Modesto City Schools. Secondary Math I. Module 1 Extra Help & Examples. Compiled by: Rubalcava, Christina

Modesto City Schools Secondary Math I Module 1 Extra Help & Examples Compiled by: Rubalcava, Christina

1.1 Ready, Set, Go! Ready Topic: Recognizing a solution to an equation. The solution to an equation is the value of the variable that makes the equation true. You can check to see if the value is a solution but substituting (plugging in the number). For example: Which value is a solution to the equation? 3x + 7 = 1; x = 2 or x = 2 To answer this question, start by substituting 2 into the equation for x: 3(2) + 7 = 1. Next, simplify the left hand side to see if it matches the right hand side: 6 + 7 = 1 The symbol means not equal to. So x = 2 is not a solution: 13 1 Next, let s do the same process to see if x = 2 is a solution: 3( 2) + 7 = 1 6 + 7 = 1 x = 2 is a solution to When we substitute 2 into the equation, the 2 sides are equal. 1 = 1 the equation 3x + 7 = 1 For equations with 2 variables like y = 3x + 7, there are many solutions (infinitely many, in fact). So how do we find these solutions? Well, for a linear equation, the solutions will be all the points on the line formed when we graph the equation. In the first example, the solution only involved 1 variable (x = 2). For an equation with 2 variables, we will write the solution as an ordered pair, (x, y). The x value is always listed first. If we are given a value for x, we can substitute this number into the equation and solve for y. Our result will be listed as an ordered pair. For example: In the equation, y = 3x + 7 if we are given (2, ) what y value will make the equation true? Based on the ordered pair, (2, ) we can see that x = 2. Substitute this into the equation: y = 3(2) + 7 We need to simplify the right hand side of the equation so see what y is equal to: y = 6 + 7 We see that when we plug in 2, we get that y = 13. y = 13 The solutions to this equation must be points on the line, so we write our solution as an ordered pair: (2, 13) Video help: https://youtu.be/fxwimt9geoc Set Topic: Using a constant rate of change to complete a table of values. In mathematics it is often useful to look for a pattern to simplify our thinking. One common type of pattern occurs when we encounter a constant rate of change. A constant rate of change means that we add (or subtract) the same amount each time. For example, in the pattern 2, 4, 6, 8, 10, we start with 2 and add 2 each time. Identifying a pattern can help us to predict what will happen next. In the pattern 2, 4, 6, 8, 10 we can predict that the next number will be 12. This same type of pattern often occurs in word problems. For example: You are saving for a Nintendo Switch. You already have $50 and you are going to mow lawns for $15/lawn. This problem can be represented with a pattern the same way as the previous example. For this situation we start with $50 and add $15 for every lawn we mow. We can use a table to organize our information: # of lawns 1 2 3 4 5 6 7 Amount of $ saved $65 $80 $95 $110 $125 $140 $155 This first box represents the original $50 plus $15 for one lawn We keep adding $15 to the previous amount because we make $15 for each additional lawn we mow. Explanation: We start with $50 and add $15 for each lawn that we mow.

Another example: Your parents loan you $300 to buy the Nintendo Switch so you don t have to wait. You will use your lawn mowing business to pay them back at a rate of $10 per week. This time the amount of money we owe is decreasing so we will be subtracting $10 each week. Here is the pattern represented in a table: # of weeks 1 2 3 4 5 6 7 Amount of $ owed $290 $280 $270 $260 $250 $240 $230 This first box represents the original $300 minus $10 for one week of payments We keep subtracting $10 from the previous amount because we pay $10 each week (our debt is decreasing by $10 each week)

1.2 Ready, Set, Go! Ready Topic: Use variables to create equations that connect with visual patterns. Video examples: https://youtu.be/dqzmwrqmgy0 Set Topic: Use variables to create equations that connect with visual patterns. There are many ways to write an expression to model this visual pattern. The way you write your expression depends on how you see the pattern changing. For example: On step 3 imagine that your friend sees 2 groups of 3 blocks on the side plus 2 groups of 3 blocks on the top and bottom plus a single block in the middle: Your friend could write their expression as 2(3) + 2(3) +1 You might see step 3 as 1 block in the center plus 4 groups of 3 blocks: You might write your expression as 1 + 4(3) Both of these expressions can be simplified to equal 13. If the number 3 in our expression changes to match the step number, we can replace the 3 in our expressions with a variable like s for step. Our friend s expression would become 2s + 2s + 1 and our expression would become 1 + 4s. If we simplify our friend s expression by adding 2s + 2s, it will become 4s + 1. Here are some helpful expressions for patterns (the variable does not have to be s it can often be n for sequences: Groups that have the same number of blocks as the step number are represented as s Groups that have one more block than the step number are represented as s + 1 Groups that have two more blocks than the step number are represented as s + 2 Groups that include one block less than the step number are represented as s 1 Groups that include two blocks less than the step number are represented as s 2 Video examples: https://youtu.be/jfy3cjhlwho

1.3 Ready, Set, Go! Ready Topic: Interpreting function notation Video examples: https://youtu.be/xdsmhh2ufbw Set Topic: Comparing explicit and recursive equations When dealing with patterns we often write two different types of equations. The recursive equation is used to describe the change from one term to the next. For arithmetic sequences we add the constant difference (or subtract for decreasing functions) to the previous term. For geometric sequences we multiply the previous term by the common ratio. The explicit equation is used to find a specific term, given the term number. It compares the term number to the term s value. In the arithmetic sequence: 3, 5, 7, 9, we have term numbers and term values. 3 is the FIRST term (term #1), the term value is 3. 5 is the SECOND term (term #2), the term value is 5. The recursive rule looks at what we do to the previous term to get the next term. Our pattern starts with 3 and adds 2 each time so the constant difference (d) is 2. Our recursive rule is: Starting term=3, now = previous term + 2. You MUST include both parts! Term # Term Value 1 3 2 5 3 7 4 9 n Previous + 2 Add 2 Add 2 Add 2 Add 2 When we think recursively, we look down the output (term value) column to get the next term. This is most useful when we want to predict the next few terms. The explicit rule compares the term number to the term value. So for the 1 st term we get 3, the 2 nd term is 5, the 3 rd term is 7, etc. To find a specific term s value, we take the starting term and add d(n starting term #). For our sequence, we would have y = 3 + 2(x 1). The 3 is our starting term, 2 is the constant difference, and we use x 1 because 3 is the FIRST term. We can simplify our explicit equation by distributing the 2 and combining like terms. The simplified equation would be y = 2x + 1. Where does the constant difference show up in this equation? Term # Term Value 1 3 2 3 + 2 = 5 3 3 + 2 + 2 = 7 4 3 + 2 + 2 + 2 = 9 we have 1 less group than the term # n 3 + 2(n 1) When we think explicitly, we look across the row to compare the term number to the term value. This is most useful when we want to find a specific term value (especially if we skip several terms like finding the 100 th term). Example: Given an arithmetic sequence with an explicit equation of y = 2x + 5, and a recursive equation starting with the 5 th term and following the rule: now = previous term + 2, we can answer the following questions. Which rule would be most useful to find the 8 th term? We can use the explicit rule because we are skipping terms. Which rule would be most useful to find the 6 th term? We can use the recursive rule because we are finding the next term.

In the geometric sequence: 3, 6, 12, 24, we have term numbers and term values. 3 is the FIRST term (term #1), the term value is 3. 6 is the SECOND term (term #2), the term value is 6. The recursive rule looks at what we do to the previous term to get the next term. Our pattern starts with 3 and multiplies by 2 each time so the common ratio is 2. Recursive rule: Starting term=3, now = previous term x 2. You MUST include both parts! Term # Term Value 1 3 2 6 3 12 Multiply by2 Multiply by 2 When we think recursively, we look down the output (term value) column to get the next term. This is most useful when we want to predict the next few terms. 4 24 Multiply by 2 Multiply by 2 n Previous x 2 The explicit rule compares the term number to the term value. So for the 1 st term we get 3, the 2 nd term is 6, the 3 rd term is 12, etc. To find a specific term s value, we take the starting term and multiply by r (n starting term #). For our sequence, we would have y = 3(2) (x 1). The 3 is our starting term, 2 is the common ratio, and we use x 1 because 3 is the FIRST term. Term # Term Value 1 3 2 3(2) = 6 3 3(2 2) = 12 4 3(2 2 2) = 24 we have 1 less group than the term # (n 1) n 3(2) When we think explicitly, we look across the row to compare the term number to the term value. This is most useful when we want to find a specific term value (especially if we skip several terms like finding the 100 th term). Example: Given a geometric sequence with an explicit equation of y = 2, a recursive equation with the 1st term=2 and following the rule: now = previous term x, we can answer the following questions. Which rule would be most useful to find the 2nd term? The recursive rule because we are finding the next term. Which rule would be most useful to find the 5 th term? We can use the explicit rule because we are skipping terms.

Video examples: https://youtu.be/yk0f7easw0a Go Topic: Evaluating exponential equations Video examples: https://youtu.be/g17gcmiu2nw

1.4 Ready, Set, Go! Task: Scott s Workout Video Example: https://youtu.be/pa9cjl7pvjw Ready Topic: Use function notation to evaluate equations Function Notation Function notation is the way a function is written. It is meant to be a precise way of giving information about the function without a rater lengthy written explanation. The most popular function notation is f(x) which is read f of x. This is NOT the multiplication of f times x. Most of the times functions are referred to by single letters, such as f, g, h and so on, but any letter(s) may be used. Video Examples: https://www.khanacademy.org/math/algebra/algebra functions/evaluating functions/v/what is a function https://www.khanacademy.org/math/algebra/algebra functions/evaluating functions/v/understanding function notation example 1 https://www.youtube.com/watch?v=y6zk_rpc290 Online practice: https://www.khanacademy.org/math/algebra/algebra functions/evaluating functions/e/functions_1

Set Topic: Finding terms for a given sequence and writing recursive and explicit functions When dealing with arithmetic and geometric sequences, we can write two types of equations. Before we can write either equation, we need to identify some key information. The first important piece of information is the starting term for the sequence or pattern. If we are given a sequence (list of numbers) the first number is the 1 st term (not the 0 term). In a word problem, or if the problem says at the beginning we start with the 0 term. The second important piece of information is the constant difference. In an arithmetic pattern, the constant difference is the number that is being added (or subtracted) each time. The first type of equation, a recursive function, compares one term to the next to see how the pattern is changing. A recursive function consists of two parts, the starting term and the rule. If we forget one of these parts, the function is incomplete. Why is the starting term important? Example 1: In the arithmetic sequence 3, 5, 7, 9, 11, We can see that the pattern increases by 2 each time. This means that our constant difference (d) is 2. To write our recursive function we need to identify the starting term and rule for the pattern. We can use the form: now = previous + constant difference for arithmetic sequences. The notation for now is f(n) and the notation for previous is f(n 1) because previous means back 1 term. Using the correct notation, the recursive rule will be in the form: f(n) = f(n 1) + d So our recursive rule would be: f(1) = 3; f(n) = f(n 1) + 2 The notation f(1) = 3 means the 1 st term is 3 The + 2 means that we add 2 to the previous term to get the next term ***If we don t have the starting term we know that we need to multiply 2 times something, but we wouldn t know where to start. Many patterns can have the rule add 2 but they wouldn t be the same sequence if they started with a different number. The sequences 2, 4, 6, 8, 10, and 3, 5, 7, 9, 11, both add 2 each time, so both would get the rule f(n) = f(n 1) + 2, without the starting term our recursive rule wouldn t tell us which pattern we would get. The second type of equation, an explicit function, allows you to take a specific input (usually n or x) and find a specific output. Rather than using the previous term to predict the next term, an explicit formula has to change the input value into the output value using a specific rule each time. In an arithmetic sequence, the constant difference describes how the pattern changes. We can think of the constant difference as the rate of change, or slope. For a geometric sequence, we don t have a constant rate of change, instead we have the common ratio. The explicit function is a rule that takes an input n value and turns it into an output f(n) value. It is sometimes easier to use a table to find the rule for the explicit function. To write the explicit function we will still need two pieces of information, the starting term and rate of change or common ratio. Example 1: Find the explicit function for the arithmetic sequence 3, 5, 7, 9, 11, Step 1: Identify the rate of change/constant difference. The pattern is + 2 so the rate of change/slope is d = 2. Step 2: Identify the starting term. In this case, the first term is 3. Answer: We can use this form for an arithmetic pattern: f(n) = starting term + d(n starting term #). Where does this pattern come from? We can use a table to see how the rule is developed.

Term #: n Term value: f(n) 1 3 2 5 = 3 + 2 3 7 = 3 + 2 + 2 4 9 = 3 + 2 + 2 +2 5 11 = 3 + 2 + 2 + 2 +2 the number of 2 s is one less than the term # n f(n) = 3 + 2(n 1) Using our information we get: f(n) = 3 + 2(n 1) or we can distribute and combine like terms to get f(n) = 2n + 1 The 3 is our starting term The 2 is the constant difference (slope) We subtract 1 because we started with the 1 st term. The number we subtract depends on which term we used as the starting term. *both forms of the explicit function are correct. The first function is set up in point slope form and the second function is in slope intercept form. The second form may be more familiar to you when graphing. Where does the constant difference appear in the recursive function? Explicit function? Given an arithmetic sequence with the recursive function: f(1) = 5; f(n) = f(n 1) + 3 and an explicit function f(n) = 5 + 3(n 1) or f(n) = 3n + 2. The constant difference in each type of function has been highlighted. For the recursive function, the constant difference is the number being added to the previous term. In the explicit function, the constant difference is the slope (rate of change). Video Examples: https://youtu.be/r9f5o4vadt4 https://learnzillion.com/lesson_plans/5940 create a recursive formula https://learnzillion.com/lesson_plans/6323 create an explicit formula Go: Topic: Reading a graph Video Examples: https://youtu.be/znu9qhesss0 https://learnzillion.com/lesson_plans/5303#additional material

1.5 Ready, Set, Go! Ready Topic: Rates of change in a table and a graph

More notes: https://mathbitsnotebook.com/algebra1/functiongraphs/fngaverageratechange.html Video Examples: https://youtu.be/gxn7xsapqju https://www.khanacademy.org/math/algebra/algebra functions/functions average rate of change/v/introduction toaverage rate of change https://www.khanacademy.org/math/algebra/algebra functions/functions average rate of change/v/average rate ofchange example 1 https://www.khanacademy.org/math/algebra/algebra functions/functions average rate of change/v/average rate ofchange example 3 Online practice: https://www.khanacademy.org/math/algebra/algebra functions/functions average rate ofchange/e/avg rate of change graphs tables Set Topic: Recursive and explicit functions of geometric sequences When dealing with geometric sequences, we can write two types of equations. Before we can write either equation, we need to identify some key information. The first important piece of information is the starting term for the sequence or pattern. If we are given a sequence (list of numbers) the first number is the 1 st term (not the 0 term). In a word problem, or if the problem says at the beginning we start with the 0 term. The second important piece of information is the common ratio. In a geometric pattern, the common ratio is a number that is multiplied (or divided) each time. The first type of equation, a recursive function, compares one term to the next to see how the pattern is changing. A recursive function consists of two parts, the starting term and the rule. If we forget one of these parts, the function is incomplete. Why is the starting term important? Example 1: In the geometric sequence 3, 6, 12, 24, 48, We can see that the pattern multiplies by 2 each time. This means that our common ratio (r) is 2. To write our recursive function we need to identify the starting term and rule for the pattern. We can use the form: now = previous x common ratio for geometric sequences. The notation for now is f(n) and the notation for previous is f(n 1) because previous means back 1 term. Using the correct notation, the recursive rule will be in the form: f(n) = f(n 1) x r

So our recursive rule would be: f(1) = 3; f(n) = f(n 1) x 2 The notation f(1) = 3 means the 1 st term is 3 The x 2 means that we multiply by 2 times the previous term to get the next term ***If we don t have the starting term we know that we need to multiply 2 times something, but we wouldn t know where to start. Many patterns can have the rule times 2 but they wouldn t be the same sequence if they started with a different number. The sequences 2, 4, 8, 16, 32, and 3, 6, 12, 24, 48, both multiply by 2 each time, so both would have a rule of f(n) = f(n 1) x 2, without the starting term our recursive rule wouldn t tell us which pattern we would get. The second type of equation, an explicit function, allows you to take a specific input (usually n or x) and find a specific output. Rather than using the previous term to predict the next term, an explicit formula has to change the input value into the output value using a specific rule each time. In a arithmetic sequence, the constant difference describes how the pattern changes. We can think of the constant difference as the rate of change, or slope. For a geometric sequence, we don t have a constant rate of change, instead we have the common ratio. The explicit function is a rule that takes an input n value and turns it into an output f(n) value. It is sometimes easier to use a table to find the rule for the explicit function. To write the explicit function we will still need two pieces of information, the starting term and rate of change or common ratio. Example 2: Find the explicit function for the geometric sequence 3, 6, 12, 24, 48, Step 1: This function does not have a constant rate of change (not adding the same amount each time) so we need to look for the common ratio. In this case, we are multiplying by 2 each time so our common ratio (r) is 2. Step 2: Identify the starting term. In this case, the first term is 3. Answer: We can use this form for an geometric pattern: f(n) = starting term(r) (n starting term #). Let s see how a table can help us develop this rule for a geometric sequence. Term #: n Term value: f(n) 1 3 2 6 = 3 x 2 or 3(2) 1 3 12 = 3 x 2 x 2 or 3(2) 2 4 24 = 3 x 2 x 2 x 2 or 3(2) 3 5 48 = 3 x 2 x 2 x 2 x2 or 3(2) 4 the number of 2 s is one less than the term #. Since we have repeated multiplication we can write the number of 2 s as the exponent. n (n 1) f(n) = 3(2) (n 1) Using our information we get: f(n) = 3(2) The 3 is our starting term The 2 is the common ratio We subtract 1 because we started with the 1 st term. The number we subtract depends on which term we used as the starting term.

Video Examples: https://youtu.be/xtstsfym3oi https://learnzillion.com/lesson_plans/5940 create a recursive formula https://learnzillion.com/lesson_plans/6323 create an explicit formula Go Topic: Recursive and explicit functions of arithmetic sequences Video Examples: https://youtu.be/o3 1SWBqDN4

1.6 Ready, Set, Go! Ready Topic: Finding the common difference When you have two consecutive numbers (numbers next to each other) in an arithmetic sequence, you can find the constant difference. You will need to subtract the numbers in the order newer older. For example, if I have the 5 th and 6 th terms I would find the constant difference by taking the 6 th term 5 th term. I can then use the constant difference to find missing terms in the sequence. Example: Given the arithmetic sequence: 5, 9,, 17, 21,, Find the constant difference and the missing terms. Step 1: We have consecutive terms (1 st term = 5 and 2 nd term = 9) so we can subtract: 9 5 = 4. Our constant difference is 4. Step 2: We can use the constant difference to find the missing terms: 5, 9, 13, 17, 21,25, What if we don t have consecutive terms? We need to see what number we would add each time to get from our starting term to our ending term. Set Example: Given the arithmetic sequence: 20,,,, 8 Find the constant difference and the missing terms. Step 1: We do NOT have consecutive terms because 20 is the 1 st term and 8 is the 5 th term. We will need to figure out what number we can add (or subtract) to get from 20 to 8. We also have to notice that we will use the constant difference 4 times (to get from the 1 st term to the 5 th term). We can set up an expression to help us: 20,,,, 8 + d + d + d + d We jump 4 times to get from 20 to 8 We can write this as: starting term + (# of jumps)(d) = ending term 20 + 4(d) = 8 4d = 12 subtract 20 from both sides d = 3 divide by 4 on both sides If we start with 20 and subtract 3 each time, we can get to 8. Step 2: We can use our constant difference of 3 to fill in the missing terms: 20, 17, 14, 11, 8 Topic: Writing the recursive function 3 3 3 3 A recursive function uses the previous term to predict what the next term will be. For both arithmetic and geometric sequences, we need to identify a starting term as the first part of our recursive function. The second part of the recursive function is written as f(n) = f(n 1)and then what is being done to the previous term. For arithmetic sequences, we will take the previous term and add the constant difference (it will look like subtraction if the common difference is negative). The recursive rule will be: starting term = a number; f(n) = f(n 1) + d.

Example: Given an arithmetic function with f(2) = 12 and f(3) = 9, write the recursive function Step 1: I know the function is arithmetic, so I will need the constant difference. These terms are consecutive (right next to each other) so I take the 3 rd term and subtract the 2 nd term: 9 12 = 3. This tells me that my constant difference is 3. Step 2: I need to identify a starting term. For this problem, the starting term is f(2) = 12. Step 3: I need to write my rule in the form: f(n) = f(n 1)and then what is being done to the previous term The rule for this problem will be: f(n) = f(n 1) 3. Answer: I need to give BOTH the starting term, and the rule: f(2) = 12; f(n) = f(n 1) 3 For a geometric sequence, we will take the previous term and multiply by the common ratio (instead of using division we will multiply by the reciprocal). The recursive rule will be in the form: f(n) = f(n 1) x r Example: Given a geometric function with f(6) = 32 and f(7) = 16, write the recursive function Key things to remember: Step 1: I know the function is geometric, so I will need the common ratio. These terms are consecutive so I take the 7 th term and divide by the 6 th term:. This tells me that my common ratio is. Step 2: I need to identify a starting term. For this problem, the starting term is f(6) = 32. Step 3: I need to write my rule in the form: f(n) = f(n 1)and then what is being done to the previous term The rule for this problem will be: f(n) = f(n 1) x. Answer: I need to give BOTH the starting term, and the rule: f(6) = 32; f(n) = f(n 1) x. To find the constant difference we SUBTRACT the terms. You will work backwards: newer term older term. To find the common ratio we DIVIDE the terms. You will work backwards: Video Examples (for set and go): https://youtu.be/n I ka9avcw https://www.khanacademy.org/math/algebra/sequences/constructing arithmetic sequences/v/recursive formula forarithmetic sequence Extra Examples: https://www.khanacademy.org/math/algebra/sequences/constructing arithmetic sequences/a/writingrecursive formulas for arithmetic sequences Go Topic: Use function notation to evaluate equations Function Notation Function notation is the way a function is written. It is meant to be a precise way of giving information about the function without a rater lengthy written explanation. The most popular function notation is f(x) which is read f of x. This is NOT the multiplication of f times x.

Most of the times functions are referred to by single letters, such as f, g, h and so on, but any letter(s) may be used. Video Examples: https://www.khanacademy.org/math/algebra/algebra functions/evaluating functions/v/what is a function https://www.khanacademy.org/math/algebra/algebra functions/evaluating functions/v/understanding function notation example 1 https://www.youtube.com/watch?v=y6zk_rpc290 Online practice: https://www.khanacademy.org/math/algebra/algebra functions/evaluating functions/e/functions_1

1.7 Ready, Set, Go! Ready Topic: Distinguishing between arithmetic and geometric sequences To determine if the sequence is arithmetic or geometric you will need to see if you have a constant difference or common ratio. An arithmetic sequence will have a pattern that adds (or subtracts) the same number each time. A geometric sequence will have a pattern that multiplies (or divides) by the same number each time. Example 1: For the given sequence, find the missing values. Underline if it has a constant difference or common ratio. State the value of the difference or ratio. Finally, indicate if the sequence is arithmetic or geometric by circling the answer. Before we can find the missing terms, we need to see if we are adding or Sequence: 15, 12,, 6, 3, multiplying. We have 2 pairs of consecutive terms, 15, 12 and 6, 3 so we use Missing terms: 15, 12, 9, 6, 3, 0 these terms to see if we have a constant difference. First we subtract (working backwards) 12 15 = 3. We want to see if the next pair of Constant difference or ratio? consecutive terms also have the same difference: 3 6 = 3. This means that Constant difference/ratio = _ 3 we have a constant difference of 3. A sequence with a constant difference Arithmetic or Geometric? is arithmetic. Finally, we can use the pattern of subtracting 3 to find the missing terms. Example 2: For the given sequence, find the missing values. Underline if it has a constant difference or common ratio. State the value of the difference or ratio. Finally, indicate if the sequence is arithmetic or geometric by circling the answer. Sequence: 1, 3,, 27, 81, Missing terms: 1, 3, 9, 27, 81, 243 Constant difference or ratio? Constant difference/ratio = _3 Arithmetic or Geometric? We begin by checking to see if we have a constant difference: 3 1 = 2 and 81 27 = 54. The answers are not the same, so we do not have a constant difference. Next, we check to see if we have a common ratio: 3 and 3. The ratios are the same, so have a common ratio of 3. A sequence with a common ratio is geometric. Finally, we can use the pattern of multiplying by 3 to find the missing terms. Video Examples: https://youtu.be/ri767qbklc4 Set Topic: Recursive and Explicit Equations We can the explicit formula in the form: f(n) = mx + b (m is constant diff. & b= 0 term or f(n) = 1 st term + d(n 1)

Organizing the sequence into a table can make it easier to identify the type of pattern and to write the recursive and explicit functions for the pattern. Example 1: Determine whether the given information represents an arithmetic or geometric sequence. Then write the recursive and explicit equation for each. 4, 12, 20, 28, 36, Step 1: Determine the type of sequence by determining if it has a constant difference or common ratio: Term #: n Term value: f(n) 1 4 2 12 3 20 4 28 5 36 n f(1)=4; f(n)=f(n 1) + 8 +8 +8 +8 +8 +8 We can see from the table that we are adding 8 each time. This means that we have an arithmetic sequence with a constant difference (d) of 8. Our recursive rule is: f(1) = 4; f(n) = f(n 1) + 8 Step 2: Write the explicit function by looking at the starting term & how many groups of 8 are added each time. Term #: n Term value: f(n) How many groups of 8? 1 4 4 + zero 8 s 2 12 = 4 + 8 4 + one 8 3 20 = 4 + 8 + 8 4 + two 8 s 4 28 = 4 + 8 + 8 + 8 4 + three 8 s 5 36 = 4 + 8 + 8 + 8 + 8 4 + four 8 s n 4 + 8 + + 8 4 + (n 1)8 s The repeated addition can be written as multiplication: ex. 8 + 8 can be written as 2(8). We always have one group less than the term number. So for the nth term, we have (n 1) groups of 8. Our explicit equation is: F(n) = 4 + 8(n 1)

Examples of phrases that signal arithmetic sequences: A taxi charges a flat rate $1.50 (this is the 0 term) and then $0.50 for each mile (this is the constant difference because we add $0.50 each mile). A student reads 5 books every 3 weeks (so the student reads part of a book each week). In this case, the student starts having read 0 books (the 0 term is 0). To find the constant difference we need to find the books per week by dividing. Example 2: Determine whether the given information represents an arithmetic or geometric sequence. Then write the recursive and explicit equation for each. Your brother invested $200 into an account that earns 10% interest each year (using simple interest). Step 1: Determine the type of sequence by determining if it has a constant difference or common ratio: # of months n $ in account: f(n) 0 200 1 220 2 242 3 266.20 4 292.82 n f(1)=4; f(n)=f(n 1) + 8 The $200 represents the amount at the beginning. This means that we are starting with the 0 term. Next, we can see that the first year the account increased by $20. The next year the account increased by $22. Since it does not increase by the same amount, this is not an arithmetic pattern. Next, we look for a common ratio by dividing the newer output by the older output:. 1.1, 1.1, 1.1. This means we have a geometric function with a common ratio (r) of 1.1. Our recursive rule is: f(0) = 200; f(n) = f(n 1) x 1.1 Step 2: Write the explicit function by looking at the starting term & how many groups of 1.1 are multiplied each time. Term #: n Term value: f(n) How many groups of 8? 1 200 200 x 1.1 0 2 220= 200 x 1.1 200 x 1.1 1 3 242= 200 x 1.1 x 1.1 200 x 1.1 2 4 266.20= 200 x 1.1 x 1.1 x 1.1 200 x 1.1 3 5 292.82= 200 x 1.1 x 1.1 x 1.1 x 1.1 200 x 1.1 4 n 200 x 1.1 x x 1.1 (n 1) 200 x 1.1 The repeated multiplication can be written using exponents: ex. 1.1 x 1.1 x 1.1 can be written as 1.1 3. We always have one group less than the term number. So for the nth term, we multiply by 1.1 (n 1) times. Our explicit equation is: f( n) 200(1.1) n 1

Examples of phrases that signal geometric sequences: An account earns 5% interest. To find percent, we multiply. This means that we will have a common ratio. A starts at a height of 10 feet (this is the 0 term) and bounces half as high (common ratio = ½) each time. We multiply each output by ½ each time so this is a geometric sequence. Video Examples: https://youtu.be/ri767qbklc4 https://youtu.be/fhx3ixwj61m https://learnzillion.com/lesson_plans/5940 https://learnzillion.com/lesson_plans/6323 create an explicit formula https://learnzillion.com/lesson_plans/4689# Go Topic: Graphing and counting slope between two points If we are given two points on a graph and the slope, we can draw the slope on the figure as a triangle using the rise and run as the sides. 1. Graph the two points and use a ruler to draw the line between them. 2. Draw the slope triangle using a vertical line for the rise (up/down) and a horizontal line for the run (left/right). Example 1: Given the points (0, 3) and (2, 2) with slope of 1. Plot the points (0, 3) and (2, 2). Connect (0, 3) and (2,2), using a ruler to draw the line. 2. We interpret the slope of 1 or down 1, right 2. We will start at the point on the left, (0, 3) and 2 draw a triangle by moving down 1 and right 2. Example 2: Given the points (0, 1) and (1, 2) with a slope of 3 1. Plot the points (0, 1 ) and (1, 2). Connect the points using a ruler to draw the line. 2. We interpret the slope of 3 as or up 3, right 1. We will start at the point on the left, (0, 1) and draw a slope triangle by moving up 3 and right 1. Video Examples: https://youtu.be/zihsqc0iud8

1.8 Ready, Set, Go! Ready Topic: Common Ratios of Geometric Sequences The common ratio of a geometric sequence is the number that we multiply by to get from one term to the next. For example, in the sequence 2, 4, 8, 16, 32, we multiply by 2 each time. If the sequence is decreasing, it may seem like we are dividing by a number each time. When this occurs we need to remember that division is the same as multiplying by the reciprocal. So if we reverse the previous sequence to get 32, 16, 8, 4, 2 we see that each number is half as big (or divided by 2). The common ratio would be ½ because dividing by 2 is the same as taking ½ of the previous number. Find the common ratio for the geometric sequences below: Ex 1: 3, 12, 48, 192, 768, We are told that the sequence is geometric, so we will be multiplying by the same number each time. We need to ask ourselves 3 times what number will give us 12? We can answer this question by diving 12 by 3. So 12 3 = 4. We can check to see if the sequence works by starting at 3 and multiplying by 4 each time. 3 x 4 = 12; 12 x 4 = 48; 48 x 4 = 192, etc. This means that 4 is the common ratio (r = 4). Ex. 2:,,1,2,4,8 We need to find out what number we are multiplying by each time. Like the previous example, we can ask ourselves what number must we multiply by to get? This seems a little more difficult to answer than the previous problem because we have fractions in the sequence. We can actually pick any 2 terms that are right next to each other in the sequence to find the common ratio. This means, that we don t have to use the fractions to find the common ratio. Instead, we can look at 1 and 2 to answer the question. To find what number we need to multiply 1 by to get 2, we will divide. So 2 1 = 2. If we check our work: x 2 = ; x 2 = 1; 1 x 2 = 2; 2 x 2 = 4, and 4 x 2 = 8. This means that 2 is the common ratio (r = 2) Ex. 3: 1, 6, 36, 216, In this example, the geometric sequence is alternating signs between negative & positive. We will need to take this into consideration when we find the common ratio. As we ve seen in the previous two examples, we can pick any two terms that are next to each other in the sequence to find the common ratio. If we use 1 and 6, we will need to figure out what number we can multiply 1 by to get a positive 6. If we do 6 1 we get 6. We need to test this number out to check that we have the correct ratio for the sequence. We get 1 x 6 = 6; 6 x 6 = 36; 36 x 6 = 216, which means 6 is the correct ratio. Ex. 4: 12, 6, 3, 1.5, 0.75 In this example, the sequence is decreasing. This means that the ratio is a proper fraction (or a decimal less than 1). In all of our examples so far, we have picked 2 terms that are next to each other in the sequence. We have then divided the terms to find the common ratio. Look back at the previous examples and pay attention to the order we divided in. You should notice that when we divided, we work backwards. So in the first example, we divided the 2 nd term by the 1 st term. In the second example, we divided the 4 th term by the 3 rd term. In example 3, we took the 2 nd term and divided by the 1 st term. This order is very important, if we divide backwards our ratio will be incorrect. We do NOT always start with the larger number.for this sequence we can divide the 2 nd term (6) by the 1 st term (12) to get the ratio. When we take 6 12 we get. If we divided the other way around, we would get a ratio of 2 which does NOT work for this sequence.

To summarize, if you are given a geometric sequence, you need to start by picking 2 terms that are next to each other in the sequence. You will then divide in the order: next term previous term. So for the sequence 243, 81, 27, 9, 3 we can pick any 2 terms in the sequence. Let s choose 9 & 3 because they are the smallest. Next, we want to divide in reverse order next previous so we will take 3 and divide by 9 (3 9) to get. This means that the common ratio is. If we start with 243 and multiply by each time, we will get 81, 27, 9, and 3. Video Examples: https://youtu.be/eacgfvn_r28 https://youtu.be/gyb8zzcp NQ Set Topic: Recursive and Explicit Equations When you are given a table to represent a sequence and asked to write the recursive and explicit formulas, you need to start by determining which type of sequence the table represents. Once you know the type of sequence, you can write the recursive and explicit rules. Step 1: Determine the type of sequence by determining if it has a constant difference or common ratio: Term #: n Term value: f(n) 1 3 2 11 3 19 4 27 5 35 n f(1)=3; f(n)=f(n 1) + 8 +8 +8 +8 +8 +8 We can see from the table that we are adding 8 each time. This means that we have an arithmetic sequence with a constant difference (d) of 8. Our recursive rule is: f(1) = 3; f(n) = f(n 1) + 8 Step 2: Write the explicit function by looking at the starting term & how many groups of 8 are added each time. Term #: n Term value: f(n) How many groups of 8? 1 3 3 + zero 8 s 2 11 = 3 + 8 3 + one 8 3 19 = 3 + 8 + 8 3 + two 8 s 4 27 = 3 + 8 + 8 + 8 3 + three 8 s 5 35 = 3 + 8 + 8 + 8 + 8 3 + four 8 s n 3 + 8 + + 8 3 + (n 1)8 s The repeated addition can be written as multiplication: ex. 8 + 8 can be written as 2(8). We always have one group less than the term number. So for the nth term, we have (n 1) groups of 8. Our explicit equation is: F(n) = 3 + 8(n 1) For an arithmetic sequence, the recursive rule is in the form: f(starting term #) = starting term value, f(n) = f(n 1) + d and the explicit equation can be written as f(n) = starting term value + d(n starting term #) where d is the common difference. For the tables above, the table starts with the 1 st term so the starting term # is 1, and the first term s value is 3. The common difference is 8.

Example 2: Determine whether the given information represents an arithmetic or geometric sequence. Then write the recursive and explicit equation for each. Your brother invested $200 into an account that earns 10% interest each year (using simple interest). Step 1: Determine the type of sequence by determining if it has a constant difference or common ratio: # of months n $ in account: f(n) 0 200 1 220 2 242 3 266.20 4 292.82 n f(1)=4; f(n)=f(n 1) + 8 The $200 represents the amount at the beginning. This means that we are starting with the 0 term. Next, we can see that the first year the account increased by $20. The next year the account increased by $22. Since it does not increase by the same amount, this is not an arithmetic pattern. Next, we look for a common ratio by dividing the newer output by the older output:. 1.1, 1.1, 1.1. This means we have a geometric function with a common ratio (r) of 1.1. Our recursive rule is: f(0) = 200; f(n) = f(n 1) x 1.1 Step 2: Write the explicit function by looking at the starting term & how many groups of 1.1 are multiplied each time. Term #: n Term value: f(n) How many groups of 8? 1 200 200 x 1.1 0 2 220= 200 x 1.1 200 x 1.1 1 3 242= 200 x 1.1 x 1.1 200 x 1.1 2 4 266.20= 200 x 1.1 x 1.1 x 1.1 200 x 1.1 3 5 292.82= 200 x 1.1 x 1.1 x 1.1 x 1.1 200 x 1.1 4 n 200 x 1.1 x x 1.1 (n 1) 200 x 1.1 The repeated multiplication can be written using exponents: ex. 1.1 x 1.1 x 1.1 can be written as 1.1 3. We always have one group less than the term number. So for the nth term, we multiply by 1.1 (n 1) times. Our explicit equation is: f( n) 200(1.1) n 1 For a geometric sequence, the recursive rule is in the form: f(starting term #) = starting term value, f(n) = f(n 1) x r and the explicit equation can be written as f(n) = starting term value x (r) (n starting term #) where r is the common ratio. For the tables above, the table starts with the 0 term so the starting term # is 0, and the starting term s value is 200. The common ratio is 1.1. Video Examples: https://learnzillion.com/lesson_plans/5940 https://learnzillion.com/lesson_plans/6323

Go Topic: Writing equations of lines given a graph The slope intercept form of a linear graph can be written in the form y = slope(x) + y intercept, or as y = mx + b where m represents the slope, and b represents the y intercept. First, let s clear up some vocabulary. The slope of the line is the ratio of the rise over run. In this graph, the line moves down 6 as it goes right 3. The rise is 6 because it moves down, the run is 3 because it moves right. The slope is equal to 2, so m = 2 In this graph, the line moves up 4 as it goes right 6. The rise is 4 because it moves up, the run is 6 because it moves right. The slope is equal to, so m = The y intercept, is the place where the graph intersects the y axis. In this graph, the line crosses the y axis at 2. This means b = 2 for this graph. In this graph, the line crosses the y axis at 1. This means b = 1 for this graph. To write the slope intercept equation from a graph, you will need to identify the slope (m) and the y intercept (b) to plug into the form: y = mx + b. In this graph, the line moves up 1 as it goes right 2. The rise is 1 because it moves up, the run is 2 because it moves right. The slope is equal to, so m =. The graph crosses the y axis at 1, so b = 1. If we plug them into slope intercept form, we get: y = x + 1 Video Examples: https://youtu.be/fhyanlhqdjk https://www.khanacademy.org/math/algebra/two var linear equations/slope intercept form/v/slope intercept form

1.9 Ready, Set, Go! Ready Topic: Comparing Arithmetic and Geometric Sequences Video Examples: https://youtu.be/ri767qbklc4 Set Topic: Finding missing terms in an Arithmetic sequence An arithmetic sequence has a common difference which means that we add (or subtract) the same number each time. For example, in the sequence 3, 6, 9, 12, 15, 18, We add 3 each time. We could keep track of this work as shown below: 3, 6, 9, 12, 15, 18 +3 +3 +3 +3 +3 We start at 3, add 3 to get to 6, add 3 again to get 9, add 3 again to get 12, add 3 again to get 15, and add 3 again to get 18. How many times do we add 3 to get from the 1 st term of 3 to the 6 th term of 18? We started with 3 and added the common difference (d) 5 times. We could write that equation like this: 3 + 5(3) = 18 Let s try to write an equation for another arithmetic sequence. 2, 7, 12, 17, 22 +5 + 5 +5 +5 1 st term # of d times we add d 6 th term Our first term, f(1) is 2 and our 5 th term, f(5) is 22. We are adding 5 each time, so our common difference is d=5. We had to add the common difference 4 times to get from the 1 st term to the 5 th term (if we take 5 th 1 st we get 4). So our equation would be: 2 + 4(5) = 22 1 st term # of d times we add d 5 th term If we generalize these equations we get: starting term + (# of times we add d)(d) = ending term For the previous examples, we were given several terms in the sequence so we could easily find the common difference. How can we use the equation we wrote to find d and missing terms if we don t have the entire sequence? Let s try an example: x (term number) 1 2 3 4 5 y(term value) 10 38 In this sequence we have the 1 st & 5 th terms, but we are missing terms 2 4. How can we find the common difference (d) and the missing terms? We can use the equation: starting term + (# of times we add d)(d) = ending term. In this case our 1 st term is 10, our 5 th term is 38 and we will have to add d 4 times. Our equation will look like this: 10 + 4d = 38

Now we can solve this equation for d to find the common difference. Once we have d, we can use it to find the missing terms. 10 + 4d = 38 4d = 28 d = 7 Starting equation Subtract 10 Divide by 4 We know now that the common difference is 7 so we can find the missing terms by adding 7. x (term number) 1 2 3 4 5 y(term value) 10 17 24 31 38 +7 +7 +7 +7 Let s try another example: x (term number) 1 2 3 4 5 y(term value) 3 13 In this sequence we have the 1 st & 3 rd terms, but we are missing terms 2, 4, and 5. How can we find the common difference (d) and the missing terms? We can use the equation: starting term + (# of times we add d)(d) = ending term. In this case our 1 st term is 3, our 3 rd term is 13 and we will have to add d 2 times. Our equation will look like this: 3 + 2d = 13 Notice, that we used the 3 rd term this time. We can use any 2 terms that are given as the starting and ending term in our equation as long as we remember to count how many times we add d. Now we can solve this equation for d to find the common difference. Once we have d, we can use it to find the missing terms. 3 + 2d = 13 2d = 10 d = 5 Starting equation Subtract 3 Divide by 2 We know now that the common difference is 5 so we can find the missing terms by adding 5. x (term number) 1 2 3 4 5 y(term value) 3 8 13 18 23 +5 +5 +5 +5 Does the process change if the sequence is decreasing? Let s take a look. x (term number) 1 2 3 4 y(term value) 24 3

In this sequence we have the 1 st & 4 th terms, but we are missing terms 2 and 3. How can we find the common difference (d) and the missing terms? We can use the equation: starting term + (# of times we add d)(d) = ending term. In this case our 1 st term is 24, our 4 th term is 3 and we will have to add d 3 times. Our equation will look like this: 24 + 3d = 3 Notice, that our 4 th term is negative. We will have to be careful of the signs as we solve the equation. Now we can solve this equation for d to find the common difference. Once we have d, we can use it to find the missing terms. 24 + 3d = 3 3d = 27 d = 9 Starting equation Subtract 24, 3 24 is 27 Divide by 3. When we divide a negative by a positive, we get a negative number. We know now that the common difference is 9 so we can find the missing terms by subtracting 9. x (term number) 1 2 3 4 y(term value) 24 15 6 3 Video Examples: https://youtu.be/gevbooy6vcs https://youtu.be/t7yh5z_9yem 9 9 9 Go Topic: Determining if a sequence is arithmetic, geometric, or neither and writing the recursive and explicit equation. Arithmetic sequences have a common difference, so you will add (or subtract) by the same number each time. Geometric sequences have a common ratio, so you will multiply (or divide) by the same amount each time. If the sequence does not have a common difference or common ratio, it is neither arithmetic or geometric. Example 1: Determine if the given sequences are arithmetic, geometric, or neither. We can first check if there is common ratio by dividing a term by the term before it. If the quotients are all the same, the sequence is geometric. If there is no common ratio, we can look for a common difference by taking a term and subtracting the term before it. If the differences are all the same, the sequence is arithmetic. If we don t have a common ratio or difference, the sequence is neither arithmetic or geometric. 1) 200, 40, 8, Common ratio? ; The ratios are the same, so the sequence is geometric. 2) 1, 4, 8, 16, Common ratio? 4; 2 the ratios are not the same, so the sequence is not geometric. Common difference? 4 1 = 3; 8 4 = 4 the differences are not the same, so the sequence is not arithmetic. This sequence is neither arithmetic, nor geometric. 3) 6, 3, 0, 3, Common ratio? ; 0 the ratios are not the same, so the sequence is not geometric. Common difference? 3 ( 6) = 3; 0 ( 3) = 3; 3 0 = 3 the differences are the same, so the sequence is arithmetic.

Example 2: Determine if the sequence is arithmetic, geometric, or neither and write the recursive and explicit equations. (If the sequence is neither, you do NOT need to write the equations). We will first need to determine if we have an arithmetic sequence, geometric sequence, or neither like we did in the previous example. Once we know which type of sequence we have, we will write the recursive and explicit equations using the following format (see tasks 1.4 & 1.5 for more detailed explanations): Recursive Explicit Example Arithmetic Always list the starting term and the rule. Rule: f(n) = f(n 1) + d ***The starting term is required as part of the recursive formula. f(n) = starting term value + d(n starting term#) For a 1 st term: f(n) = starting term value +d(n 1) 3, 7, 11, 15, 19, d = 4; the 1 st term is 3 so f(1) = 3 Recursive: f(1) = 3; f(n) = f(n 1) + 4 Explicit: f(n) = 3 + 4(n 1) **can be simplified to: f(n) = 4n 1 Geometric Always list the starting term and the rule. Rule: f(n) = f(n 1) x r ***The starting term is required as part of the recursive formula. (n starting term#) f(n) = starting term value x r For a 1 st (n 1) term: f(n) = starting term value x r 3, 12, 48, 192, r = 4; the 1 st term is 3 so f(1) = 3 Recursive: f(1) = 3; f(n) = f(n 1) x 4 Explicit: f(n) = 3 x (4) n 1 **you CANNOT multiply 3 by 4 due to the order of operations!!! 1) 20, 10, 5, 2.5, Common ratio? ; ;. The ratios are the same, so the sequence is geometric. Next, we need to identify the starting term. In this sequence, the 1 st term is 20, so f(1) = 20, and the common ratio is r=. Recursive Rule: f(1) = 20; f(n) = f(n 1) x Explicit Rule: f(n) = 20 x Our starting term is 1 (first term) so our exponent is n 1 2) 3, 6, 10, 15, 21, Common ratio? 2; the ratios are not the same, so the sequence is not geometric. Common difference? 6 3 = 3; 10 6 = 4 the differences are not the same, so the sequence is not arithmetic. This sequence is neither arithmetic, nor geometric. We will not write a recursive or explicit rule for this sequence. 3) 14, 11, 8, 5, 2, Common ratio? ; 0 the ratios are not the same, so the sequence is not geometric. Common difference? 11 14 = 3; 8 11 = 3; 5 8 = 3 the differences are the same, so the sequence is arithmetic. The 1 st term is 14, so f(1) = 14 & d = 3. Recursive Rule: f(1) = 14; f(n) = f(n 1) 3 Explicit Rule: f(n) = 14 3(n 1) or simplified as f(n) = 3n + 17 Video Examples: https://youtu.be/e0qkym5lrbi https://learnzillion.com/lesson_plans/5940 https://learnzillion.com/lesson_plans/6323 create an explicit formula

1.10 Ready, Set, Go! Ready Topic: Arithmetic and Geometric Sequences To find terms in an arithmetic sequence, you will add the common difference to the previous term to get the next term. To find terms in a geometric sequence, you will multiply each term by the common ratio to get the next term. Example 1: Find the first 5 terms for the given arithmetic & geometric sequences using the given information. Arithmetic sequence: f(1) = 4; common difference d = 6 Geometric sequence: g(1) = 4; common ratio r = 6 Arithmetic f(n) f(1) = 4 *this was given to us g(1) = 4 *this was given to us Geometric g(n) f(2) = 10 *we take the previous term, 4, and add 6 g(2) = 24 *we take the previous term, 4, and multiply by 6 f(3) = 16 *we take the previous term, 10, and add 6 g(3) = 144 *we take the previous term, 24, and multiply by 6 f(4) = 22 *we take the previous term, 16, and add 6 g(4) = 864 *we take the previous term, 144, and multiply by 6 f(5) = 28 *we take the previous term, 22, and add 6 g(5) = 5184 *we take the previous term, 864, and multiply by 6 Which sequence is growing faster? Why? Example 2: Find the first 5 terms for the given arithmetic & geometric sequences using the given information. Arithmetic sequence: f(1) = 4; common difference d = 2 Geometric sequence: g(1) = 24; common ratio r = Arithmetic f(n) Geometric g(n) f(1) = 4 *this was given to us g(1) = 24 *this was given to us f(2) = 6 *we take the previous term, 4, and add 2 g(2) = 12 *we take the previous term, 24, and multiply by f(3) = 8 *we take the previous term, 6, and add 2 f(4) = 10 *we take the previous term, 8, and add 2 f(5) = 12 *we take the previous term, 10, and add 2 g(3) = 6 *we take the previous term, 12, and multiply by g(4) = 3 *we take the previous term, 6, and multiply by g(5) = 1.5 *we take the previous term, 3, and multiply by Which sequence will have a bigger value for the 100 th term. Why? In an arithmetic sequence, the numbers get larger if the common difference is positive, and smaller if the common difference is negative. In a geometric sequence, the numbers grow if the common ratio is larger than 1, and the numbers get smaller (decay) if the ratio is between 0 and 1. If we have a common ratio like, the numbers in the sequence will decrease. When we compare arithmetic and geometric sequences, we see that if d is positive and r is greater than 1, the geometric sequence will grow faster (see the first example). Video Examples: https://youtu.be/r1hvjsqfghe

Set Topic: Finding missing terms in a geometric sequence A geometric sequence has a common ratio which means that we multiply (or divide) by the same number each time. For example, in the sequence 3, 6, 12, 24, 48, We multiply by 2 each time. We could keep track of this work as shown below: 3, 6, 12, 24, 48 x2 x2 x2 x2 We start at 3, multiply by 2 to get to 6, multiply by 2 again to get 12, multiply by 2 again to get 24, and multiply by 2 again to get 48. How many times do we multiply by 2 to get from the 1 st term of 3 to the 5 th term of 48? We started with 3 and multiplied by the common ratio (r) 4 times. We could write that equation like this: 3 x (2) 4 = 48 *** remember 2 x 2 x 2 x 2 can be written as 2 4 Let s try to write an equation for another geometric sequence. 1, 3, 9, 27, 81 x3 x3 x3 x3 1 st term Our first term, f(1) is 1 and our 5 th term, f(5) is 81. We are multiplying by 3 each time, so our common ratio is r=3. We had to multiply by the common difference 4 times to get from the 1 st term to the 5 th term (if we take 5 th 1 st we get 4). So our equation would be: 1 x (3) 4 = 22 r # of 5 th term times we multiply by r 1 st term r 5 # of th term times we multiply by r If we generalize these equations we get: starting term value x (r) n starting term # = ending term For the previous examples, we were given several terms in the sequence so we could easily find the common ratio. How can we use the equation we wrote to find r and missing terms if we don t have the entire sequence? Let s try an example: x (term number) 1 2 3 4 y(term value) 2 128 In this sequence we have the 1 st & 4 th terms, but we are missing terms 2 & 3. How can we find the common ratio (r) and the missing terms? We can use the equation: starting term value x (r) n starting term # = ending term. In this case our 1 st term is 2, our 4 th term is 128 and we will have to multiply by r 3 times. Our equation will look like this: 2 x r x r x r = 128 or 2(r) 3 = 128 Now we can solve this equation for r to find the common difference. Once we have r, we can use it to find the missing terms. 2(r) 3 = 128 Starting equation (r) 3 = 64 r = 4 Divide by 2 Use mental math, a calculator, or roots to answer the question what number to the 3 rd power is equal to 64?

We know now that the common ratio is 4, so we can find the missing terms by multiplying by 4. x (term number) 1 2 3 5 y(term value) 2 8 32 128 x4 x4 x4 Let s try another example: x (term number) 1 2 3 4 5 y(term value) 16 2 In this sequence we have the 1 st & 4 th terms, but we are missing terms 2, 3, and 5. How can we find the common ratio (r) and the missing terms? We can use the equation: starting term value x (r) n starting term # = ending term. In this case our 1 st term is 16, our 4 th term is 2 and we will have to multiply by r 3 times. Our equation will look like this: 16 x r x r x r = 2 or 16 x (r) 3 = 2 Notice, that we used the 4 th term this time. We can use any 2 terms as the starting and ending term in our equation as long as we remember to count how many times we multiply by r. Now we can solve this equation for r to find the common ratio. Once we have r, we can use it to find the missing terms. 16 x (r) 3 = 2 (r) 3 = Starting equation Divide by 16 r = Use mental math, a calculator, or roots to answer the question what number to the 3 rd power is equal to? It is helpful to look at the top and bottom of the fraction separately. We know now that the common difference is 5 so we can find the missing terms by adding 5. x (term number) 1 2 3 4 5 y(term value) 16 8 4 2 1 x x x x In the previous two examples, we had an odd number jumps between terms. Does the process change if we have an even number of jumps? Let s take a look. x (term number) 1 2 3 4 5 y(term value) 2 162 In this sequence we have the 1 st & 5 th terms, but we are missing terms 2, 3, and 4. How can we find the common ratio (r) and the missing terms? We can use the equation: starting term value x (r) n starting term # = ending term. In this case, our 1 st term is 2, our 5 th term is 162 and we will have to multiply by r 4 times. Our equation will look like this: 2 x r x r x r x r = 162 or 2(r) 4 = 162

Solve for r: 2(r) 4 = 162 (r) 4 = 81 r = 3 Starting equation Divide by 2. Use mental math, calculators, or roots to answer the question what number to the 4 th power is equal to 81? ***Notice there are 2 answers. This is because 3 x 3 x 3 x 3 = 81 AND ( 3) x ( 3) x ( 3) x ( 3) = 81 If we use r = 3 the missing terms would be: x (term number) 1 2 3 4 5 y(term value) 2 6 18 54 162 x3 x3 x3 x3 If we use r = 3 the missing terms would be: x (term number) 1 2 3 4 5 y(term value) 2 6 18 54 162 x ( 3) x ( 3) x ( 3) x ( 3) *** When r is raised to an even power, we will have 2 possible values for the common ratio. Let s take a look at some even exponents to see what happens: (2) 2 = 2 x 2 = 4 vs. ( 2) 2 = ( 2) x ( 2) = 4 ***A negative times a negative gives a positive answer (5) 4 = 5 x 5 x 5 x 5 = 625 vs. ( 5) 4 = ( 5) x ( 5) x ( 5) x ( 5) = 625 A negative times a negative = positive A negative times a negative = positive Video Examples: https://youtu.be/yfqdcfmdnma https://youtu.be/rdfvyozo2j8 Go Topic: Writing the explicit equation of a geometric sequence To write the explicit equation of a geometric sequence, we use the form: (starting term value) x (r) n starting term # = ending term The key components of this equation are a starting term number, the value of that starting term, and the common ratio.

For example, if we knew that the first term of a sequence was 20, f(1) = 20 and that the common ratio was 3, we could write the following equation: f(n) = 20(3) n 1 In this equation, 20 is the starting term s value, r = 3, and we use the 1 st term (starting term # is 1). Here is another example of a geometric pattern: In this case, we start at the beginning, we are using the 0 term. The 0 term has 3 dots, this means that f(0) = 3. If we make a table to represent this pattern it would look like this: We know that this pattern is Term # Term Value 0 min (beginning) 3 1 min 6 2 min 12 geometric, so we need to find the common ratio. To find r we divide a term by the previous term: r = 2 We also know that 0 is our starting term number, and 3 is our starting term value. Using this information, we can write n starting term # the explicit equation of the geometric sequence: ending term = (starting term value) x (r) Our equation would be: f(n) = 3(2) n 0 which simplifies to f(n) = 3(2) n Can we write the explicit equation for a geometric sequence if we are given the starting term and common ratio? Given: f(1) = 4, r = 5 In this case we are given all the information we need to write the explicit equation. We will use 4 as the starting term value, 5 for r, and 1 as the starting term number. We will then plug these values into our rule: n starting term # ending term = (starting term value) x (r) f(n) = 4(5) n 1 This is our explicit equation. One last example: Given: f(1) = 6, f(n) = 3f(n 1) This one may seem a bit tricky because we are not told the common ratio. However, we can use what we know about recursive equations to find it. In a recursive equation for a geometric sequence, the ratio is the number that is multiplied by the previous term. In the given recursive rule, we see that 3 is being multiplied by f(n 1) which represents the previous term. This means that the common ratio is 3. We also know that the starting term number is 1, and the starting term value is 6. We can use this information to write the explicit equation: f(n) = 6(3) n 1 Video Examples: http://virtualnerd.com/algebra 1/exponents exponential functions/geometric sequences/geometricsequence terms/geometric sequence find rule (only need to watch until 3:12)

1.11 Ready, Set, Go! Ready Topic: Comparing linear equations and arithmetic sequences When we explore multiple representations of arithmetic sequences, we see that the sequence has a constant rate of change called the common difference. This rate of change is also the slope of the line formed when we graph the sequence. Example 1: Create multiple representations of the following sequence Sequence: 2, 5, 8, 11, 14, Recursive Rule: f(1) = 2; f(n) = f(n 1) + 3 Table: Explicit Equation: f(n) = 2 + 3(n 1) or f(n) = 3n 1 n (term #) 1 2 2 5 3 8 f(n) 4 11 5 14 +3 +3 +3 +3 Graph: 14 12 10 8 6 4 2-2 -1 1 2 3 4 5 6 7 8-2 +3 +1 This is a sequence, so the graph is discrete. The slope is 3, this is the same as the common difference d = 3 Example 2: Create multiple representations of the linear equation Equation: f(n) = 4n + 1 Table: x y 2 7 1 3 0 1 1 5 2 9 +4 +4 +4 +4 Graph: 10 +1 8 +4 6-10 -8-6 -4-2 2 4 6 8 10-2 -10 What similarities do you see between the sequence in example 1 and the equation in example 2? 4 2-4 -6-8 This is a linear function, so the graph is continuous. This means our inputs can be negative numbers, zero, and positive numbers. We can also find the value of the function for fractions. Video Examples: https://youtu.be/7sg8h0y8ozk https://youtu.be/r4cdr_jbnmq https://youtu.be/ma_sofvrehq

Set Topic: Representations of arithmetic sequences In this module, your teacher asked to create multiple representations of arithmetic and geometric sequences. These representations include a context, a table, a graph, and recursive and explicit equations. Here are some examples of creating multiple representations. Example 1: Using the given representation for the arithmetic pattern, create the other representations. Recursive Equation: Explicit Equation: Table: Graph Context: Amanda is going to earn $5 per week for helping her grandma with some chores around the house. She plans to save her money. At the end of first week she has $5, at the end of the second week, she has a total of $10, and so on. In this example, we have the context and we need to create the other representations. In this case, it may be easiest to make a table to organize our information. Table: week total $ 1 5 2 10 3 15 4 20 We can look down the table to find our common difference. In this case, we are adding $5 each week so d = 5.We also know that our first term is 5 or f(1) = 5. Next, we can use the information from the table to write a recursive equation. We will use the starting term of 5 and the common difference of 5. Recursive Equation: f(starting term #) = starting term value; f(n) = f(n 1) + d f(1) = 5; f(n) = f(n 1) + 5 Our starting term # is 1, the starting value is 5, and d = 5 We can also use the starting term f(1) = 5 and common difference d = 5 to write an explicit equation. Explicit Equation: f(n) = starting term value + d(n starting term #) f(n) = 5 + 5(n 1) Our starting term # is 1, the starting value is 5, and d = 5 We can distribute and simplify our explicit equation to get f(n) = 5n, but this is not required.

Finally, we can graph the values from our table. Since a sequence can only have whole number inputs (1 st term, 2 nd term, 3 rd term, etc.) our graph will be discrete. This means that we will not connect the points with a line. week total $ 1 5 2 10 3 15 4 20 45 40 35 30 25 20 10 15 5 1 2 3 4 5 6 7 8 9 Example 2: Using the given representation for the arithmetic pattern, create the other representations. Recursive Equation: f(1) = 20; f(n) = f(n 1) 3 Explicit Equation: Table: Graph Context: In this example, we have a recursive equation. In this case, it may be easiest to make a table to organize our information. We are told that our first term is 20, f(1) = 20 and that the common difference is d = 3. This means we will start with 20 and subtract 3 each time. Table: term n value f(n) 1 20 2 17 3 14 4 11 We can also use the starting term f(1) = 20 and common difference d = 3 to write an explicit equation. Explicit Equation: f(n) = starting term value + d(n starting term #) f(n) = 20 3(n 1) Our starting term # is 1, the starting value is 20, and d = 3 We can distribute and simplify our explicit equation to get f(n) = 3n + 23, but this is not required.