lesson 8 homework 1.6

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Lesson 8.1.1, lesson 8.1.2, lesson 8.1.3, lesson 8.1.4, lesson 8.1.5, lesson 8.2.1, lesson 8.2.2, lesson 8.2.3, lesson 8.2.4, lesson 8.2.5.

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Solving Rational Functions (Lesson 8.6)

Unit 1: sequences and linear functions, day 1: recursive sequences, day 2: applications of arithmetic sequences, day 3: sum of an arithmetic sequence, day 4: applications of geometric sequences, day 5: sequences review, day 6: quiz 1.1 to 1.4, day 7: linear relationships, day 8: point-slope form of a line, day 9: standard form of a linear equation, day 10: quiz 1.5 to 1.7, day 11: unit 1 review, day 12: unit 1 test, unit 2: linear systems, day 1: linear systems, day 2: number of solutions, day 3: elimination, day 4: larger systems of equations, day 5: quiz 2.1 to 2.4, day 6: systems of inequalities, day 7: optimization using systems of inequalities, day 8: quiz 2.5 to 2.6, day 9: unit 2 review, day 10: unit 2 test, unit 3: function families and transformations, day 1: interpreting graphs, day 2: what is a function, day 3: translating functions, day 4: quiz 3.1 to 3.3, day 5: quadratic functions and translations, day 6: square root functions and reflections, day 7: absolute value functions and dilations, day 8: equations of circles, day 9: quiz 3.4 to 3.7, day 10: unit 3 review, day 11: unit 3 test, unit 4: working with functions, day 1: using multiple strategies to solve equations, day 2: solving equations, day 3: solving nonlinear systems, day 4: quiz 4.1 to 4.3, day 5: combining functions, day 6: composition of functions, day 7: inverse relationships, day 8: graphs of inverses, day 9: quiz 4.4 to 4.7, day 10: unit 4 review, day 11: unit 4 test, unit 5: exponential functions and logarithms, day 1: writing exponential functions, day 2: graphs of exponential functions, day 3: applications of exponential functions, day 4: quiz 5.1 to 5.3, day 5: building exponential models, day 6: logarithms, day 7: graphs of logarithmic functions, day 8: quiz 5.4 to 5.6, day 9: unit 5 review, day 10: unit 5 test, unit 6: quadratics, day 1: forms of quadratic equations, day 2: writing equations for quadratic functions, day 3: factoring quadratics, day 4: factoring quadratics. part 2., day 5: solving using the zero product property, day 6: quiz 6.1 to 6.4, day 7: completing the square, day 8: completing the square for circles, day 9: quadratic formula, day 10: complex numbers, day 11: the discriminant and types of solutions, day 12: quiz 6.5 to 6.9, day 13: unit 6 review, day 14: unit 6 test, unit 7: higher degree functions, day 1: what is a polynomial, day 2: forms of polynomial equations, day 3: polynomial function behavior, day 4: repeating zeros, day 5: quiz 7.1 to 7.4, day 6: multiplying and dividing polynomials, day 7: factoring polynomials, day 8: solving polynomials, day 9: quiz 7.5 to 7.7, day 10: unit 7 review, day 11: unit 7 test, unit 8: rational functions, day 1: intro to rational functions, day 2: graphs of rational functions, day 3: key features of graphs of rational functions, day 4: quiz 8.1 to 8.3, day 5: adding and subtracting rational functions, day 6: multiplying and dividing rational functions, day 7: solving rational functions, day 8: quiz 8.4 to 8.6, day 9: unit 8 review, day 10: unit 8 test, unit 9: trigonometry, day 1: right triangle trigonometry, day 2: solving for missing sides using trig ratios, day 3: inverse trig functions for missing angles, day 4: quiz 9.1 to 9.3, day 5: special right triangles, day 6: angles on the coordinate plane, day 7: the unit circle, day 8: quiz 9.4 to 9.6, day 9: radians, day 10: radians and the unit circle, day 11: arc length and area of a sector, day 12: quiz 9.7 to 9.9, day 13: unit 9 review, day 14: unit 9 test, learning targets.

Solve equations with rational functions using a variety of methods.

Identify extraneous solutions.

Activity: One Last HW Assignment

Lesson handouts, media locked.

Our Teaching Philosophy:

Experience first, formalize later (effl), experience first.

One last time, we're going to help out Reese and Brody with their homework. The goal is that as students solve the problems in Reese's homework, they will note the process they're using and will apply it to Brody's homework. The problems were paired according to the strategy used to solve the question. For example, in the first question of Reese's homework, the denominators are the same so we just need to solve the numerator. This is the same in Brody's homework.

Give groups plenty of time to just work through the problems. There isn't one set way to solve each problem so let them try out different methods. The intention of today's lesson is that students use a variety of methods to solve depending on what they notice in the structure of the equations. As you are checking in with groups, look for multiple solution paths to the same problem and ask all of them to put their work on the board. Some methods may be more efficient but they are all equally valid. We want to celebrate the students' thinking and problem solving approaches.

You do need to look for a group that solved Brody's 2nd question in a way that created an extraneous solution. Using cross products does this. Ask that group to share their work so you will be able to debrief extraneous solutions.

Formalize Later

After many solution paths have been added to the board, ask groups to explain their work. Make sure to affirm each student who put up their work. We don't want to give the impression that one method is better than another. As groups are explaining their work, try to summarize the method they used as a margin note. We've put some examples of these methods on the answer key but your students may come up with more.

When going over Brody's 2nd homework question, point out how some groups found a solution of only 9 while some groups found solutions of 9 and 0. Do both solutions work? Can we check our answers? Students will see that 0 cannot be a solution because when we plug it into our original equation, it becomes undefined. This means 0 is an extraneous solution. Extraneous solutions often show up when we take a linear equation and turn it into a quadratic by multiplying by x . Often the extraneous solution makes a denominator zero or gives the opposite sign when plugged in.

After going over all of the problems and summarizing the different methods used, create a list of these methods in the QuickNotes. Feel free to add to the list on the answer key if your students came up with more methods. Groups can work on the Check Your Understanding problems. Question #2 has an extraneous solution.

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Eureka Math Grade 8 Module 5 Lesson 8 Answer Key

Engage ny eureka math 8th grade module 5 lesson 8 answer key, eureka math grade 8 module 5 lesson 8 exploratory challenge/exercise answer key.

Exercise 1. Consider the function that assigns to each number x the value x 2 . a. Do you think the function is linear or nonlinear? Explain. Answer: I think the function is nonlinear. The equation describing the function is not of the form y = mx + b.

d. What shape does the graph of the points appear to take? Answer: It appears to take the shape of a curve.

e. Find the rate of change using rows 1 and 2 from the table above. Answer: \(\frac{25 – 16}{ – 5 – ( – 4)}\) = \(\frac{9}{ – 1}\) = – 9

f. Find the rate of change using rows 2 and 3 from the table above. Answer: \(\frac{16 – 9}{ – 4 – ( – 3)} = \) = \(\frac{7}{ – 1}\) = – 7

g. Find the rate of change using any two other rows from the table above. Answer: Student work will vary. \(\frac{16 – 25}{4 – 5}\) = \(\frac{ – 9}{ – 1}\) = 9

h. Return to your initial claim about the function. Is it linear or nonlinear? Justify your answer with as many pieces of evidence as possible. Answer: This is definitely a nonlinear function because the rate of change is not a constant for different intervals of inputs. Also, we would expect the graph of a linear function to be a set of points in a line, and this graph is not a line. As was stated before, the expression x 2 is nonlinear.

Exercise 2. Consider the function that assigns to each number x the value x 3 . a. Do you think the function is linear or nonlinear? Explain. Answer: I think the function is nonlinear. The equation describing the function is not of the form y = mx + b.

e. Find the rate of change using rows 2 and 3 from the table above. Answer: \(\frac{ – 8 – ( – 3.375)}{ – 2 – ( – 1.5)}\) = \(\frac{ – 4.625}{ – 0.5}\) = 9.25

f. Find the rate of change using rows 3 and 4 from the table above. Answer: \(\frac{ – 3.375 – ( – 1)}{ – 1.5 – ( – 1)}\) = \(\frac{ – 2.375}{ – 0.5}\) = 4.75

g. Find the rate of change using rows 8 and 9 from the table above. Answer: \(\frac{1 – 3.375}{1 – 1.5}\) = \(\frac{ – 2.375}{ – 0.5}\) = 4.75

h. Return to your initial claim about the function. Is it linear or nonlinear? Justify your answer with as many pieces of evidence as possible. Answer: This is definitely a nonlinear function because the rate of change is not a constant for any interval of inputs. Also, we would expect the graph of a linear function to be a line, and this graph is not a line. As was stated before, the expression x 3 is nonlinear.

Exercise 3. Consider the function that assigns to each positive number x the value \(\frac{1}{x}\). a. Do you think the function is linear or nonlinear? Explain. Answer: I think the function is nonlinear. The equation describing the function is not of the form y = mx + b.

e. Find the rate of change using rows 1 and 2 from the table above. Answer: \(\frac{10 – 5}{0.1 – 0.2}\) = \(\frac{5}{ – 0.1}\) = – 50

f. Find the rate of change using rows 2 and 3 from the table above. Answer: \(\frac{5 – 2.5}{0.2 – 0.4}\) = \(\frac{2.5}{ – 0.2}\) = – 12.5

g. Find the rate of change using any two other rows from the table above. Answer: Student work will vary. \(\frac{1 – 0.625}{1 – 1.6}\) = \(\frac{0.375}{ – 0.6}\) = – 0.625

h. Return to your initial claim about the function. Is it linear or nonlinear? Justify your answer with as many pieces of evidence as possible. Answer: This is definitely a nonlinear function because the rate of change is not a constant for any interval of inputs. Also, we would expect the graph of a linear function to be a line, and this graph is not a line. As was stated before, the expression \(\frac{1}{x}\) is nonlinear.

Exercises 4–10 In each of Exercises 4–10, an equation describing a rule for a function is given, and a question is asked about it. If necessary, use a table to organize pairs of inputs and outputs, and then plot each on a coordinate plane to help answer the question.

Eureka Math Grade 8 Module 5 Lesson 8 Problem Set Answer Key

Question 1. Consider the function that assigns to each number x the value x 2 – 4. a. Do you think the function is linear or nonlinear? Explain. Answer: The equation describing the function is not of the form y = mx + b. It is not linear.

b. Do you expect the graph of this function to be a straight line? Answer: No

Question 2. Consider the function that assigns to each number x greater than – 3 the value \(\frac{1}{x + 3}\). a. Is the function linear or nonlinear? Explain. Answer: The equation describing the function is not of the form y = mx + b. It is not linear.

d. Was your prediction to (b) correct? Answer: Yes, the graph appears to be taking the shape of some type of curve.

b. What is the average rate of change for this function from an input of x = – 2 to an input of x = – 1? Answer: \(\frac{ – 2 – 1}{ – 2 – ( – 1)}\) = \(\frac{ – 3}{ – 1}\) = 3

Eureka Math Grade 8 Module 5 Lesson 8 Exit Ticket Answer Key

Question 2. Consider the function that assigns to each number x the value \(\frac{1}{2}\) x 2 . Do you expect the graph of this function to be a straight line? Briefly justify your answer. Answer: The equation is nonlinear (not of the form y = mx + b), so the function is nonlinear. Its graph will not be a straight line.

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