intervals bounded by these critical values result in a solution. The solution to the inequality will consist of the set of all points contained by the solution intervals. 

 

Method To Solve Linear, Polynomial, or Absolute Value Inequalities:

 

  • Move all terms to one side of the inequality sign by applying the Addition, Subtraction, Multiplication, and Division Properties of Inequalities. You should have only zero on one side of the inequality sign.
  • Solve the associated equation using an appropriate method. This solution or solutions will make up the set of critical values.  At these values, sign changes occur in the inequality.
  • Plot the critical values on a number line. Use closed circles for and inequalities, and use open circles ο for < and > inequalities.
  • Test each interval defined by the critical values. If an interval satisfies the inequality, then it is part of the solution.  If it does not satisfy the inequality, then it is not part of the solution.

 

 

Example:  Solve 3x + 5(x + 1)    4x – 1 and graph the solution

 

3x + 5(x + 1)    4x – 1 Given

3x + 5x + 5    4x – 1 Distributive Property

8x + 5    4x – 1 Combine Like Terms

4x + 6    0 Subtract 4x from both sides, add 1 to both sides using Addition and Subtraction Properties of Inequality

Now, solve 4x+6 = 0

4x = -6 Addition Property of Equality

x = – 6/4 = -3/2 Division Property of Equality

 

Plot the critical value

Test arbitrary values of each interval. You may test the value in the original inequality or it’s simplified form.  

 

In Interval 1, let x=1 in 4x + 6    0. 

4(1) + 6 0 is FALSE.

 

In Interval 2, let x = -2 in 4x + 6   0. 

4(-2) + 6 0 is TRUE.  So Interval 2 is the solution.

 

The solution is x – 3/2, or in interval notation, (- , – 3/2].  The graph is 

Note: It is more convenient to use the circle notation for endpoints of the graph rather than the interval bracket notation since we do not know which way the brackets will point until the inequality intervals are tested.

 

Example:  Solve 3x3 + 5x2  >  4x2 and graph the solution

 

3x3 + 5x2  >  4x2 Given

3x3 + x2  >  0 Subtraction Property of Inequalities

  

Now, solve 3x3 + x2  =  0

x2(3x + 1) = 0 Distributive Property

x2 = 0 or 3x + 1 = 0 Zero Product Law

 

Solve x2 = 0

x=±√0 = 0 Extract Square Roots  

 

Solve 3x + 1 = 0

3x = -1 Add –1 to  using Addition Property of Equality

x = -1/3 Division Property of Equality

 

Plot the critical values, x = 0 and x =-1/3. Use open circles this time!

Test arbitrary values of each interval. You may test the value in the original inequality or it’s simplified form.  

 

In Interval 1, let x=1 in x2(3x + 1) > 0.

12(31 + 1) > 0 is TRUE, so Interval 1 is part of the solution.

 

In Interval 2, let x=-1/4 in x2(3x + 1) > 0.

(-1/4)2(3(-1/4) + 1) > 0 is TRUE,  since simplified, we get (1/16)(1/4) > 0,

so Interval 2 is part of the solution.

 

In Interval 3, let x=-1 in x2(3x + 1) > 0.

(-1)2(3(-1) + 1) > 0 is FALSE, since simplified, we get (1)(-2) > 0,

so Interval 3 is NOT part of the solution.

 

We shade Interval 1 and Interval 2, but do not include the endpoints.

The solution is x>0 or –1/3 < x < 0.  

The interval notation of this solution is (0 , ) U (-1/3, 0).

 

A COMMON MISTAKE TO AVOID! 

Students often get to the equation x2(3x + 1) = 0 and then divide both sides by x2 and solve 

3x + 1 = 0, thus losing the zero solution. Whenever you divide both sides of an algebraic equation by x or a power of x, you incorrectly lose the zero solution.

 

Example:  Solve 2x + 5  > 1 and graph the solution

 

2x + 5  > 1 Given

Solve the related equation 2x + 5  = 1

 

To solve absolute value equations, you must solve two cases:

2x + 5 = 1  and   -(2x + 5) = 1 

 

Solve 2x + 5 = 1

2x = -4 Add –5 to both sides using Addition Property of Equality

x = -4/2 = -2 Division Property of Equality

 

Solve –(2x + 5) = 1

-2x – 5 = 1 Multiply through by –1 using Distributive Property

-2x = 6 Add 5 to both sides using Addition Property of Equality

x = 6/(-2) = -3 Division Property of Equality

 

Plot the critical values, x = 0 and x =-1/3. Use open circles since this is >.

 

Plot the critical values, x = 0 and x =-1/3. Use open circles since this is >.

 

Test arbitrary values of each interval. You may test the value in the original inequality or it’s simplified form.  

 

In Interval 1, let x=0 in2x + 5 > 1.

20 + 5  > 1  is TRUE, so Interval 1 is part of the solution.

 

In Interval 2, let x = -2.5 in 2x + 5 > 1.

2(-2.5) + 5  > 1  is FALSE, so Interval 2 is NOT part of the solution.

 

In Interval 3, let x=- 4 in 2x + 5 > 1.

2(-4) + 5  > 1  is TRUE, so Interval 3 is part of the solution.

 

We shade Interval 1 and Interval 3, but do not include the endpoints.

 

The solution is x > -2 or x < -3. In interval notation, we would write this as

(- , -3) U (-2, ).

 

Method To Solve Rational Inequalities:

 

  • Move all terms to one side of the inequality sign by applying the Addition, Subtraction, Multiplication, and Division Properties of Inequalities. You should have only zero on one side of the inequality sign.
  • Solve the associated equation using an appropriate method. This solution or solutions will make up the set of critical values.  At these values, sign changes occur in the inequality. 
  • Find all values that result in Division By Zero. These are also critical values for rational inequalities.
  • Plot the critical values on a number line. Use closed circles for and unless the value results in division by zero – always use open circles for values resulting in division by zero since this value can not be part of the solution!   Always use open circles ο for < and > inequalities.
  • Test each interval defined by the critical values. If an interval satisfies the inequality, then it is part of the solution.  If it does not satisfy the inequality, then it is not part of the solution.

 

 

In summary, the only difference between solving a rational inequality and a polynomial inequality is that there are additional critical values that result in division by zero, and you never include these additional values as part of the solution, even if it is a   or inequality. 

 

Solve the related equations (-2x + 10)/(x – 3) = 0  and x – 3 = 0.

 

(-2x + 10)/(x – 3) = 0  

-2x + 10 = 0          Clear fractions by multiplying both sides by (x-3)

-2x = -10         Add –10 to both sides using Addition Property of Equality

x = -10/(-2) = 5         Division Property of Equality

 

x – 3 = 0

x = 3         Addition Property of Equality

 

Plot the critical numbers.  Use a closed circle for x=5 but an open circle for x=3.

 

In Interval 1, we let x = 6. This results in (-2 6 + 10)/(6 – 3) 0  or 

–2/3 0 which is TRUE.  So Interval 1 is part of the solution.

 

In Interval 2, we let x = 4. This results in (-2 4 + 10)/(4 – 3) 0  or 

2/1 0 which is FALSE.  So Interval 2 is NOT part of the solution.

 

In Interval 3, we let x = 2. This results in (-2 2 + 10)/(2 – 3) 0  or 

6/(-1) 0 which is TRUE.  So Interval 3 is part of the solution.

 

This results in a graph of

with solution x < 3 or x 5.  Interval notation is (- , 3) U [5, ).

 

Absolute Value Inequalities 

 

In the previous section we solved equations that contained absolute values. In this section we want to look at inequalities that contain absolute values. We will need to examine two separate cases.

Inequalities Involving < and 

As we did with equations let’s start off by looking at a fairly simple case.

This says that no matter what p is it must have a distance of no more than 4 from the origin. This means that p must be somewhere in the range,

 

We could have a similar inequality with the < and get a similar result.

In general, we have the following formulas to use here,

Example: 

 

There really isn’t much to do other than plug into the formula. As with equations pp simply represents whatever is inside the absolute value bars. So, with this first one we have,

 

Now, this is nothing more than a fairly simple double inequality to solve so let’s do that.

 

The interval notation for this solution is 

 

Inequalities Involving > and 

Once again let’s start off with a simple number example.

 

This says that whatever p is it must be at least a distance of 4 from the origin and so p must be in one of the following two ranges,

 

These solutions must be written as two inequalities.

Here is the general formula for these.

 

Example:

 

Again, p represents the quantity inside the absolute value bars so all we need to do here is plug into the formula and then solve the two linear inequalities.

 

The interval notation for these are 

 

Graphing and Functions

 

Graphing

 

We will start off with the Rectangular or Cartesian coordinate system. This is just the standard axis system that we use when sketching our graphs. Here is the Cartesian coordinate system with a few points plotted.

 

Each point in the coordinate system is defined by an ordered pair of the form (x,y). The first number listed is the x-coordinate of the point and the second number listed is the y-coordinate of the point. The ordered pair for any given point, (x,y), is called the coordinates for the point.

 


If you are graphing equations for the first time and you have no specialized method to use, you can graph many equations by doing the following:

 

  1. Find and plot all y-intercepts. To find these let x = 0 and solve for y. Then find and plot all x-intercepts. To find these, let y=0 and solve for all values of x.
  2. Find and plot more points on in between the intercepts you plotted and on each side of the intercepts you plotted.
  3. Draw a smooth curve through your plotted points from left to right. 

 

Example

 

It helps to have an idea of what this graph is liable to look like when picking values of x. Here is a table of values for this equation.

 

Let’s verify the first one and we’ll leave the rest to you to verify. For the first one we simply plug x= −2 into the equation and compute y.

 

Here is the graph of this equation.

 

Graphing Lines

 

Let’s start this section off with a quick mathematical definition of a line. Any equation that can be written in the form,

 

where we can’t have both A and B be zero simultaneously is a line. It is okay if one of them is zero, we just can’t have both be zero. Note that this is sometimes called the standard form of the line.

 

Slopes

 

The slope of a line is a measure of the steepness of a line and it can also be used to measure whether a line is increasing or decreasing as we move from left to right. Here is the precise definition of the slope of a line.

Given any two points on the line say, (x1,y1) and (x2,y2), the slope of the line is given by, 

You will often hear the slope as being defined as follows,

 

Example

 

Determine the slope of the following line. Sketch the graph of the line.

 

So, here is the slope for this part.

 

Be careful with minus signs in these computations. It is easy to lose track of them. Also, when the slope is a fraction, as it is here, leave it as a fraction. Do not convert to a decimal unless you absolutely have to.

Here is a sketch of the line.

 

Point-slope form

 

The next special form of the line that we need to look at is the point-slope form of the line. This form is very useful for writing down the equation of a line. If we know that a line passes through the point (x1,y1) and has a slope of m then the point-slope form of the equation of the line is,

 

Slope-intercept form

 

The final special form of the equation of the line is probably the one that most people are familiar with. It is the slope-intercept form. In this case if we know that a line has slope m and has a y-intercept of (0,b) then the slope-intercept form of the equation of the line is,

 

Example

 

Determine the slope of the following equation and sketch the graph of the line.

 

First solve the equation for y.

 

 

 

So, the slope for this line is 3 and the y-intercept is the point (0,−1).

 

Now, to find the second point we usually like the slope written as a fraction to make it clear what the rise and run are. So,

 

The second point is then,

 

Here is a sketch of the graph of the line.

 

Parallel and Perpendicular lines

 

Suppose that the slope of Line 1 is m1 and the slope of Line 2 is m2. We can relate the slopes of parallel lines and we can relate slopes of perpendicular lines as follows.

 

 

Graphing Circles

 

This is the standard form of the equation of a circle with radius r and center (h,k).

 

Example: 

 

If a circle has a center at (3,-1) with radius 2, what is its equation? 

 

Here, h=3, k = -1, and r = 2. We substitute these values into the circle formula to get (x – 3)2 + (y – (-1))2 = 22 which simplifies to 

 

(x – 3)2 + (y +1)2 = 4

 

Example 2:

 

Determine the center and radius of the following circle and sketch the graph of the circle.

 

It looks like the x coordinate of the center is zero. However, with the y term, we can see that the y coordinate of the center must be 3. The center and radius of this circle is then,

 

Here is a sketch of the circle. The center is marked with a red cross in this graph.

 

Functions

 

A function from the set of x-values to the set of y-values is a rule or formula where xvalues are input and the function assigns exactly 1 y-value to each x-value that is input. 

 

The set of all x-values that are input is called the DOMAIN.

The set of all y-values that correspond to the given x-values is called the RANGE.

 

Examples of equations that represent y as a function of x, and some that don’t:

 

  • y = 2x + 1 is a function of x since each x-value input results in only 1 y-value.
  • |y| =x is NOT a function of x since x=9 corresponds to both y = 9 and y = -9.
  • y = x2 is a function of x since each x-value input results in only 1 y-value.
  • y2 = x is NOT a function of x since x = 4 results in y = 2 and y = -2.
  • y = ±√x is NOT a function of x since x = 4 results in y = 2 and y = -2.

Function Notation

 

We may write a formula that defines a function with what is called “function notation”.

 

We replace y with f(x).

 

We call f(x) “f of x” and it means “the y=value when x is input.”

 

Example: 

 

If f(x) = 2x + 1, what is f (3), f (- 5), and f(A)?

 

f(3) = 2(3) + 1 = 7

 

f(- 5) = 2(- 5) + 1 = – 9

 

f(A) = 2(A) + 1 or 2A + 1

 

To use Function Notation, just remember to replace each x in the original formula with whatever is after f within the parentheses!

 

Domain and Range

 

The domain of an equation is the set of all x’s that we can plug into the equation and get back a real number for y. The range of an equation is the set of all y’s that we can ever get out of the equation.

 

Example 2

 

Determine the domain of the following function.

 

So, in this case there are no square roots so we don’t need to worry about the square root of a negative number. There is however a possibility that we’ll have a division by zero error. To determine if we will we’ll need to set the denominator equal to zero and solve.

 

So, we will get division by zero if we plug in x=−5 or x=2. That means that we’ll need to avoid those two numbers. However, all the other values of x will work since they don’t give division by zero. The domain is then,

 

Polynomial Functions

 

Dividing Polynomials

 

To divide a polynomial by another polynomial, you use the Division Algorithm in the same way you would divide 162 ÷ 5 .  You Divide, Multiply, Subtract, and Carry Down, repeatedly as shown below.

 

Divide 162 ÷ 5  = 3 (tens) 

Multiply 3 (tens) 5 = 30 5 = 15 (tens)

Subtract 16 (tens) – 15 (tens) = 1 (ten)

Carry Down 2

 

Divide 12 ÷ 5  = 2

Multiply 2 5 =  10

Subtract 12 – 10 = 2 

Nothing left to carry down, so we stop.

Our answer is 32 r 2  or 32 2/5

 

You can apply the same procedure to the division of two polynomials,

(x4 – 2x2 + x –2)  ÷  (x2 + x – 4)

 

First, however, we must insert zero placeholders for missing terms and rewrite as

(x4 + 0x3 – 2x2 + x –2)  ÷  (x2 + x – 4)

 

Now, set up as a standard division problem and repeat the steps Divide, Multiply, Subtract, Carry Down over and over until the divisor no longer may be divided into the result at the bottom.

 

Step 1 – Divide leading term of dividend x4 by leading term of divisor x2.  The result is x2 and this is the first part of the answer. 

 

Step 2 – Multiply your answer x2 by divisor (x2 + x – 4) using the Distributive Property to get x4 + x3 – 4x2.  Place under dividend.

 

Step 3 – Subtract x4 + x3 – 4x2 from divisor. Remember to correctly distribute the negative through the polynomial and add the opposite of each term!

 

Step 4 – Carry Down the x. 

 

REPEAT these steps.  Divide leading term at bottom –x3 by leading term of divisor x2 to get –x. This is the next part of your answer. Multiply your answer –x by the divisor  

(x2 + x – 4) using the Distributive Property to get –x3 + 2x2 + x.  Place result below dividend. Subtract to get 3x2 – 3x. Carry Down –2.  Divide 3x2 by x2 to get 3. Multiply 3 by (x2 + x – 4) and place below. Subtract to get –6x + 10.  This is your remainder.  You can write the remainder as a fraction as shown.

 

Synthetic Division – The Shortcut for Dividing by (x – c)

When dividing a polynomial f(x) by a linear factor (x-c), we can use a shorthand notation. saving steps and space. Here is the procedure:

 

Procedure For Synthetic Division of f(x) by (x – c): 

  1. Write the value of “c” and the coefficients of f(x) in a row. For example, if we divided f(x) = 3x3 + 2x – 1 by (x – 4) we would write

 

  1. Carry down the first coefficient. In this case carry down the 3.
  2. Multiply this carried down coefficient by the value of c. 

In this case, multiply 3 4 = 12. Place this result in the next column.  

  1. Add the column entries and place result at bottom. In this case you add 0+12 to get 12.  Multiply this addition result by “c” and place in next column. In this case you multiply 12 4 = 48. 

 

  1. Repeat Step 4 for all columns.  In this example, you get

 

  1. The bottom row of numbers reveals the answer along with the remainder. In this case, the numbers 3  12  50  199 indicate an answer of  

3x2 + 12x + 50   r 199   or   3x2 + 12x + 50  + 199/(x – 4) 

 

TIP: The answer will always have degree one less than the dividend. Always!

 

Also, when dividing by x plus something, c will be negative.  For example, if you divide by (x + 5), this is the same as dividing by (x – – 5).  So c = – 5.

 

Why does Synthetic Division work?  If you compare long division side-by-side with synthetic division, you can fairly easily see why this shortcut works every time.  This is left as a group exercise.

 

The Division Algorithm, Applied To Polynomials

If polynomial f(x) divided by polynomial D(x) results in quotient Q(x) with remainder R(x), then we may write f(x) = D(x)●Q(x) + R(x). 

 

In other words, if we divide a polynomial by another polynomial, resulting in an answer, we can multiply that answer by our divisor, add the remainder, and we should get back our dividend. 

 

Example: Divide f(x) = (x2 –3x + 3) by (x –1).  Then apply the Division Algorithm to rewrite f(x) as a product plus a remainder. Verify that the product with remainder added does indeed equal f(x).

 

The synthetic division for this results in   

which means f(x) ÷ (x – 1) = 1x – 2  r 1.  So we can rewrite f(x) as

f(x) = (x2 –3x + 3) = (x – 1)(x – 2) + 1

 

Verifying this, we multiply out (x – 1)(x – 2) + 1 to get

x2 –2x –x + 2   + 1 by the Distributive Property, which equals

x2 – 3x + 3 after combining like terms. 

 

NOTE: This is the same thing you did when you “checked your work” after dividing two numbers when first learning division. For example, if you divide 43 by 7 to get 6 r1, you checked your answer by multiply your quotient 6 by 7 and then adding the remainder 1 to get 6 7 + 1 = 43.

 

Graphing Polynomials

 

Graphing polynomials can be easy if you know what all the x-intercepts are.  Or graphing polynomials, by hand without a graphing calculator, can be only accomplished using calculus.  We will look at methods of graphing more “manageable” polynomials as well as some methods of quickly predicting behavior of the less-manageable.

 

Using Function Shift Rules to Plot Even Powers

 

You can easily plot even powers of x if they are in a function-shift form since all even powers of x like y=x2, y=x4, y=x6, y=x8, etc have a similar “U” shape containing points (0,0), (1,1), and (-1,1), as shown below. The higher the power on x, the more “flattened” out the curve will be between x=1 and x=-1 and the steeper the curve will be for x>1 and x<-1.

 

Example: Graph y = (x –3)10 + 1 by using function shift rules.

This will be a shift of y=x10 right 3 and up 1.  So, we get a flattened out “U” shaped curve with the points (0,0), (1,1), and (-1,1) shifted right 3 and up 1 as shown below.

 

Using Function Shift Rules to Plot Odd Powers

You can easily plot odd powers of x if they are in a function-shift form since all even powers of x like y=x3, y=x5, y=x7, y=x9, etc have a similar “S” shape containing points (0,0), (1,1), and (-1,-1), as shown below. The higher the power on x, the more “flattened” out the curve will be between x=1 and x=-1 and the steeper the curve will be for x>1 and x<-1, as was the case with the even powers.

 

Example: Graph y = (x + 2)11 – 1 by using function shift rules.

This will be a shift of y=x11 left 2 and down 1.  So, we get a flattened out “S” shaped curve with the points (0,0), (1,1), and (-1,-1) shifted left 2 and down 1 as shown below.

 

Leading Term Test For Polynomials

 

Often, your polynomial does not fit the simple form of the previous two examples, and function shift rules do not apply. We can still make predictions about the behavior of the graph by using the Leading Term Test, which in summary, states:

 

The leading term of a polynomial will predict the behavior of the graph on the far right side and far left side.  

 

Example:

 

Use the Leading Term Test to predict the behavior of the graph of  f(x) = -x5 + 2x + 1.

 

The leading term is –x5.  For positive x-values, this term is negative. So the graph will fall on the far right. Why does this happen? The leading term has the highest power of x and thus “dominates” the function in determining what the graph does as x gets larger.

 

For negative values of x,  –x5  is positive. So the graph will rise on the far left.  Again, the leading term has the highest power of x and thus “dominates” the function in determining what the graph does as x takes on larger negative values.

 

The leading term test tells us what happens on the far right and far left, and our conclusions are verified below in the actual graph.

 

Graphing Polynomials With Known Zeros

 

If you know the zeros of a polynomial, or they may be determined by factoring, then you can use the procedure covered back in graphs of functions. The method and example are given below.

 

Intercepts Method For Graphing Functions

 

  1. Find and plot all intercepts. To find y-intercepts, let x=0 and solve for y. To find x-intercepts, let y=0 and solve for x. 
  2. Plot points on each side of each x-intercept. Find and plot at least one point between each two x-intercepts and one point on each side of the largest and smallest x-intercept.
  3. Draw a smooth curve through the points from left to right.

 

Example: Graph f(x) = x3 – x.

First, rewrite as y = x3 – x.  Now, find intercepts.

 

When x = 0, we get y = 03 – 0 = 0.  So our y-intercept is (0,0)

 

When y=0, we get 0 = x3 – x.  We solve this equation. 

0 = x3 – x Given

0 = x(x2 – 1) Use Distributive Property to factor out x.

0 = x(x + 1)(x – 1) Use Distributive Property to factor x2 – 1.   

CONTINUED ON NEXT PAGE

 

x = 0 The Zero Product Law allows us to let each factor = 0

x + 1 = 0  

x – 1 = 0

x = -1, x = 1, x = 0 Solve each equation using Addition Property of Equality

 

So the x-intercepts are (-1,0), (1,0), and (0,0)

 

We plot these points to get

 

Now, find and plot points on each side of each of these x-intercepts by letting x= -2,      x=-1/2, x=1/2, and x=2. Draw a smooth curve through the points.  Note that you could also use the Leading Term Test to predict that the graph rises to the far right since x3 is positive for x>0 and falls to the far left since x3 is negative for x<0.  The Leading Term Test would allow us to skip plotting the points (2,6) and (-2,-6).

 

We get the points 

 

 x         y

-2     -6

 2      6

-1/2 3/8

 1/2 -3/8

 

Multiplicity Rules For Polynomial Zeros

 

The multiplicity of a zero x=c refers to the power N on the factor (x – c)N where (x – c)N is part of the factorization of the polynomial. 

 

For example, if f(x) = (x – 2)3(x – 5)2, then f(x) will have zero x=2 with multiplicity 3 and zero x = 5 with multiplicity 2.   We know x=2 and x=5 are zeros since f(2) = 0 and 

f(5) = 0.  

 

If a real zero has even multiplicity, then the graph will “bounce off” the x-intercept corresponding to the zero. (Remember that x-intercepts occur at real zeros).

 

If a real zero has odd multiplicity, then the graph will “pass through” the x-intercept corresponding to the zero. (Remember that x-intercepts occur at real zeros).

So, using these rules, we can say that for f(x) = (x – 2)3(x – 5)2, the graph will (bounce off) the point (5,0) since x=5 has even multiplicity and the graph will pass through the point (2,0) since x=2 has odd multiplicity. These conclusions are verified below in the graph of f(x).

 

Why Do These Multiplicity Rules Work?  

When a factor is raised to an even power, then values above and below the value of the zero do no result in a sign change.  For example, in the function f(x) = (x – 2)3(x – 5)2, the x-value x = 4.9 results in 

f(4.9) = (4.9 – 2)3(4.9 – 5)= 2.93 (-.1)=  24.389 0.01 = 0.24389 

and when x=5.1, we get 

f(5.1) = (5.1 – 2)3(5.1 – 5)= 3.13 (.1)=  29.791 0.01 = 0.29791

The even power resulted in a positive result each time, so the graph “bounces off”. 

 

When a factor is raised to an odd power, then values above and below the value of the zero do result in a sign change.  For example, in the function f(x) = (x – 2)3(x – 5)2, the x-value x = 2.1 results in 

f(2.1) = (2.1 – 2)3(2.1 – 5)= 0.13 (-2.9)=  0.001 8.41 = 0.00841   

and when x=1.9, we get 

f(1.9) = (1.9 – 2)3(1.9 – 5)= (-0.1)3 (-3.1)= -0.001 9.61 = -0.00961 

The odd power resulted in a negative result at x=1.9 and a positive result at x=2.1. 

 

Example: Given f(x) = x3 – x2 – 5x – 3, use the fact that x=3 is a zero to write f(x) in completely factored form.  Then use this form to find all the zeros and then graph.  Show all x-intercepts and make use of the Multiplicity Rules.   

 

Given that x=3 is a zero, this means that (x – 3) is a factor.  So if we divide f(x) by

(x – 3), we are able to write f(x) as a product of (x-3) and this result.

 

We can divide f(x) using synthetic division. Synthetic division results in a zero remainder (as it should be!) 

 

 

 

So this means f(x) = (x – 3)(x2 + 2x + 1).  We can factor this some more to get

f(x) = (x – 3)(x +1)(x + 1) or 

f(x) = (x – 3)(x+1)2

 

The factor (x-3) is raised to the 1st power so it has odd multiplicity 1.

The factor (x+1) is raised to the 2nd power so is has even multiplicity 2.

 

So, with this information, we know that :

  • The graph has x-intercepts (3,0) and (-1,0).
  • The graph “bounces off” the point (-1,0) and passes through the point (3,0).
  • Also, the Leading Term test predicts that the graph rises up to the right since leading term x3 is positive for x>0 and falls to the left since leading term x3 is negative for x<0.

 

Putting all this together, we conclude that the graph bounces off (-1,0) from below since it is negative on the far left.  Also, we conclude that it must rise up again and pass through (3,0) in order to rise up on the far right. Finding a few more points like (0,-3) and (1,-8) also help to give us a little better picture of the graph. The graph is shown below.

 

Exponential and Logarithm Functions

 

Exponential Functions 

 

Exponential functions are perhaps the most important class of functions in mathematics. We use this type of function to calculate interest on investments, growth and decline rates of populations, forensics investigations, as well as in many other applications. 

 

Let’s start off this section with the definition of an exponential function.

 

If b is any number such that b>0 and b≠1 then an exponential function is a function in the form,

f(x)=bx

where b is called the base and x can be any real number.

 

Example

 

Since we don’t have any knowledge on what these graphs look like we’re going to have to pick some values of x and do some function evaluations. 

Here are some evaluations for these two functions,

 

Here is the sketch of the two graphs.

 

As a final topic in this section we need to discuss a special exponential function. In fact this is so special that for many people this is THE exponential function. Here it is,

f(x)=ex

where e=2.718281828…. Note the difference between f(x)=bx and f(x)=ex. In the first case b is any number that meets the restrictions given above while e is a very specific number. Also note that e is not a terminating decimal.

This special exponential function is very important and arises naturally in many areas.

Example 

 

Sketch the graph of f(x)=ex.

 

Let’s first build up a table of values for this function.

 

To get these evaluation (with the exception x=0) you will need to use a calculator. In fact, that is part of the point of this example. Make sure that you can run your calculator and verify these numbers.

Here is a sketch of this graph.

Logarithm Functions 

 

Logarithmic functions, or for short, log functions, serve as the inverse functions of exponential functions. 

 

Here is the definition of the logarithm function.

If b is any number such that b>0 and b≠1 and x>0 then,

y=logbx is equivalent to by=x

We usually read this as “log base b of x”.

In this definition y=logbx is called the logarithm form and by=x is called the exponential form.

Example 

Evaluate the following logarithm.

As suggested above, let’s convert this to exponential form.

So, since,

 

we must have the following value of the logarithm.

 

Example 2

 

Evaluate the following logarithm.

 

Now, this one looks different from the previous one, but it really isn’t any different. As always let’s first convert to exponential form.

 

First, notice that the only way that we can raise an integer to an integer power and get a fraction as an answer is for the exponent to be negative. So, we know that the exponent has to be negative.

Now, let’s ignore the fraction for a second and ask 5?=125. In this case if we cube 5 we will get 125.

So, it looks like we have the following,

 

Now, let’s introduce some special logarithms that occur on a very regular basis. They are the common logarithm and the natural logarithm. Here are the definitions and notations that we will be using for these two logarithms.

 

Example

 

Evaluate the following logarithm.

 

Answer

 

Example 2

 

Evaluate the following logarithm.

 

Answer

 

Properties of Logarithms

 

Example 

 

Simplify the following logarithm.

 

In this case we’ve got a product and a quotient in the logarithm. In these cases it is almost always best to deal with the quotient before dealing with the product. Here is the first step in this part.

 

 

Now, we’ll break up the product in the first term and once we’ve done that we’ll take care of the exponents on the terms.

 

 

The final topic that we need to discuss in this section is the change of base formula.

 

where we can choose b to be anything we want it to be. In order to use this to help us evaluate logarithms this is usually the common or natural logarithm. Here is the change of base formula using both the common logarithm and the natural logarithm.

 

Example 

 

Evaluate log57.

 

Now, we have to use the change of base formula.

 

Exponential and Logarithm Equations

 

Exponential Equations

 

In this section we will look at solving exponential equations

There are two methods for solving exponential equations. One method is fairly simple but requires a very special form of the exponential equation. The other will work on more complicated exponential equations but can be a little messy at times.

Let’s start off by looking at the simpler method. This method will use the following fact about exponential functions.

Note that this fact does require that the base in both exponentials to be the same. If it isn’t then this fact will do us no good.

 

Example

 

Solve the following:

 

We have the same base on both exponentials so there really isn’t much to do other than to set the two exponents equal to each other and solve for x.

 

Now, the equation in the example relied upon the fact that we were able to get the same base on both exponentials, but that just isn’t always possible.

 

Consider the following equation.

 

This is a fairly simple equation however the method we used in the previous examples just won’t work because we don’t know how to write 9 as a power of 7.  So, the method we used in the first set of examples won’t work. The problem here is that the x is in the exponent. We need a way to get the x out of the exponent and luckily for us we have a way to do that. Recall the following logarithm property:

 

So, if we had,

 

we could use this property as follows

 

Example 

 

Solve the following equation.

 

We can use any logarithm that we’d like to so let’s try the natural logarithm.

 

Now, we need to solve for x. This is easier than it looks. If we had 7x=9 then we could all solve for x simply by dividing both sides by 7. It works in exactly the same manner here.

 

Now, that is technically the exact answer. However, in this case it’s usually best to get a decimal answer so let’s go one step further.

 

Logarithm Equations 

 

Let’s start off by looking at equations in which each term is a logarithm and all the bases on the logarithms are the same. In this case we will use the fact that,

If we’ve got two logs in the problem, one on either side of an equal sign and both with a coefficient of one, then we can just drop the logarithms.

 

Example 1 

 

Solve the following equation.


 First deal with the coefficient in front of the first term.

 

Now that we’ve got two logarithms with the same base and coefficients of 1 on either side of the equal sign we can drop the logs and solve

 

Now let’s take a look at the second kind of logarithmic equation. This equation will have all the terms but one be a logarithm and the one term that doesn’t have a logarithm will be a constant.

 

In order to solve these kinds of equations we will need to remember the exponential form of the logarithm.

 

Example

 

Solve the following equation.

 

The exponential form of this equation is,

 

Notice that this is an equation that we can easily solve.

 

Applications for Exponential and Logarithm Equations

 

Compound Interest

 

If we were to put P dollars into an account that earns interest at a rate of r (written as a decimal) for t years (yes, it must be years), the equation would be:

 

Exponential Growth and Decay

 

Earthquake Intensity

 

The Richter scale is commonly used to measure the intensity of an earthquake.

If E is the energy released, measured in joules, during an earthquake then the magnitude of the earthquake is given by,

 

Example 

 

If 8×1014 joules of energy is released during an earthquake what was the magnitude of the earthquake?