Tag: Eureka Math Algebra 1 Module 5

Engage NY Eureka Math Algebra 1 Module 5 Lesson 9 Answer Key

Eureka Math Algebra 1 Module 5 Lesson 9 Example Answer Key

Example 1.
Marymount Township secured the construction of a power plant, which opened in 1990. Once the power plant opened in 1990, the population of Marymount increased by about 20% each year for the first ten years and then increased by 5% each year after that.
a. If the population was 150, 000 people in 2010, what was the population in 2000?
Answer:
Sample Response: We can tell that this problem involves a geometric sequence because we are multiplying each term by either 1.2 or 1.05.
This is also a piecewise function since the first ten years the population grows at one rate, and after that it grows at a different rate.
We need to start backward from 2010 to 2000, since we know the size of the population for 2010.
Geometric sequence: a_n = a₁ rⁿ
2000 to 2010
150 000 = a₁ (1.05)¹⁰ → \(\frac{150000}{(1.05)^{10}}\) = a₁ → a₁ = 92 086.99 ≈ 92 087

b. How should you round your answer? Explain.
Answer:
The 2010 value appears to be rounded to the nearest thousand (150, 000). We will use a similar level of precision in our result: 92, 000.

c. What was the population in 1990?
Answer:
Sample Response: For 1990 to 2000 we know the final population from our answer to part (a), so we can use that to find the initial population in 1990. Note: For this sample response, we rounded off to 92, 000 people.
92 087 = a₁ (1.2)¹⁰ → \(\frac{92087}{(1.2)^{10}}\) = a₁ → a₁ = 14 872.56, so 15 000

Example 2.
If the trend continued, what would the population be in 2009?
Answer:
Sample response: Since that is one year before the end of our sample, we can divide 150, 000 by 1.05 to find the value before it.
\(\frac{150000}{1.05}\) ≈ 143 000

Eureka Math Algebra 1 Module 5 Lesson 9 Exercise Answer Key

Opening Exercise
What does it mean to attend to precision when modeling in mathematics?
Answer:
Using clear and accurate language when interpreting the meaning and discussing the results of the mathematical model; labeling axes on a graph; specifying the meaning of a variable in a particular function; including units of measure when appropriate; showing mathematical work clearly using correct mathematical symbols and notations; making accurate calculations and checking for the reasonableness of the calculations.

Exercises

Exercise 1.
A tortoise and a hare are having a race. The tortoise moves at 4 miles per hour. The hare travels at 10 miles per hour. Halfway through the race, the hare decides to take a 5 – hour nap and then gets up and continues at 10 miles per hour.
a. If the race is 40 miles long, who won the race? Support your answer with mathematical evidence.
Answer:
The time for the tortoise to finish is \(\frac{40 \text { miles }}{4 \mathrm{mph}}\), or 10 hours. The time for the hare to finish is \(\frac{40 \text { miles }}{4 \mathrm{mph}}\) + 5 hours , or 9 hours. So, the hare beat the tortoise by one hour.

b. How long (in miles) would the race have to be for there to be a tie between the two creatures, if the same situation (as described in Exercise 1) happened?
Answer:
Sample solution: Let the tortoise’s time be T_t and the hare’s time be T_h.
T_t = \(\frac{D}{4}\)
T_h = \(\frac{D}{10}\) + 5 (racing + napping)
Since both creatures finished the same distance D in the same TOTAL time,
T_t = T_h
\(\frac{D4latex] = [latex]\frac{D}{10}\) + 5
D = 33 \(\frac{1}{3}\)
The race would need to be 33 \(\frac{1}{3}\) miles in order for there to be a tie.
Check: T_t = \(\frac{33 \frac{1}{3}}{4}\) = 8 \(\frac{1}{3}\) T_h = \(\frac{33 \frac{1}{3}}{10}\) + 5 = 8 \(\frac{1}{3}\)

Exercise 2.
The graph on the right represents the value V of a popular stock. Its initial value was $12/share on day 0.
Note: The calculator uses X to represent t, and Y to represent V.

a. How many days after its initial value at time t = 0 did the stock price return to $12 per share?
Answer:
By the symmetry of quadratic equations, the stock must return to its initial value after 6 more days, at t = 12.

b. Write a quadratic equation representing the value of this stock over time.
Answer:
Since the quadratic equation reaches a minimum at (6, 3), use vertex form to write V = a(t – 6)² + 3. The initial value at t = 0 was 12. So, by substitution, 12 = a(0 – 6)² + 3 → a = \(\frac{1}{4}\). Therefore, the final equation is V = \(\frac{1}{4}\) (t – 6)² + 3.

c. Use this quadratic equation to predict the stock’s value after 15 days.
Answer:
V = \(\frac{1}{4}\) (15 – 6)² + 3 → V = 23.25
The predicted value of the stock after 15 days is $23.25.

Eureka Math Algebra 1 Module 5 Lesson 9 Problem Set Answer Key

Question 1.
According to the Center for Disease Control and Prevention, the breast cancer rate for women has decreased at 0.9% per year between 2000 and 2009.
a. If 192, 370 women were diagnosed with invasive breast cancer in 2009, how many were diagnosed in 2005? For this problem, assume that there is no change in population from 2005 and 2009.
Answer:
Geometric sequence: Common ratio is (1 – 0.009) = 0.991.
a_n = a₁ (common ratio)^{n – 1}
192 370 = a₁ (0.991)⁴
\(\frac{192370}{(0.991)^{4}}\) = a₁
a₁ = 199 453.98…
a₁ = 199, 454

b. According to the American Cancer Society, in 2005 there were 211, 240 people diagnosed with breast cancer. In a written response, communicate how precise and accurate your solution in part (a) is, and explain why.
Answer:
My solution is precise because my classmates and I followed the same protocols and our values were similar to each other. Since the model we used did not take into account the population increase, our values were off by 11, 786 people, which is 5.6%. I believe that being off by only 5.6% is still pretty close to the actual value. My solution was precise and accurate; it could have been more accurate if the population growth was taken into account in the exponential model.

Question 2.
The functions f and g represent the population of two different kinds of bacteria, where x is the time (in hours) and f and g are the number of bacteria (in thousands). f(x) = 2x² + 7 and g(x) = 2^x.
a. Between the third and sixth hour, which bacteria had a faster rate of growth?
Answer:
Looking at the graphs of the two functions, the graph of y = g(x) is steeper over the interval [3, 6] than the graph of y = f(x). Therefore, the bacteria population represented by y = g(x) has a faster rate of growth on this interval.

Using the functions to find the average rate of change over the interval [3, 6],
\(\frac{f(6) – f(3)}{6 – 3}\) = \(\frac{79 – 25}{3}\) = 18
\(\frac{g(6) – g(3)}{6 – 3}\) = \(\frac{64 – 8}{3}\) ≈ 18.7
g has a slightly higher average rate of change on this interval.

b. Will the population of g ever exceed the population of f? If so, at what hour?
Answer:
Since the question asks for the hour (not the exact time), I made a table starting at the 6th hour and compared the two functions. Once g exceeded f, I stopped. So, sometime in the 7 th hour, the population of g exceeds the population of f.

Eureka Math Algebra 1 Module 5 Lesson 9 Exit Ticket Answer Key

The distance a car travels before coming to a stop once a driver hits the brakes is related to the speed of the car when the brakes were applied. The graph of f (shown) is a model of the stopping distance (in feet) of a car traveling at different speeds (in miles per hour).

Question 1.
One data point on the graph of f appears to be (80, 1000). What do you think this point represents in the context of this problem? Explain your reasoning.
Answer:
In this problem, 80 would represent the speed in miles per hour and 1, 000 would represent the stopping distance in feet. It does not make sense for a car to be traveling at 1, 000 mph, much less only take 80 ft. to come to a stop.

Question 2.
Estimate the stopping distance of the car if the driver is traveling at 65 mph when she hits the brakes. Explain how you got your answer.
Answer:
I can estimate the stopping distance by sketching a curve to connect the data points and locating the y – coordinate of a point on this curve when its x – coordinate is 65. My estimate is approximately 670 ft.
Alternately, assume this is a quadratic function in the form f(x) = kx², where f is the stopping distance, in feet, and x is the speed, in miles per hour. Using the point (80, 1000), k = \(\frac{1000}{80^{2}}\) = 0.15625, so f(x) = 0.15625x². Using this model, f(65) = 660.15625. My estimate is approximately 660 ft.

Question 3.
Estimate the average rate of change of f between x = 50 and x = 60. What is the meaning of the rate of change in the context of this problem?
Answer:
f(50) ≈ 400 and f(60) ≈ 580. \(\frac{580 – 400}{10}\) = 18. The average rate of change between x = 50 and x = 60 is approximately 18 ft/mph. This means that between 50 and 60 mph, the stopping distance is increasing by approximately 18 ft. for each additional mph increase in speed.

Question 4.
What information would help you make a better prediction about stopping distance and average rate of change for this situation?
Answer:
A table of data or an algebraic function would help in making better predictions.

Engage NY Eureka Math Algebra 1 Module 5 Lesson 8 Answer Key

Eureka Math Algebra 1 Module 5 Lesson 8 Example Answer Key

Example 1.
Christine has $500 to deposit in a savings account, and she is trying to decide between two banks. Bank A offers 10% annual interest compounded quarterly. Rather than compounding interest for smaller accounts, Bank B offers to add $15 quarterly to any account with a balance of less than $1,000 for every quarter, as long as there are no withdrawals. Christine has decided that she will neither withdraw, nor make a deposit for a number of years.
Develop a model that will help Christine decide which bank to use.
Answer:
Students may decide to use a table, graph, or equation to model this situation. For this situation, a table allows comparison of the balances at the two banks. It appears that Bank B has a higher balance until the fourth year.

It is tempting to just say that at more than 3 years, Bank A is better. If students reach this conclusion, use the following line of questioning:
Can we give a more exact answer? Since the interest is compounded quarterly, we may want to consider the quarters of year 3. If after choosing a bank, Christine wanted to make sure her money was earning as much as possible, after which quarter would she make the withdrawal?
Since the balances intersect in year 4, we can look at the balances for each quarter of that year:

We can see from this table that it is not until the 4th quarter that Bank A begins to make more money for Christine. If she chooses Bank A, she should leave her money there for more than 3 years and 9 months. If she chooses Bank B, she should withdraw before the 4th quarter of Year 3.

Bank A: A(t) = 500(1 + \(\frac{0.10}{4}\))^4t or 500(1.025)^4t
Bank B: B(t) = 500 + 60t (Since her money earns $15 quarterly, then $15(4), or $60 per year.)

Example 2.
Alex designed a new snowboard. He wants to market it and make a profit. The total initial cost for manufacturing set – up, advertising, etc. is $500,000, and the materials to make the snowboards cost $100 per board.
The demand function for selling a similar snowboard is D(p) = 50 000 – 100p, where p represents the selling price (in dollars) of each snowboard.
a. Write an expression for each of the following in terms of p.
Demand Function (number of units that will sell)
Answer:
This is given as 50 000 – 100p.

Revenue (number of units that will sell)(price per unit, p)
Answer:
(50 000 – 100p)p = 50 000p – 100p²

Total Cost (cost for producing the snowboards)
Answer:
This is the total overhead costs plus the cost per snowboard times the number of snowboards:
500 000 + 100(50 000 – 100p)
500 000 + 5 000 000 – 10 000p
5 500 000 – 10 000p

b. Write an expression to represent the profit.
Answer:
Profit = Total Revenue – Total Cost
= (50 000p – 100p² ) – (5 500 000 – 10 000p)
Therefore, the profit function is f(p) = – 100p² + 60 000p – 5 500 000.

c. What is the selling price of the snowboard that will give the maximum profit?
Answer:
Solve for the vertex of the quadratic function using completing the square.
(f(p) = – 100p² + 60 000p – 5 500 000
= – 100(p² – 600p + ) – 5 500 000
= – 100(p² – 600p + 300² ) – 5 500 000 + 100(300)²
= – 100(p – 300)² – 5 500 000 + 9 000 000
= – 100(p – 300)² + 3 500 000)
The maximum point is (300,3 500 000). Therefore, the selling price that will yield the maximum profit is $300.

d. What is the maximum profit Alex can make?
Answer:
According to the vertex form, the maximum profit would be $3,500,000 for selling at a price of $300 for each snowboard.

Eureka Math Algebra 1 Module 5 Lesson 8 Exercise Answer Key

Exercises
Alvin just turned 16 years old. His grandmother told him that she will give him $10,000 to buy any car he wants whenever he is ready. Alvin wants to be able to buy his dream car by his 21st birthday, and he wants a 2009 Avatar Z, which he could purchase today for $25,000. The car depreciates (reduces in value) at a rate is 15% per year. He wants to figure out how long it would take for his $10,000 to be enough to buy the car, without investing the $10,000.

Exercise 1.
Write the function that models the depreciated value of the car after n number of years.

Answer:
f(n) = 25 000(1 – 0.15)ⁿ or f(n) = 25 000(0.85)ⁿ

a. Will he be able to afford to buy the car when he turns 21? Explain why or why not.
Answer:
f(n) = 25 000(0.85)ⁿ
f(5) = 25 000(0.85)⁵ = 11 092.63
No, he will not be able to afford to buy the car in 5 years because the value of the car will still be $11,092.63, and he has only $10,000.

b. Given the same rate of depreciation, after how many years will the value of the car be less than $5,000?
Answer:
f(10) = 25 000(0.85)¹⁰ = 4921.86
In 10 years, the value of the car will be less than $5,000.

c. If the same rate of depreciation were to continue indefinitely, after how many years would the value of the car be approximately $1?
Answer:
f(n) = 25 000(0.85)ⁿ = 1
In about 62 years, the value of the car is approximately $1.

Exercise 2.
Sophia plans to invest $1,000 in each of three banks.
Bank A offers an annual interest rate of 12%, compounded annually.
Bank B offers an annual interest rate of 12%, compounded quarterly.
Bank C offers an annual interest rate of 12%, compounded monthly.
a. Write the function that describes the growth of investment for each bank in n years.
Answer:
Bank A A(n) = 1000(1.12)ⁿ
Bank B B(n) = 1000(1 + \(\frac{0.14}{4}\))⁴ⁿ or 1000(1.03)⁴ⁿ
Bank C C(n) = 1000(1 + \(\frac{0.12}{12}\))¹²ⁿ or 1000(1.01)¹²ⁿ

b. How many years will it take to double her initial investment for each bank? (Round to the nearest whole dollar.)

Answer:

For Bank A, her money will double after 7 years.
For Banks B and C, her investment will double its value during the 6^th year.

c. Sophia went to Bank D. The bank offers a “double your money” program for an initial investment of $1,000 in five years, compounded annually. What is the annual interest rate for Bank D?
Answer:
Given information for Bank D:
Initial investment = $1,000
Number of Years = 5 (She would have $2,000 in 5 years.)
Compounded annually:
2×1000 = 1000(1 + r)⁵
2000 = 1000(1 + r)⁵
\(\frac{2000}{1000}\) = \(\frac{1000(1 + r)^{5}}{1000}\)
2 = (1 + r)⁵
Since we see that 2 is the 5th power of a number, we must take the 5th root of 2.
\(\sqrt [ 5 ]{ 2 }\) = (1 + r)
1.1487 = 1 + r
1.1487 – 1 = r
r = 0.1487 or 14.87% (annual interest rate for Bank D)

Eureka Math Algebra 1 Module 5 Lesson 8 Problem Set Answer Key

Question 1.
Maria invested $10,000 in the stock market. Unfortunately, the value of her investment has been dropping at an average rate of 3% each year.
a. Write the function that best models the situation.
Answer:
f(n) = 10 000(1 – 0.03)ⁿ or 10 000(0.97)ⁿ
where n represents the number of years since the initial investment

b. If the trend continues, how much will her investment be worth in 5 years?
Answer:
f(n) = 10 000(0.97)ⁿ
f(5) = 10 000(0.97)⁵
f(5) = 8587.34 or $8,587.34

c. Given the situation, what should she do with her investment?
Answer:
There are multiple answers, and there is no wrong answer. Sample Responses:
Answer 1: After two years, she should pull out her money and invest it in another company.
Answer 2: According to experts, she should not touch her investment and wait for it to bounce back.

Question 2.
The half – life of the radioactive material in Z – Med, a medication used for certain types of therapy, is 2 days. A patient receives a 16 mCi dose (millicuries, a measure of radiation) in his treatment. (Half – life means that the radioactive material decays to the point where only half is left.)
a. Make a table to show the level of Z – Med in the patient’s body after n days.

Answer:

b. Write a formula to model the half – life of Z – Med for n days. (Be careful here. Make sure that the formula works for both odd and even numbers of days.)
Answer:
f(n) = 16(\(\frac{1}{2}\))^{\(\frac{n}{2}\)}
where n represents the number of days after the initial measurement of 16 mCi

c. How much radioactive material from Z – Med is left in the patient’s body after 20 days of receiving the medicine?
Answer:
For n = 20
f(20) = 16(0.5)¹⁰
f(20) = 0.015625
After ten days, there is 0.015625 mCi of the radioactive material in Z – Med left in the patient’s body.

Question 3.
Suppose a male and a female of a certain species of animal were taken to a deserted island. The population of this species quadruples (multiplies by 4) every year. Assume that the animals have an abundant food supply and that there are no predators on the island.
a. What is an equation that can be used to model the population of the species?
Answer:
f(n) = 2(4)ⁿ
where n represents the number of years after their arrival at the island

b. What will the population of the species be after 5 years?

Answer:

In 5 years, the population of the animals will reach 2,048.

c. Write an equation to find how many years it will take for the population of the animals to exceed 1 million. Find the number of years, either by using the equation or a table

Answer:

Using the equation: 2(4)ⁿ = 1 000 000
4ⁿ = 500 000
Note: Students will likely be unable to solve this equation without using trial and error (educated guessing). They may come up with 2(4)^9.5 = 1 048 576 using this method.

Question 4.
The revenue of a company for a given month is represented as R(x) = 1,500x – x² and its costs as C(x) = 1,500 + 1,000x. What is the selling price, x, of its product that would yield the maximum profit? Show or explain your answer.
Answer:
P(x) = Revenue – Cost
P(x) = R(x) – C(x)
P(x) = (1500x – x²) – (1500 + 1000x)
P(x) = – x² + 500x – 1500
Profit Function

To find the vertex, we can complete the square for the function:
P(x) = – x² + 500x – 1500
= – 1(x² – 500x + ) – 1500 Group the x – terms and factor out the – 1.
= – 1(x² – 500x + 62 500) – 1500 + 62 500 Complete the square.
P(x) = – 1(x – 250)² + 61 000
So, the maximum point will be (250,61 000), and the selling price should be $250 per unit to yield a maximum profit of $61,000.

Eureka Math Algebra 1 Module 5 Lesson 8 Exit Ticket Answer Key

Answer the following question. Look back at this (or other) lessons if you need help with the business formulas.

Jerry and Carlos each have $1,000 and are trying to increase their savings. Jerry will keep his money at home and add $50 per month from his part – time job. Carlos will put his money in a bank account that earns a 4% yearly interest rate, compounded monthly. Who has a better plan for increasing his savings?
Answer:
Jerry’s savings: J(n) = 1000 + 10(n), where n represents the number of months
Carlos’s savings: C(n) = 1000(1 + \(\frac{0.06}{12}\))ⁿ, where n represents the number of months

In the short term, Jerry’s plan is better, but I know that Carlos’s plan will eventually exceed Jerry’s because an increasing exponential model will always outgain an increasing linear model eventually. If they continued saving for 253 months, Carlos’s plan would be better.

Engage NY Eureka Math Algebra 1 Module 5 Lesson 7 Answer Key

Eureka Math Algebra 1 Module 5 Lesson 7 Exercise Answer Key

Opening Exercise
What is this data table telling us?

Answer:
The relationship between the age of runners in the NYC Marathon and their running time.

Exercises

Exercise 1.
Use the following data table to construct a regression model, and then answer the questions.

a. What function type appears to be the best fit for this data? Explain how you know.
Answer:
The relationship between frying time and moisture content is best modeled by an exponential regression. Using the calculator yields the function f(x) = 13.895(0.957)^x with a coefficient of determination of approximately 0.904. (r = -0.95077…, so r² = 0.9039 ….)

b. A student chooses a quadratic regression to model this data. Is he right or wrong? Why or why not?
Answer:
This data cannot be modeled by a quadratic regression because as cooking time increases, moisture content will always decrease and never begin to increase again. Also, in looking at the longer-term trend, we see that for a quadratic model the values are decreasing initially but will eventually begin to increase. This makes the quadratic model less reliable for larger x-values.

c. Will the moisture content for this product ever reach 0%? Why or why not?
Answer:
The moisture content will never reach 0% because exponential decay functions get smaller and smaller but never disappear entirely.

d. Based on this model, what would you expect the moisture content to be of a chicken breast fried for 50 minutes?
Answer:
f(50) = 13.895(0.957⁵⁰) = 1.5
The expected moisture content for a chicken breast fried for 50 minutes would be 1.5%.

Exercise 2.
Use the following data table to construct a regression model, then answer the questions based on your model.

Answer:
Using technology to find a linear regression model, we find that the best-fit line is y = -0.3988x + 31.517, with a correlation coefficient of -0.952.

a. What trends do you see in this collection of data?
Answer:
The data seem to be dropping gradually over the years but at a fairly constant, though small, negative rate. The correlation coefficient of nearly -1 indicates that this model has a strong negative linear relationship.

b. How do you interpret this trend?
Answer:
The rate of leisure-time physical activity in the U.S. has slowly declined over the years since 1988 and is likely to continue to do so.

c. If the trend continues, what would we expect the percentage of people in the U.S. who report no leisure-time physical activity to be in 2020?
Answer:
The year 2020 is year 32, since 1988, so f(32) = -0.3988(32) + 31.517 = 18.755. So, if this trend continues, we would expect about 19% of the population to report no leisure-time physical activity in 2020.

Eureka Math Algebra 1 Module 5 Lesson 7 Problem Set Answer Key

Question 1.
Use the following data tables to write a regression model, and then answer the questions:
Prescription Drug Sales in the United States Since 1995

a. What is the best model for this data?
Answer:
The best model for this data would be an exponential regression, given by the function f(t) = 65.736(1.161)^t with a correlation coefficient of 0.987 and a correlation of determination of 0.975.

b. Based on your model, what were prescription drug sales in 2002? 2005?
Answer:
2002: f(7) = 186.9, and 2005: f(10) = 292.5

c. For this model, would it make sense to input negative values for t into your regression? Why or why not?
Answer:
Because there were prescription drug sales in the years prior to 1995, it would make sense to use negative numbers with this model (unless some drastic change in drug sales in 1995 makes this model inaccurate for preceding years).

Question 2.
Use the data below to answer the questions that follow:
Per Capita Ready-to-Eat Cereal Consumption in the United States per Year Since 1980

a. What is the best model for this data?
Answer:
The best regression fit here is the quadratic f(t) = -0.018t² + 0.637t + 10.797 with correlation coefficient of 0.92 and a coefficient of determination (r²) of 0.85.

b. Based on your model, what would you expect per capita cereal consumption to be in 2002? 2005?
Answer:
According to the model, f(22) = 16.1, and f(25) = 15.5.
Based on the model, the expected per capita cereal consumption in 2002 would be 16.1 lb. In 2005, it would be 15.5 lb.
(Note: Because this model has a little lower coefficient of determination (0.85), these predictions may not seem to fit well with the given data table.)

c. For this model, will it make sense to input t-values that return negative output values? Why or why not?
Answer:
No. Negative output values for this model would correspond to negative pounds of cereal consumed, which is impossible. Therefore, this model would only be useful over the domain where f is positive.

Eureka Math Algebra 1 Module 5 Lesson 7 Exit Ticket Answer Key

Question 1.
Use the following data table to construct a regression model, and then answer the questions.

a. What is the best regression model for the data?
Answer:
The best model for regression here is linear, modeled using the calculator as f(x) = 3.657x + 25.277 with a correlation coefficient of 0.6547.

b. Based on your regression model, what height would you expect a person with a shoe length of 13.4 inches to be?
Answer:
f(13.4) = 74 → a person with shoes 13.4 inches long might be 74 inches tall.

c. Interpret the value of your correlation coefficient in the context of the problem.
Answer:
Based on the correlation coefficient, there is a moderate positive linear relationship between shoe length and height.

Engage NY Eureka Math Algebra 1 Module 5 Lesson 6 Answer Key

Eureka Math Algebra 1 Module 5 Lesson 6 Example Answer Key

Example 1.
Enrique is a biologist who has been monitoring the population of a rare fish in Lake Placid. He has tracked the population for 5 years and has come up with the following estimates:

Create a graph and a function to model this situation, and use it to predict (assuming the trend continues) when the fish population will be gone from the Lake Placid ecosystem. Verify your results, and explain the limitations of each model.
Answer:
When we plot this data in a reasonable viewing window, the data appear to be linear. One limitation of using a graph is that it is difficult to get a viewing window that allows us to see all of the key features of the function represented. When we look at the first differences, we can see that this is not a linear relationship. To determine when the fish population will be gone, we need to find the function to model this situation.

Answer:

Looking at differences may help us identify the function type. The first differences are 101, 103, 105, 107, and so on. These are not the same but are at regular intervals. That tells us that there will be a common second difference, in this case 2. This data can be modeled with a quadratic function.
Creating a symbolic representation:
The standard form for a quadratic function is f(x) = ax² + bx + c.
We know that (0,1000) is the y – intercept. That tells us c = 1000. Now we have f(x) = ax² + bx + 1000.

Substitute (1,899): 899 = a + b + 1000→a + b = – 101
Substitute (2,796): 796 = 4a + 2b + 1000→4a + 2b = – 204
Now solve the linear system:
202 = – 2a – 2b
– 204 = 4a + 2b
– 2 = 2a → a = – 1
Now solve for b: – 1 + b = – 101 → b = – 100
The function model is f(x) = – x² – 100x + 1000.
So, now we need to determine when the fish population will be 0.
Solve 0 = – x² – 100x + 1000.
Using the quadratic formula: x = \(\frac{100 \pm \sqrt{( – 100)^{2} – 4( – 1)(1000)}}{2( – 1)}\) ≈ – 109.16 or 9.16
Since only the positive number of years is reasonable in this situation, we will say that x = 9.16 years after 2002.
Depending on whether the initial measurement is taken at the beginning of 2002 or at the end, we could say that the fish population will be gone during 2011 or that it will be gone by 2012.
The function model has fewer limitations than the graph, but it provides results that need to be interpreted in the context of the problem.

Eureka Math Algebra 1 Module 5 Lesson 6 Exercise Answer Key

Opening Exercise

a. Identify the type of function that each table appears to represent (e.g., quadratic, linear, exponential, square root).
Answer:
A: Linear
B: Exponential
C: Quadratic

b. Explain how you were able to identify the function.
Answer:
I looked at the difference in each output.
A is constant so that it is linear (common difference is 2),
B is a geometric sequence (common ratio is 1.5), and for
C the difference of the differences is the same number (6), so it is quadratic.

c. Find the symbolic representation of the function.
Answer:
A: f(x) = 2x + 3
B: f(x) = 4(1.5^x)
C: f(x) = 3x²

d. Plot the graphs of your data.

Answer:

Exercises

Exercise 1.
Bella is a BMX bike racer and wants to identify the relationship between her bike’s weight and the height of jumps (a category she gets judged on when racing). On a practice course, she tests out 7 bike models with different weights and comes up with the following data.

a. Bella is sponsored by Twilight Bikes and must ride a 32 lb. bike. What can she expect her jump height to be?
Answer:
By looking at the differences between the terms, we can identify that this is a linear relationship (or an arithmetic sequence) that decreases by 0.08 for every 1 lb. increase in weight.
8.9 = a₀ – 0.08(20) → 8.9 = a₀ – 1.6 → a₀ = 10.5
h(w) = 10.5 – 0.08w
h(32) = 10.5 – 0.08(32) = 7.94
For a 32 lb. bike, her expected jump height would be 7.94 ft.

b. Bella asks the bike engineers at Twilight to make the lightest bike possible. They tell her the lightest functional bike they could make is 10 lb. Based on this data, what is the highest she should expect to jump if she only uses Twilight bikes?
Answer:
h(w) = 10.5 – 0.08w
h(10) = 10.5 – 0.08(10) = 9.7
If she uses the Twilight bike, her expected jump height would be 9.7 ft.

c. What is the maximum weight of a bike if Bella’s jumps have to be at least 2 ft. high during a race?
Answer:
h(w) = 10.5 – 0.08w
2 = 10.5 – 0.08(w) → w = 106.25
The maximum bike weight would be 106.25 lb. in order to jump at least 2 ft.

Exercise 2.
The concentration of medicine in a patient’s blood as time passes is recorded in the table below.

a. The patient cannot be active while the medicine is in his blood. How long, to the nearest minute, must the patient remain inactive? What are the limitations of your model(s)?
Answer:
If we plot the points, we can see a general shape for the graph.

This might be a quadratic relationship, but we cannot be sure based on just the graph. If we look at the differences, we see that the first differences are 55.5, 27.5, – 0.5, and – 28.5. Then, the second differences are consistently – 28. Yes, this is a quadratic relationship.
Since we know (0,0) is on the graph, the constant of the quadratic function is 0. Now we can use a system of two linear equations to find the symbolic representation:
M(t) = at² + bt + 0
Substituting (0.5,55.5) → 55.5 = 0.25a + 0.5b
→ 222 = a + 2b (Multiply both sides by 4 to eliminate the decimals.)
Substituting (1,83) → 83 = a + b
a + 2b = 222
– a – b = – 83
b = 139)
so, a + 139 = 83 → a = – 56
M(t) = – 56t² + 139t
We set the function equal to zero and find the other zero of the function to be x = 0 or 2.48 hours. For this context, 2.48 hours makes sense. In hours and minutes, the patient must remain inactive for 2 hours and 29 minutes.

The function is a good model to use for answering this question since the graph requires some speculation and estimation.
Alternatively, students may generate a model by estimating the location of the vertex at about (1.25,84), and then using the vertex form of a quadratic equation. Using this method results in an equation of approximately f(x) = – 50(x – 1.25)² + 84. A useful extension activity could be to compare these two methods and models.

b. What is the highest concentration of medicine in the patient’s blood?
Answer:
Using the symmetry of the zeros of the function, we find the vertex to be at the midpoint between them or at (1.24,86.25). So the highest concentration will be 86.25 ml.

Exercise 3.
A student is conducting an experiment, and as time passes, the number of cells in the experiment decreases. How many cells will there be after 16 minutes?

Answer:
If we divide each term by the one before it, we see that this is a geometric sequence where the rate is 0.55 and the initial value is 5,000,000. Therefore, we can represent this sequence symbolically by
c(m) = 5 000 000(0.55)^m.
c(16) = 5 000 000(0.55)¹⁶ → c(16) = 350.57
After 16 minutes, there will be approximately 351 cells.

Eureka Math Algebra 1 Module 5 Lesson 6 Problem Set Answer Key

Question 1.
Research linear, quadratic, and exponential functions using the Internet. For each of the three types of functions, provide an example of a problem/situation you found on the Internet where that function was used to model the situation or answer the problem. Include the actual function used in the example and web page where you found the example.
Answer:
Answers will vary.

Eureka Math Algebra 1 Module 5 Lesson 6 Exit Ticket Answer Key

Lewis’s dad put $1,000 in a money market fund with a fixed interest rate when he was 16. Lewis cannot touch the money until he is 26, but he gets updates on the balance of his account.

a. Develop a model for this situation.
Answer:
We might try graphing this data. However, in the viewing window that shows our data points (see graph below), it appears that the function might be linear. Let’s try zooming out to see more of the key features of this graph. (See graph below.)
x: [0,5] y: [975,1500]

This viewing window gives us a close – up of the data points and their relation to each other. However, we cannot really see the features of the graph that represent the data.

x: [ – 25,25] y: [0,9700]

In this version of the graph, you can see how the data from our table is grouped on a very short section of the graph. From this view, we can see the exponential nature of the graph.

Using the data table to find a function model: The first and second differences have no commonalities; therefore, this is not linear or quadratic. Checking to see if there is a common ratio, we see that this is an exponential relationship (or a geometric sequence) where the common ratio is 1.1, and the initial value is 1000. Check: Since on this table time starts at t = 0, using t as the exponent will yield $1,000 for the initial balance.
Therefore, we can represent this sequence symbolically by A(t) = 1000(1.1)^t

b. Use your model to determine how much Lewis will have when he turns 26 years old.
Answer:
Using the function: Since Lewis will be 26 ten years after he turns 16, we will need to evaluate A(10):
A(10) = 1000(1.1)¹⁰ → A(10) = 2593.74 or $2,593.74.
We might also try extending our data table to verify this result. There are a couple of precision decisions to make: Shall we use 1.1 as the common ratio? How soon should we begin rounding numbers off? For this table, we decided to use 1.1 as the common ratio and rounded to the nearest cent.

We might also try to answer this question using our graph. Below is another view of the graph. Can you estimate the balance at t = 10?

c. Comment on the limitations/validity of your model.
Answer:
As we saw in the first and second versions of the graph, there are limitations to the graphic model because we cannot always see the key features of the graph in a window that lets us see all the data points clearly. Being able to see the graph using both windows was more helpful. Then, in part (b), we saw how difficult it was to estimate the value of A(10). We were also able to extend the table without too much difficulty, after deciding what level of precision we needed to use.

The equation was most helpful but requires interpretation of the data (noticing that the common ratio was very close but not absolutely perfect, and making sure we started with t = 0).

Regardless of whether we use a graphical, numerical, or algebraic model, one limitation is that we are assuming the growth rate will remain constant until he is 26.

Engage NY Eureka Math Algebra 1 Module 5 Lesson 5 Answer Key

Eureka Math Algebra 1 Module 5 Lesson 5 Example Answer Key

Example 1.
Determine whether the sequence below is arithmetic or geometric, and find the function that will produce any given term in the sequence:
16, 24, 36, 54, 81, …
Is this sequence arithmetic?
Answer:
The differences are 8, 12, 18, 27, … so right away we can tell this is not arithmetic; there is no common difference.

Is the sequence geometric?
Answer:
The ratios are all 1.5, so this is geometric.

What is the analytical representation of the sequence?
Answer:
Since the first term is 16 and the common ratio is 1.5, we have f(n) = 16(1.5)^{(n – 1)}.

Eureka Math Algebra 1 Module 5 Lesson 5 Exercise Answer Key

Exercises
Look at the sequence and determine the analytical representation of the sequence. Show your work and reasoning.
Exercise 1.
A decorating consultant charges $50 for the first hour and $2 for each additional whole hour. How much would 1, 000 hours of consultation cost?

Answer:
By subtracting, we see that this is an arithmetic sequence where we are adding 2 but starting at 50.
f(n) = 50 + 2(n – 1) = 48 + 2n
f(1000) = 48 + 2(1000) = 2048
1, 000 hours of consultation would cost $2, 048.

Exercise 2.
The sequence below represents the area of a square whose side length is the diagonal of a square with integer side length n. What would be the area for the 100th square? Hint: You can use the square below to find the function model, but you can also just use the terms of the sequence.

Answer:
Looking at first differences, we see that they are not the same (no common difference):
6, 10, 14, 18, ….
When we look for a common ratio, we find that the quotients of any two consecutive terms in the sequence are not the same: \(\frac{8}{2}\) ≠ \(\frac{18}{8}\) ≠ \(\frac{32}{18}\) ≠⋯
However, I noticed that the first difference increases by 4. This is an indication of a quadratic sequence, and the function equation must have a n². But since for n = 1 we would have n² = 1, we must need to multiply that by 2 to get the first term. Now, check to see if 2n² will work for the other terms.
f(n) = 2n²
Checking: (f(2) = 2(2² ) = 8
f(3) = 2(3² ) = 18
Yes! It works. So, f(100) = 2(100)² = 20 000.
Therefore, the area of the 100th square is 20, 000 square units.

Exercise 3.
What would be the tenth term in the sequence?

Answer:
There is no common difference. But the ratios are as follows: \(\frac{6}{3}\) = 2, \(\frac{12}{6}\) = 2, \(\frac{24}{12}\) = 2, …. This is a geometric sequence with a common ratio of 2. And the terms can be written as shown below.

The tenth term in the sequence is 3(512) or 1536.

Eureka Math Algebra 1 Module 5 Lesson 5 Problem Set Answer Key

Solve the following problems by finding the function/formula that represents the nth term of the sequence.
Question 1.
After a knee injury, a jogger is told he can jog 10 minutes every day and that he can increase his jogging time by
2 minutes every two weeks. How long will it take for him to be able to jog one hour a day?

Answer:
This is an arithmetic sequence where the minutes increase by 2 every two weeks. (Note: We can either let 2n represent the number of weeks or let n represent a two – week period. Either way, we will end up having to compensate after we solve.) Let’s try it with n representing a two – week period:

f(n) = initial time + (n – 1)(common difference)
60 = 10 + (n – 1)(2) → 60 = 10 + 2n – 2 → 60 = 2n + 8
2n + 8 = 60 → 2n = 52 → n = 26
At the beginning of the 26^th 2 – week period, the jogger will be able to jog for 60 minutes. This will occur after 25 weeks ⋅2, or 50 weeks, or at the beginning of the 51st week.

Question 2.
A ball is dropped from a height of 10 feet. The ball then bounces to 80% of its previous height with each subsequent bounce.
a. Explain how this situation can be modeled with a sequence.
Answer:
According to the problem, to find the next height, multiply the current height by 0.8. This means the sequence is geometric.

b. How high (to the nearest tenth of a foot) does the ball bounce on the fifth bounce?
Answer:
f(n) = (initial height) (common ratio)^n for n bounces.
f(5) = 10(0.8)⁵ = 3.2768
The ball bounces approximately 3.3 feet on the fifth bounce.

Question 3.
Consider the following sequence:
8, 17, 32, 53, 80, 113, …

a. What pattern do you see, and what does that pattern mean for the analytical representation of the function?
Answer:
Difference of the differences is 6. Since the second difference is a nonzero constant, then the pattern must be quadratic.

b. What is the symbolic representation of the sequence?
Answer:
Sample response: 3n² does not work by itself. (If n = 1, then 3n² would be 3, but we have an 8 for the first term.) So, there must be a constant that is being added to it. Let’s test that theory:
(f(n) = 3n² + b
f(1) = 3(1)² + b = 8
3 + b = 8
+ b = 5
So, the terms of the sequence can be found using the number of the term, as follows:
f(n) = 3n² + 5
We can easily check to see if this function generates the sequence, and it does.

Question 4.
Arnold wants to be able to complete 100 military – style pull – ups. His trainer puts him on a workout regimen designed to improve his pull – up strength. The following chart shows how many pull – ups Arnold can complete after each month of training. How many months will it take Arnold to achieve his goal if this pattern continues?

Answer:
This pattern does not have a common difference or a common ratio. When we look at the first differences (3, 5, 7, 9, 11, …), we see that the second differences would be constant (2, 2, 2, …). That means this is a quadratic sequence with n² in the nth term formula. For n = 1 we have 1² = 1, so we need to add 1 to get the first term to be 2. So, in general, we have the function f(n) = n² + 1. Let’s test that on the other terms:

2² + 1 = 5, 3² + 1 = 10, …. Yes, it works.
Now we need to find out which month (n) will produce 100 as the resulting number of pull – ups:
n² + 1 = 100 → n = \(\sqrt{99}\) ≈ 9.9
So, if this trend continues, at 10 months, Arnold will be able to complete 100 pull – ups.

Eureka Math Algebra 1 Module 5 Lesson 5 Exit Ticket Answer Key

Question 1.
A culture of bacteria doubles every 2 hours.
a. Explain how this situation can be modeled with a sequence.
Answer:
To find the next number of bacteria, multiply the previous number by 2. This situation can be represented by a geometric sequence. There will be a common ratio between each term of the sequence.

b. If there are 500 bacteria at the beginning, how many bacteria will there be after 24 hours?
Answer:
Using the exponential function f(n) = a ∙ bⁿ, where n represents the number of times the bacteria culture doubles, a represents the amount when n = 0, so a = 500. b is the growth rate, so b = 2.
f(n) = 500 ∙ 2ⁿ
f(n) = 500 ∙ 2¹² = 2 048 000
After 24 hours, there will be 2,048,000 bacteria.

Engage NY Eureka Math Algebra 1 Module 5 Lesson 4 Answer Key

Eureka Math Algebra 1 Module 5 Lesson 4 Example Answer Key

Example 1.
Read the problem below. Your teacher will walk you through the process of using the steps in the modeling cycle to guide your solution.
The relationship between the length of one of the legs, in feet, of an animal and its walking speed, in feet per second, can be modeled by the graph below. Note: This function applies to walking not running speed. Obviously, a cheetah has shorter legs than a giraffe but can run much faster. However, in a walking race, the giraffe has the advantage.

A T – Rex’s leg length was 20 ft. What was the T – Rex’s speed in ft/sec?
Answer:
→ What are the units involved in this problem? Define the quantities and variables you would use to model this graph.
Leg length is in feet, and speed is measured in feet per second, so time is measured in seconds. Use x to represent the number of feet in leg length and f to represent the speed based on leg length.

→ What type of function does this graph represent? What clues in the graph helped you recognize the function?
There are only positive values and it starts at (0,0), so this is probably a square root function, but it could be the right half of a cube root function. I need to check a few points to be sure.
After you get the correct response to the first question, draw or project the following three transformations of the square root function on the board or screen: f(x) = \(\sqrt{x}\), g(x) = \(\sqrt{ax}\), h(x) = \(\sqrt{(x – b)}\).

→ Which transformation of the function does this graph represent? How can we determine that?
Sample Response: f cannot be the form because the square root of 2 is not 8, so it is either g or h. We can test both forms by substituting the x – and y – values from the points provided to us.
First, check the form: g(x) = \(\sqrt{ax}\).
(2,8)→8 = \(\sqrt{2a}\) → 64 = 2a→a = 32
(8,16)→16 = \(\sqrt{8a}\) → 256 = 8a→a = 32
Since we got a = 32 for both ordered pairs, it appears that g is the correct form and that g(x) = \(\sqrt{32x}\).
To make sure, we will also check the form: h(x) = \(\sqrt{(x – b)}\).
(2,8) → 8 = \(\sqrt{(2 – b)}\) → 8² = 2 – b → 64 = 2 – b → b = – 62
(8,16) → 16 = \(\sqrt{(8 – b)}\) → 16² = 8 – b → 256 = 8 – b → b = – 248
The function cannot be h.

→ Is the problem solved?
We are not finished. Now we need to use the function we found to calculate the speed of the T – Rex with a 20 – foot leg using g(x) = \(\sqrt{32x}\).
What is the walking speed of the T – Rex?
g(20) = \(\sqrt{(32)(20)}\) → g(20) = \(\sqrt{640}\) → g(20) = 8\(\sqrt{10}\), or about 25 feet per second.

→ What if we doubled the length of T – Rex’s legs? Would the T – Rex walk twice as fast?
A square root function does not double if the input is doubled. The graph shows that when the input went from 2 to 8, the output does not quadruple. A proportional relationship would have a double – the – input – double – the – output effect, but a square root function is not proportional. A proportional relationship between two quantities must be linear. If we look at the Eduardo graph (Exercise 1), we can see a proportional relationship.

Eureka Math Algebra 1 Module 5 Lesson 4 Exercise Answer Key

Exercises
Now practice using the modeling cycle with these problems:

Exercise 1.
Eduardo has a summer job that pays him a certain rate for the first 40 hours per week and time and a half for any overtime. The graph below is a representation of how much money he earns as a function of the hours he works in one week.

Eduardo’s employers want to make him a salaried employee, which means he does not get overtime. If they want to pay him $480 per week but have him commit to 50 hours a week, should he agree to the salary change? Justify your answer mathematically.
a. Formulate (recall this step from Lesson 1).
i. What type of function can be represented by a graph like this (e.g., quadratic, linear, exponential, piecewise, square root, or cube root)?
Answer:
The graph is of a function that is piecewise defined and made up of two linear functions.

ii. How would you describe the end behavior of the graph in the context of this problem?
Answer:
As x gets infinitely large, so does the function. However, realistically there is a limit to how many hours per week he can work, so there is a maximum value for this function.

iii. How does this affect the equation of our function?
Answer:
The slope for each graph is positive. We may need to state restrictions on the domain and also on the range.

b. Compute
i. What strategy do you plan to use to come up with the model for this context?
Answer:
Identify the domains for each piece (linear function) in the piecewise function. Find the slope and y – intercept for each linear piece, and then put the equations in slope – intercept form.

ii. Find the function of this graph. Show all your work.
Answer:
For the lower piece, domain: 0 ≤ x ≤ 40.
The y – intercept is 0, and the given points in that domain are (0,0) and (22,198). To calculate the slope \(\frac{198}{22}\) = 9.
f(x) = 9x if 0 ≤ x ≤ 40
For the upper piece, domain: 40 < x. This part of the problem requires using not only the graph but also the context from the problem. Up to this point, Eduardo has worked 40 hours and has earned 9 dollars an hour. 40∙9 = 360. This is the y – value when x = 40 for the second linear equation. Again, we find the slope of the equation by selecting two points (70,765) and (60,630). \(\frac{765 – 630}{70 – 60}\) = \(\frac{135}{10}\) = 13.50. Looking at the graph, we also see that the second line was moved 40 units to the right (translated), so x is (x – 40).
f(x) = 13.5(x – 40) + 360 if x > 40

c. Interpret
i. How much does Eduardo make an hour?
Answer:
Eduardo makes an hourly rate of $9 if 0 ≤ x ≤ 40 and $13.50 if x > 40.

ii. By looking only at the graphs, which interval has a greater average rate of change: x<20 or x>45? Justify your answer by making connections to the graph and its verbal description.
Answer:
The verbal description states that Eduardo gets paid more after 40 hours of work. If you look at how steep the graph is after 45 hours, you can see that it is increasing at a faster rate than when it was 20 hours or less.

iii. Eduardo’s employers want to make Eduardo a salaried employee, which means he does not get overtime. If they want to pay him $480 per week but have him commit to 50 hours a week, should he agree to the salary change? Justify your answer mathematically.
Answer:
f(50) = 13.5(50 – 40) + 360→13.5(10) + 360→135 + 360 = $495. He would get paid more as an hourly employee if he worked 50 hours a week. It seems that it would be in his best interest to keep the hourly agreement rather than the salary. Students may come up with other responses but should at least understand that Eduardo earns more money for 50 hours of work as an hourly employee than he would as a salaried employee. If students show that understanding, then they have answered the question correctly.

d. Validate
How can you check to make sure your function models the graph accurately?
Answer:
For the values in our graph, we can substitute the x – values into the function to see if the same given y – value is a result.

Exercise 2.
The cross – section view of a deep river gorge is modeled by the graph shown below where both height and distance are measured in miles. How long is a bridge that spans the gorge from the point labeled (1,0) to the other side? How high above the bottom of the gorge is the bridge?

a. Formulate
i. What type of function can be represented by a graph like this (e.g., quadratic, linear, exponential, piecewise, square root, or cube root)?
Answer:
Quadratic

ii. What are the quantities in this problem?
Answer:
The quantities are height relative to the location of the bridge (f) and horizontal distance from the highest point in the gorge (x).

iii. How would you describe the end behavior of the graph?
Answer:
Opening upward

iv. What is a general form for this function type?
Answer:
f(x) = ax² + bx + c or f(x) = a(x – h)² + k or f(x) = a(x – m)(x – n)

v. How does knowing the function type and end behavior affect the equation of the function for this graph?
Answer:
The equation is a second – degree polynomial where the leading coefficient is positive.

vi. What is the equation we would use to model this graph?
Answer:
Using the three ordered pairs in the graph:
(0,5): From this pair, we see that c = 5 and f(x) = ax² + bx + 5
Now substitute the other two ordered pairs to form a linear system:
(1,0): a(1)² + b(1) + 5 = 0 ⇒ a + b = – 5 ⇒ – 3a – 3b = 15
(3,2): a(3)² + b(3) + 5 = 2⇒9a + 3b = – 3
From here we see that 6a = 12 and a = 2. So, substituting into the first linear equation, we have b = – 7.
So, the equation is f(x) = 2x² – 7x + 5.

b. Compute
i. What are the key features of the graph that can be used to determine the equation?
Answer:
The y – intercept and the other two given ordered pairs

ii. Which key features of the function must be determined?
Answer:
The zeros and the vertex must be determined.

iii. Calculate the missing key features, and check for accuracy with your graph.
Answer:
f(x) = 2x² – 7x + 5
Writing the function in factored form, we find f(x) = (2x – 5)(x – 1). So, the other end of the bridge is at the point (2.5,0). This appears accurate based on the x – intercept of the graph. The vertex is located midway between the x – intercepts. Therefore, the vertex will occur at x = 1.75. Evaluating f at x = 1.75, we find that f(1.75 ) = – 1.125. The vertex is (1.75, – 1.125), which looks to be accurate based on the graph.

c. Interpret
i. What domain makes sense for this context? Explain.
Answer:
Since x is the horizontal distance from (0,5), the domain that makes sense is x>0. In order to know the upper limits of the domain, we would need to know the height of the highest point on the other side of the gorge, which cannot be determined with much accuracy from the given graph. We could assume it is between 3 and 3.5 miles.

ii. How wide is the bridge with one side located at (1,0)?
Answer:
It would be the difference between the x – coordinates of the x – intercepts.
2.5 – 1 = 1.5
The bridge is 1.5 miles wide.

iii. How high is the bridge above the bottom of the gorge?
Answer:
The minimum value of the function represents the lowest point. Since the height shown on the graph is relative to the location of the bridge, the bridge would be 1.125 miles above the bottom of the gorge. The y – coordinate of the minimum point represents that height below the bridge.

iv. Suppose the gorge is exactly 3.5 miles wide from its two highest points. Find the average rate of change for the interval from x = 0 to x = 3.5,[0,3.5]. Explain this phenomenon. Are there other intervals that behave similarly?
Answer:
The average rate of change for that interval is 0. The two points are points of symmetry straight across the curve from each other. The line passing through them is horizontal. Any interval with endpoints that are symmetry points will be horizontal (0 slope).

d. Validate
How can you check to make sure that your function models the graph accurately?
Answer:
We can substitute any x – value from the graph into the function to see if the resulting y – value is reasonable for the graph.

Exercise 3.
Now compare four representations that may be involved in the modeling process. How is each useful for each phase of the modeling cycle? Explain the advantages and disadvantages of each.
Answer:
The verbal description helps us define the quantities and, in this case, write the factored form of the equation for the function. We always need to find another representation to analyze or interpret the situation.

The graph is visual and allows us to see the overall shape and end behavior of the function. We can “see” some integer values of the function but must estimate any non – integer values. The graph is a good way to check calculations to make sure results are reasonable.
The table, in general, does not help us to see shapes (patterns in x – y) any better than the graph does. However, a table with equal x – intervals does help us to see the patterns in y – values very well.

The equation allows us to accurately calculate values of the function for any real number. It can be rewritten in various forms to help us see features of the function (vertex form, standard form, factored form). It allows deep analysis and is sometimes referred to as an analytical model.

Eureka Math Algebra 1 Module 5 Lesson 4 Problem Set Answer Key

Question 1.
During tryouts for the track team, Bob is running 90 – foot wind sprints by running from a starting line to the far wall of the gym and back. At time t = 0, he is at the starting line and ready to accelerate toward the opposite wall. As t approaches 6 seconds he must slow down, stop for just an instant to touch the wall, then turn around, and sprint back to the starting line. His distance, in feet, from the starting line with respect to the number of seconds that has passed for one repetition is modeled by the graph below.

(Note: You may refer to Lesson 1, Problem Set 1 to help answer this question.)
How far was Bob from the starting line at 2 seconds? 6.5 seconds? (Distances, in feet, should be represented to the nearest tenth.)
Answer:
So far we know: f(t) = a(t – 6)² + 90 (Now we need to find a.)
Substitute (0,0):
a(0 – 6)² + 90 = 0
a( – 6)² + 90 = 0
36a = – 90
a = \(\frac{ – 90}{36}\) = \(\frac{ – 5}{2}\)
The final function is f(t) = \(\frac{ – 5}{2}\) (t – 6)² + 90.
f(2) = \(\frac{ – 5}{2}\) (2 – 6)² + 90 = 50
f(6.5) = \(\frac{ – 5}{2}\) (6.5 – 6)² + 90 = 89.4
Bob was 50 ft and 89.4 ft from the starting line at 2 seconds and 6.5 seconds, respectively.

Question 2.
Kyle and Abed each threw a baseball across a field. The height of the balls (in feet) is described by functions A and K, where t is the number of seconds the baseball is in the air. K models the height of Kyle’s baseball (equation below), and A models the height of Abed’s baseball (graph below).
K(t) = – 16t² + 66t + 6

a. Which ball was in the air for a longer period of time?
Answer:
We need to find the time when the ball is back on the ground (i.e., when the height function equals 0).
First Abed’s:
We can see from the graph that the ball is in the air from t = 0 to somewhere between 4 and 6. So, just by reading the graph, we might estimate the time in the air to be a little more than 5 seconds. We can get a more accurate number by formulating a model and then doing some calculations.
Using the vertex form: A(t) = a(t – 2.75)² + 126
If we substitute (1,77):
a(1 – 2.75)² + 126 = 77
3.0625a = – 49
a = – 49/3.0625 = – 16
So, A(t) = – 16(t – 2.75)² + 126 = 0.
– 16(t – 2.75)² = – 126
(t – 2.75)² = – \(\frac{126}{ – 16}\)
(t – 2.75)² = 7.875
t – 2.75 = ±\(\sqrt{7.875}\)
t = 2.75±\(\sqrt{7.875}\)
t ≈ – 0.06 or 5.6
Only the positive value makes sense in this context, so Abed’s ball was in the air for 5.6 seconds.
Now Kyle’s: K(t) = – 16t² + 66t + 6 = 0
Using the quadratic formula to find the zeros:
t = \(\frac{ – 66 \pm \sqrt{66^{2} – 4( – 16)(6)}}{2( – 16)} = \frac{ – 66 \pm \sqrt{4740}}{ – 32}\) ≈ – 0.09 or 4.2
Only the positive value makes sense in this context. So Kyle’s ball was in the air for 4.2 seconds.
Abed’s ball was in the air longer.

b. Whose ball goes higher?
ans;
We see in the graph of y = A(t) that the highest point for Abed’s ball is 126 ft. Now we need to find the maximum value for K. One way to find the maximum is to put the equation in vertex form:
K(t) = – 16(t² – 4.125t + ) + 6
= – 16(t – 2.0625)² + 6 + 16(2.0625)²
= – 16(t – 2.0625)² + 6 + 68.0625
= – 16(t – 2.0625)² + 74.0625
So, the vertex for K is (2.0625,74.0625), and Kyle’s ball went about 74 ft into the air. Abed’s ball went higher.

c. How high was Abed’s ball when he threw it?
Answer:
Now we are looking for the y – intercept for y = A(t) (i.e., the height when time is 0). We can either use the vertex form of the equation, or we can rewrite the equation in standard form.
A(t) = – 16(t – 2.75)² + 126
= – 16(t² – 5.5t + 7.5625) + 126
= – 16t² + 88t – 121 + 126
= – 16t² + 88t + 5
That means (0,5) is the vertical – intercept, and the ball was at 5 ft when thrown by Abed.
Or we can substitute t = 0 into the vertex form:
A(t) = – 16(t – 2.75)² + 126
= – 16(0 – 2.75)² + 126
= – 16(7.5625) + 126
= – 121 + 126
= 5
So, the ball left Abed’s hand at a height of 5 ft.

Eureka Math Algebra 1 Module 5 Lesson 4 Exit Ticket Answer Key

Question 1.
Why might we want to represent a graph of a function in analytical form?
Answer:
Graphs require estimation for many values, and for most we can calculate exact values using the function equation. Some key features that may not be visible or clear on a graph can be seen in the symbolic representation.

Question 2.
Why might we want to represent a graph as a table of values?
Answer:
In a table of values, we can sometimes better see patterns in the relationship between the x – and y – values.

Engage NY Eureka Math Algebra 1 Module 5 Lesson 3 Answer Key

Eureka Math Algebra 1 Module 5 Lesson 3 Example Answer Key

Example 1.
Gregory plans to purchase a video game player. He has $500 in his savings account and plans to save $20 per week from his allowance until he has enough money to buy the player. He needs to figure out how long it will take. What type of function should he use to model this problem? Justify your answer mathematically.
Answer:
→ Gregory decides that the exponential function can best represent the situation. Do you agree or disagree? Why? Support your answer mathematically.
I disagree because Gregory’s money increases at a constant rate of $20 a week with a starting balance of $500. The graph of the amount of money that Gregory saves over a period of time is a linear model.

→ What are the variables and quantities of this problem?
The rate is $20 per week, w is the number of weeks, and the initial value is $500.

→ What function represents the amount of Gregory’s money over a period of time (in weeks)?
The model is f(w) = 20w + 500. If the function is graphed, the slope of the line is 20 to reflect the constant rate of $20/week, and the y – intercept is $500, the initial amount.

Example 2.
One of the highlights in a car show event is a car driving up a ramp and flying over approximately five cars placed end – to – end. The ramp is 8 ft. at its highest point, and there is an upward speed of 88 ft/sec before it leaves the top of the ramp. What type of function can best model the height, h, in feet, of the car t seconds after leaving the end of the ramp? Justify your answer mathematically.
Answer:
→ What type of function can best model the height, h, in feet, of the car t seconds after leaving the end of the ramp? What were your clues? Justify your answer mathematically.
Quadratic function. This is an object in motion problem. The car would leave the ramp in an upward and forward motion and then, after travelling higher for a short time, would begin the fall due to the force of gravity.

→ What form would the equation take?
Since the distance is measured in feet and the time in seconds, we would use
h(t) = – 16t² + v₀ t + h₀, and the equation would be h(t) = – 16t² + 88t + 8.

Example 3.
Margie got $1,000 from her grandmother to start her college fund. She is opening a new savings account and finds out that her bank offers a 2% annual interest rate, compounded monthly. What type of function would best represent the amount of money in Margie’s account? Justify your answer mathematically.
Answer:
→ What type of function would best represent the amount of money in Margie’s account? Justify your answer mathematically.
Exponential Function. The amount of deposited money grows over time at a constant rate, and the pattern can be best described by an exponential function, f(x) = ab^x, where a represents the initial investment, and b is the expression (1 + \(\frac{r}{n}\)) as defined in the compounded interest formula:
P(t) = P₀ (1 + \(\frac{r}{n}\))^nt.

→ What function represents the amount of money deposited in the bank compounded monthly at the rate of 2%, if the initial amount of deposit was $1,000?
A(n) = 1000(1 + \(\frac{0.02}{12}\))^12t
Note that we do not know how long Margie plans to leave the money in her account, so we do not know what the value of t is yet.
Remind students that percentages, in most cases, must be changed to decimals when used in exponential expressions.

Eureka Math Algebra 1 Module 5 Lesson 3 Exercise Answer Key

Exercises

Exercise 1.
City workers recorded the number of squirrels in a park over a period of time. At the first count, there were 15 pairs of male and female squirrels (30 squirrels total). After 6 months, the city workers recorded a total of 60 squirrels, and after a year, there were 120.
a. What type of function can best model the population of squirrels recorded over a period of time, assuming the same growth rate and that no squirrel dies?
Answer:
Exponential function

b. Write a function that represents the population of squirrels recorded over x number of years. Explain how you determined your function.
Answer:
Students may use the general exponential function f(x) = ab^x, by figuring out that this is a doubling exponential problem (in this case, the number of squirrels doubles every 6 months). So, the function would be f(x) = 30(2)^2x because the squirrel population would double twice each year.

Exercise 2.
A rectangular photograph measuring 8 in by 10 in is surrounded by a frame with a uniform width, x.
a. What type of function can best represent the area of the picture and the frame in terms of x (the unknown frame’s width)? Explain mathematically how you know.
Answer:
This function can be best represented by a quadratic function because this is an area problem where the product of two linear measurements results in a quadratic.

b. Write an equation in standard form representing the area of the picture and the frame. Explain how you arrive at your equation.
Answer:
The dimensions of the picture are 8 in by 10 in. Taking into consideration the width of the frame, we have to add 2x to both the width and the length of the picture. Doing so results in (8 + 2x) and (10 + 2x). So, the area of the picture and the frame is A(x) = (8 + 2x)(10 + 2x) or A(x) = 4x² + 36x + 80.

Exercise 3.
A ball is tossed up in the air at an initial rate of 50 ft/sec from 5 ft. off the ground.
a. What type of function models the height (h, in feet) of the ball after t seconds?
Answer:
Quadratic function

b. Explain what is happening to the height of the ball as it travels over a period of time (in t seconds).
Answer:
The initial height of the ball is 5 ft., and it travels upward with an initial velocity of 50 ft/sec. As time increases, the ball continues to travel upward, with the force of gravity slowing it down, until it reaches the maximum height and falls back to the ground.

c. What function models the height, h (in feet), of the ball over a period of time (in t seconds)?
Answer:
h(t) = – 16t² + 50t + 5

Exercise 4.
A population of insects is known to triple in size every month. At the beginning of a scientific research project, there were 200 insects.
a. What type of function models the population of the insects after t years?
Answer:
Exponential function

b. Write a function that models the population growth of the insects after t years.
Answer:
Using the general form for exponential growth (b = 3), we have the initial population of 200 and 12t is the number of growth cycles over t years. The function would be f(t) = 200(3)^12t.
P = P₀ (1 + r)^nt
P(t) = 200(1 + 2)^12t, where r represents the growth rate at 200% and n = 12.
So, P(t) = 200(3)^12t.

Eureka Math Algebra 1 Module 5 Lesson 3 Problem Set Answer Key

Question 1.
The costs to purchase school spirit posters are as follows: two posters for $5, four posters for $9, six posters for $13, eight posters for $17, and so on.
a. What type of function would best represent the cost of the total number of posters purchased?
Answer:
Linear function

b. What function represents the cost of the total number of posters purchased? How did you know? Justify your reasoning.
Answer:
Let x represent the number of school spirit posters. The four ordered pairs indicate a constant rate of change, (m = 2), so the equation is y = 2x + b. To find b, we need to substitute any ordered pair, say (2,5): 5 = 2(2) + b, so b = 1. The final equation for the function is f(x) = 2x + 1.

c. If you have $40 to spend, write an inequality to find the maximum number of posters you could buy.
Answer:
2x + 1 ≤ 40

Question 2.
NYC Sports Gym had 425 members in 2011. Based on statistics, the total number of memberships increases by 2% annually.
a. What type of function models the total number of memberships in this situation?
Answer:
Exponential function

b. If the trend continues, what function represents the total number of memberships in n years? How did you know? Justify your reasoning.
Answer:
f(n) = 425(1 + 0.02)ⁿ
The initial number of members is 425. The yearly growth rate of 2% means I have to multiply by 1.02 for each year. So, 1.02 will be the common ratio for this exponential function.

Question 3.
Derek throws a baseball upward from an initial height of 3 ft . The baseball hits the ground after 2 seconds.
a. What was the initial velocity of the baseball?
Answer:
h(t) = – 16t² + v₀t + 3
h(2) = – 16(2)² + 2v₀ + 3 = 0
v₀ = 30.5
The initial velocity of the baseball was 0.5 ft/sec.

b. What is the function that models the height, h (in feet), of the baseball over a period of time t (in seconds)?
Answer:
h(t) = – 16t² + 30.5t + 3

c. At what time did the baseball reach its maximum height? What was the maximum height of the baseball?
Answer:
The baseball reached its maximum height after 0.953125 seconds. The maximum height of the baseball was approximately 17.535 feet.

Eureka Math Algebra 1 Module 5 Lesson 3 Exit Ticket Answer Key

Question 1.
Create a model to compare these two texting plans :
a. Plan A costs $15 a month, including 200 free texts. After 200, they cost $0.15 each.
b. Plan B costs $20 a month, including 250 free texts. After 250, they cost $0.10 each.
Answer:
Monthly cost of Plan A: where t represents the number of texts per month

Monthly cost of Plan B: where t represents the number of texts per month

Engage NY Eureka Math Algebra 1 Module 5 Lesson 2 Answer Key

Eureka Math Algebra 1 Module 5 Lesson 2 Example Answer Key

Example 1.
Noam and Athena had an argument about whether it would take longer to get from NYC to Boston and back by car or by train. To settle their differences, they made separate, nonstop round trips from NYC to Boston. On the trip, at the end of each hour, both recorded the number of miles they had traveled from their starting points in NYC. The tables below show their travel times, in hours, and the distances from their starting points, in miles. The first table shows Noam’s travel time and distance from the starting point, and the second represents Athena’s. Use both data sets to justify your answers to the questions below.

a. Who do you think is driving, and who is riding the train? Explain your answer in the context of the problem.
Answer:
This is an open – ended question with no right or wrong answer. Check for mathematically sound reasoning based on the data.
Sample Response: It appears that Athena is riding the train since she was able to go 81 miles in the first hour, and Noam was able to go only 55, which is a typical highway speed. Also, Athena made the round trip in 10 hours, while Noam made it in 8, so it looks like the train may have had to stop at stations when it was near Boston, slowing its progress considerably during the 2 hours when the train was outside of Boston.

b. According to the data, how far apart are Boston and New York City? Explain mathematically.
Answer:
This is an open – ended question with no right or wrong answer. Check for mathematically sound reasoning based on the data.
Sample Response: Based on the symmetry of the values in the table, Noam’s maximum distance was 220 miles (at 4 hours). Using the same symmetry, Athena’s maximum distance was 225 miles. They may have started or ended at different places, but it is more likely that the route for the train was slightly different from that of the car.

c. How long did it take each of them to make the round trip?
Answer:
Let’s call Noam’s distance N and Athena’s distance A. Noam was traveling for 8 hours and Athena for 10 hours. The zeros for N are (0,0) and (8,0) and for A are (0,0) and (10,0). It took Noam 8 hours; it took Athena 10 hours.

d. According to their collected data, which method of travel was faster?
Answer:
If we assume that Noam was traveling by car, then the car was faster by 2 hours, overall. However, the speed of the train was faster for the first and the last hour and then slowed.

e. What was the average rate of change for Athena for the interval from 3 to 4 hours? How might you explain that in the context of the problem?
Answer:
She only traveled 27 miles per hour: \(\frac{216 – 189}{4 – 3}\). Since she was likely on the train, there may have been stops during that time period.

f. Noam believes a quadratic function can be used as a model for both data sets. Do you agree? Use and describe the key features of the functions represented by the data sets to support your answer.
Answer:
The two data sets have several things in common. Both are symmetric, both have a vertex, and both have (0,0) as one of the x – intercepts and the y – intercept. The other x – intercept is (8,0) for Noam and (10,0) for Athena. However, Noam’s data set has a positive constant difference of + 55 on one side of the vertex, which then changes to – 55 on the other side. Noam’s data set can be best modeled by an absolute value or other piecewise function. Athena’s data does not have a constant first difference, so it is not linear. When we check the second differences, we find it to be constant ( – 18). Therefore, Athena’s trip can be modeled with a quadratic function.

Eureka Math Algebra 1 Module 5 Lesson 2 Exercise Answer Key

Opening Exercise
When tables are used to model functions, we typically have just a few sample values of the function and therefore have to do some detective work to figure out what the function might be. Look at these three tables:

Answer:
→ What do you notice about the three sets of data? Can you identify the type of function they represent?
Students may observe some of the examples listed below. It is also possible that they will not notice much of anything, or they may have ideas that are incorrect. Just let them brainstorm, and then come back to their ideas later to see how many were helpful.

→ The orange set is growing much faster than the first two. Many of the function values in the blue and green are the same. It is difficult to know the function by looking at only these numbers for the first two. However, I notice that the function values in the green set were growing steadily and then decreased at the end. It might help if we had more data points.

→ Have students make conjectures about the type of function they believe each data set represents and support their conjectures with evidence, which might include a graph.
Now show a data plot for each:

→ Now can you see the trends more clearly?
→ Yes, it is obvious that the first is linear. It is clearer that the second is quadratic. And the third looks to be exponential.
→ We learned to determine the type of function by looking at the shapes of graphs. Can we also determine the type of function by just using its table of values? Analyze each table and find any special patterns that may help determine the type of function it represents.

→ The table of values for the linear function has a constant first difference of 6, indicating the graph will have a slope of 6. The quadratic function does not have a constant difference like a linear function. However, the difference of differences (second difference) is constant. The exponential function does not have constant first or second differences. However, the y – value is multiplied by a constant value of 3. It should be noted that the x – values are all increasing by one. Not all data are presented where the x – values are increasing by a constant value, so it is necessary to interpolate values to produce equal – sized intervals.

Exercises

Exercise 1.
Explain why each function can or cannot be used to model the given data set.

a. f(x) = 3x + 5
Answer:
This function cannot be used to model the data set. The y – intercept is 5, but the first difference is not constant, and the data set is not a linear function.

b. f(x) = – (x – 2)² + 9
Answer:
This function can be used to model the data set. The second difference has a constant value of – 2; therefore, it is a quadratic function. The vertex is (2,9), and it is a maximum since the leading coefficient is negative.

c. f(x) = – x² + 4x –5
Answer:
This function cannot be used to model the data set. The y – intercept for this equation is – 5 instead of 5.

d. f(x) = 3^x + 4
Answer:
This function cannot be use used to model the data set. The y – intercept is 5, and f(x) values are not being multiplied by a constant value. It is not an exponential function.

e. f(x) = (x – 2)² + 9
Answer:
This function cannot be used to model the data set. The vertex is (2,9); however, it is a minimum in this function.

f. f(x) = – (x + 1)(x – 5)
Answer:
This function can be used to model the data set. One of the x – intercepts is x = 5, and the second x – intercept is x = – 1 by following the pattern of the data. The function equation indicates x – intercepts of – 1 and 5, where the vertex is a maximum value of the function.

Exercise 2.
Match each table below to the function and the context, and explain how you made your decision.

Answer:

Equations:
f(x) = 12x

h(x) = – 9|x – 3| + 27

g(x) = – (x)(x – 6)

p(x) = 2^x

q(x) = – 16x² + 30x + 160

Contexts:
1. The population of bacteria doubled every month, and the total population vs. time was recorded.
2. A ball was launched upward from the top of a building, and the vertical distance of the ball from the ground vs. time was recorded.
3. The height of a certain animal’s vertical leap was recorded at regular time intervals of one second; the animal returned to ground level after six seconds.
4. Melvin saves the same amount of money every month. The total amount saved after each month was recorded.
5. Chris ran at a constant rate on a straight – line path and then returned at the same rate. His distance from his starting point was recorded at regular time intervals.

Eureka Math Algebra 1 Module 5 Lesson 2 Problem Set Answer Key

Question 1.

Answer:

a. Determine the function type that could be used to model the data set at the right, and explain why.
Answer:
Quadratic—the second difference is 4.

b. Complete the data set using the special pattern of the function you described in part (a).
Answer:
See the completed table for answers.

c. If it exists, find the minimum or maximum value for the function model. If there is no minimum or maximum, explain why.
Answer:
Minimum value occurs at (4, – 8).

Question 2.

Answer:

a. Determine the function type that could be used to model the data set, and explain why.
Answer:
Exponential—y – value is being multiplied by constant value 4.

b. Complete the data set using the special pattern of the function you described in part (a).
Answer:
See responses in the completed table.

c. If it exists, find the minimum or maximum value for the function model. If there is no minimum or maximum, explain why.
Answer:
Since this is an exponential function, the y – values increase as x – values increase, and there is no maximum value. As x gets smaller and smaller (i.e., moves further to the left on the number line), the y – values approach 0 but never reach 0 and never become negative. Therefore, there is no minimum value of this function.

Question 3.

Answer:

a. Determine the function type that could be used to model the data set, and explain why.
Answer:
Linear—the first difference is 6.

b. Complete the data set using the special pattern of the function you described in part (a).
Answer:
See responses in the completed table.

c. If it exists, find the minimum or maximum value for the function model. If there is no minimum or maximum, explain why.
Answer:
There is no minimum or maximum value for a linear function (except for a horizontal line). For this function, the y – values decrease as the x – values decrease, and the y – values increase as the x – values increase.

Question 4.
Circle all the function types that could possibly be used to model a context if the given statement applies.
a. When x – values are at regular intervals, the first difference of y – values is not constant.
Linear Function           Quadratic Function          Exponential Function           Absolute Value Function
Answer:

b. When x – values are at regular intervals, the second difference of y – values is not constant.
Linear Function           Quadratic Function          Exponential Function          Absolute Value Function
Answer:

c. When x – values are at regular intervals, the quotient of any two consecutive y – values is a constant that is not equal to 0 or 1.
Linear Function          Quadratic Function          Exponential Function           Absolute Value Function
Answer:

d. There may be up to two different x – values for y = 0.
Linear Function          Quadratic Function            Exponential Function          Absolute Value Function
Answer:

Eureka Math Algebra 1 Module 5 Lesson 2 Exit Ticket Answer Key

Question 1.
Analyze these data sets, recognizing the unique pattern and key feature(s) for each relationship. Then use your findings to fill in the missing data, match to the correct function from the list on the right, and describe the key feature(s) that helped you choose the function.

Equations:
f(x) = 6^x
h(x) = – 3(x – 2)² + 18
g(x) = – 2(x + 1)(x – 3)
r(x) = 4x + 6
Table A: __________________ Key Feature(s): ________________________________________________________

Table B: __________________ Key Feature(s): ________________________________________________________

Table C: __________________ Key Feature(s): ________________________________________________________

Table D: __________________ Key Feature(s): ________________________________________________________
Answer:

Table A: r(x) = 4x + 6                     Key Feature(s): Constant slope (first difference) is 4

Table B: h(x) = – 3(x – 2)² + 18      Key Feature(s): (2, 18) is the maximum point; Second difference is –6

Table C: f(x) = 6^x                           Key Feature(s): Exponential growth; Common ratio is 6

Table D: g(x) = – 2(x + 1)(x – 3)     Key Feature(s): x – intercepts ( – 1, 0) and (3, 0); Second difference is –4

Engage NY Eureka Math Algebra 1 Module 5 Lesson 1 Answer Key

Eureka Math Algebra 1 Module 5 Lesson 1 Example Answer Key

Example 1.
Eduardo has a summer job that pays him a certain rate for the first 40 hours each week and time – and – a – half for any overtime hours. The graph below shows how much money he earns as a function of the hours he works in one week.

Answer:
Start by asking students to consider the function equation for this graph, and ask them to justify their choices. If students are unable to come up with viable options, consider using this scaffolding suggestion. Otherwise skip to the questions that follow, and use them to guide the discussion. Try to use as little scaffolding as possible in this section so that students have an experience closer to a true modeling situation.
→ Right now we are in the formulate stage of the modeling cycle. This means we are starting with a problem and selecting a model (symbolic, analytical, tabular, and/or graphic) that can represent the relationship between the variables used in the context. What are the variables in this problem? What are the units?
→ Time worked (in hours); earnings (in dollars)
→ We have identified the variables. Now let’s think about how the problem defines the relationship between the variables.
→ The number of dollars earned is dependent on the number of hours worked. The relationship is piecewise linear because the average rate of change is constant for each of the intervals (pieces), as depicted in the graph.

→ So what does this graph tell you about Eduardo’s pay for his summer job?
He has a constant pay rate up to 40 hours, and then the rate changes to a higher amount. (Students may notice that his pay rate from 0 to 40 hours is $9, and from 40 hours on is $13.50.)

→ The graph shows us the relationship. In fact, it is an important part of the formulating step because it helps us to better understand the relationship. Why would it be important to find the analytical representation of the function as well?
The equation captures the essence of the relationship succinctly and allows us to find or estimate values that are not shown on the graph.

→ How did you choose the function type? What were the clues in the graph?
Visually, the graph looks like two straight line segments stitched together. So, we can use a linear function to model each straight line segment. The presence of a sharp corner usually indicates a need for a piecewise defined function.

→ There are four points given on the graph. Is that enough to determine the function?
In this case, yes. Each linear piece of the function has two points, so we could determine the equation for each.
What do you notice about the pieces of the graph?
The second piece is steeper than the first; they meet where x = 40; the first goes through the origin; there are two known points for each piece

Eureka Math Algebra 1 Module 5 Lesson 1 Exercise Answer Key

Opening Exercise
The graphs below give examples for each parent function we have studied this year. For each graph, identify the function type and the general form of the parent function’s equation; then offer general observations on the key features of the graph that helped you identify the function type. (Function types include linear, quadratic, exponential, square root, cube root, cubic, absolute value, and other piecewise functions. Key features may include the overall shape of the graph,
x – and y – intercepts, symmetry, a vertex, end behavior, domain and range values or restrictions, and average rates of change over an interval.)




Answer:

Exercises

Exercise 1.
Write the function in analytical (symbolic) form for the graph in Example 1.
a. What is the equation for the first piece of the graph?
Answer:
The two points we know are (0, 0) and (22, 198). The slope of the line is 9 (or $9/hour), and the equation is f(x) = 9x.

b. What is the equation for the second piece of the graph?
Answer:
The second piece has the points (60, 630) and (70, 765). The slope of the line is 13.5 (or $13.50/hour), and the equation in point – slope form would be either y – 630 = 13.5(x – 60) or y – 765 = 13.5(x – 70), with both leading to the function, f(x) = 13.5x – 180.

c. What are the domain restrictions for the context?
Answer:
The graph is restricted to one week of work with the first piece starting at x = 0 and stopping at x = 40. The second piece applies to x – values greater than 40. Since there are 168 hours in one week, the absolute upper limit should be 168 hours. However, no one can work nonstop, so setting 80 hours as an upper limit would be reasonable. Beyond 168 hours, Eduardo would be starting the next week and would start over with $9/hour for the next 40 hours.

d. Explain the domain in the context of the problem.
Answer:
The first piece starts at x = 0 and stops at x = 40. The second piece starts at x>40. From 0 to 40 hours the rate is the same: $9/hour. Then, the rate changes to $13.50/hour at x>40. After 80 hours, it is undefined since Eduardo would need to sleep.

For each graph below, use the questions and identified ordered pairs to help you formulate an equation to represent it.

Exercise 2.
Function type:
Parent function:
Transformations:
Equation:
Answer:
Function type: Exponential
Parent function: f(x) = a^x
Transformations: It appears that the graph could be that of a parent function because it passes through (0, 1), and the x – axis is a horizontal asymptote.
Equation: The fact that the graph passes through the point (0, 1) and the x – axis is a horizontal asymptote indicates there is no stretch factor or translation.
Finding a using (1, 3):
3 = a¹
3 = a
f(x) = 3^x

Exercise 3.
Function type:
Parent function:
Transformations:
Equation:
Answer:
Function type: Square root
Parent function: f(x) = \(\sqrt{x}\)
Transformations: Appears to be a stretch
Equation: f(x) = a\(\sqrt{x}\)
Checking for stretch or shrink factor using (4, 4):
4 = a\(\sqrt{4}\)
4 = a(2)
2 = a
Checking a = 2 with (1, 2):
2 = 2\(\sqrt{1}\)
2 = 2 Yes
f(x) = 2\(\sqrt{x}\)
Note: Students may need a hint for this parent function since they have not worked much with square root functions. Additionally, the stretch factor could be inside or outside the radical. You might ask students who finish early to try it both ways and verify that the results are the same (you could use f(x) = a\(\sqrt{x}\) or f(x) = \(\sqrt{bx}\)).

Exercise 4.
Function type:
Parent function:
Transformations:
Equation:
Answer:
Function type: Cubic
Parent function: f(x) = x³
Transformations: Appears to be a vertical shift of 2 with no horizontal shift
Equation: f(x) = ax³ + 2
Checking for stretch or shrink with ( – 1, 1):
1 = a( – 1)³ + 2
1 = a (no stretch or shrink)
Checking with (2, 10):
10 = (2)³ + 2
10 = 8 + 2
10 = 10 Yes
f(x) = x³ + 2

Exercise 5.
Function type:
Parent function:
Transformations:
Equation:
Answer:
Parent function: f(x) = \(\sqrt [ 3 ]{ x }\)
Transformations: Appears to be a shift to the right of 1
Equation: f(x) = a\(\sqrt [ 3 ]{ x – 1 }\)
Checking for possible stretch or shrink using (9, 2):
2 = a\(\sqrt [ 3 ]{ 9 – 1 }\)
1 = a (no stretch or shrink)
Now check (0, – 1):
– 1 = \(\sqrt [ 3 ]{ 0 – 1 }\)
– 1 = – 1 Yes
f(x) = \(\sqrt [ 3 ]{ x – 1 }\)

Exercise 6.
Function type:
Parent function:
Transformations:
Equation:
Answer:
Function type: Quadratic
Parent function: f(x) = x²
Transformations: Shift up 2 units and to the right 1 unit
Equation: Using the vertex form with (1, 2):
f(x) = a(x – 1)² + 2
Finding the stretch or shrink factor using (0, 5):
5 = a(0 – 1)² + 2
3 = a(1)
a = 3
Checking with (2, 5):
5 = 3(2 – 1)² + 2
3 = 3(2 – 1)
3 = 3(1) Yes. There is a stretch factor of 3.
f(x) = 3(x – 1)² + 2

Eureka Math Algebra 1 Module 5 Lesson 1 Problem Set Answer Key

Question 1.
During tryouts for the track team, Bob is running 90 – foot wind sprints by running from a starting line to the far wall of the gym and back. At time t = 0, he is at the starting line and ready to accelerate toward the opposite wall. As t approaches 6 seconds, he must slow down, stop for just an instant to touch the wall, turn around, and sprint back to the starting line. His distance, in feet, from the starting line with respect to the number of seconds that has passed for one repetition is modeled by the graph below.

a. What are the key features of this graph?
Answer:
The graph appears to represent a quadratic function. The maximum point is at (6, 90). The zeros are at (0, 0) and (12, 0).

b. What are the units involved?
Answer:
Distance is measured in feet and time in seconds.

c. What is the parent function of this graph?
Answer:
We will attempt to model the graph with a quadratic function. The parent function could be f(t) = t².

d. Were any transformations made to the parent function to get this graph?
Answer:
It has a negative leading coefficient, and it appears to shift up 90 units and to the right 6 units.

e. What general analytical representation would you expect to model this context?
Answer:
f(t) = a(t – h)² + k

f. What do you already know about the parameters of the equation?
Answer:
a < 0, h = 6, k = 90

g. Use the ordered pairs you know to replace the parameters in the general form of your equation with constants so that the equation will model this context. Check your answer using the graph.
Answer:
To find a, substitute (0, 0) for (x, y) and (6, 90) for (h, k):
0 = a(0 – 6)² + 90
– 90 = a(36)
a = – \(\frac{90}{36}\) = – 2.5
f(t) = – 2.5(t – 6)² + 90
Now check it with (12, 0):
0 = – 2.5(12 – 6)² + 90
– 90 = – 2.5(36)
– 90 = – 90 Yes

Question 2..
Spencer and McKenna are on a long – distance bicycle ride. Spencer leaves one hour before McKenna. The graph below shows each rider’s distance in miles from his or her house as a function of time since McKenna left on her bicycle to catch up with Spencer. (Note: Parts (e), (f), and (g) are challenge problems.)

a. Which function represents Spencer’s distance? Which function represents McKenna’s distance? Explain your reasoning.
Answer:
The function that starts at (0, 20) represents Spencer’s distance since he had a 1 – hour head start. The function that starts at (0, 0) represents McKenna’s distance since the graph is described as showing distance since she started riding. That means at the time she started riding (t = 0 hours), her distance would need to be 0 miles.

b. Estimate when McKenna catches up to Spencer. How far have they traveled at that point in time?
Answer:
McKenna will catch up with Spencer after about 3.25 hours. They will have traveled approximately 41 miles at that point.

c. One rider is speeding up as time passes and the other one is slowing down. Which one is which, and how can you tell from the graphs?
Answer:
I know that Spencer is slowing down because his graph is getting less steep as time passes. I know that McKenna is speeding up because her graph is getting steeper as time passes.

d. According to the graphs, what type of function would best model each rider’s distance?
Answer:
Spencer’s graph appears to be modeled by a square root function. McKenna’s graph appears to be quadratic.

e. Create a function to model each rider’s distance as a function of the time since McKenna started riding her bicycle. Use the data points labeled on the graph to create a precise model for each rider’s distance.
Answer:
If Spencer started 1 hour before McKenna, then ( – 1, 0) would be a point on his graph. Using a square root function in the form f(x) = k\(\sqrt{x + 1}\) would be appropriate. To find k, substitute (0, 20) into the function.
20 = k\(\sqrt{0 + 1}\)
20 = k
So, f(x) = 20\(\sqrt{x + 1}\). Check with the other point (3, 40):
f(3) = 20\(\sqrt{3 + 1}\)
f(3) = 20\(\sqrt{4}\) = 40
For McKenna, using a quadratic model would mean the vertex must be at (0, 0). A quadratic function in the form g(x) = kx² would be appropriate. To find k, substitute (1, 4) into the function.
4 = k(1)²
4 = k
So, g(x) = 4x². Check with the other point (3, 36): g(3) = 4(3)² = 36.

f. What is the meaning of the x – and y – intercepts of each rider in the context of this problem?
Answer:
Spencer’s x – intercept ( – 1, 0) shows that he starts riding one hour before McKenna. McKenna’s x – intercept shows that at time 0, her distance from home is 0, which makes sense in this problem. Spencer’s y – intercept (0, 20) means that when McKenna starts riding one hour after he begins, he has already traveled 20 miles.

g. Estimate which rider is traveling faster 30 minutes after McKenna started riding. Show work to support your answer.
Answer:
Spencer:
\(\frac{f(0.5) – f(0.4)}{0.5 – 0.4}\) ≈ 8.3
McKenna:
\(\frac{g(0.5) – g(0.4)}{0.5 – 0.4}\) = 3.6
On the time interval from [0.4, 0.5], Spencer’s average rate of change was approximately 8.3 mph, and McKenna’s average rate of change was 3.6 mph. Therefore, Spencer is traveling faster than McKenna
30 minutes after McKenna begins riding because his average rate of change is greater than McKenna’s average rate of change.

Eureka Math Algebra 1 Module 5 Lesson 1 Exit Ticket Answer Key

Read the problem description, and answer the questions below. Use a separate piece of paper if needed.
A library posted a graph in its display case to illustrate the relationship between the fee for any given late day for a borrowed book and the total number of days the book is overdue. The graph, shown below, includes a few data points for reference. Rikki has forgotten this policy and wants to know what her fine would be for a given number of late days. The ordered pairs on the graph are (1, 0.1), (10, 1), (11, 1.5), and (14, 3).

Question 1.
What type of function is this?
Answer:
Piecewise linear

Question 2.
What is the general form of the parent function(s) of this graph?
Answer:

Question 3.
What equations would you expect to use to model this context?
Answer:

Students may be more informal in their descriptions of the function equation and might choose to make the domain restriction of the second piece inclusive rather than the first piece since both pieces are joined at the same point.

Question 4.
Describe verbally what this graph is telling you about the library fees.
Answer:
The overdue fee is a flat rate of $0.10 per day for the first 10 days and then increases to $0.50 per day after 10 days. The fee for each of the first 10 days is $0.10, so the fee for 10 full days is $0.10(10) = $1.00. Then, the fee for 11 full days of late fees is $1.00 + $0.50 = $1.50, etc. (From then on, the fee increases to $0.50 for each additional day.)

Question 5.
Compare the advantages and disadvantages of the graph versus the equation as a model for this relationship. What would be the advantage of using a verbal description in this context? How might you use a table of values?
Answer:
Graphs are visual and allow us to see the general shape and direction of the function. However, equations allow us to determine more exact values since graphs only allow for estimates for any non – integer values. The late – fee scenario depends on integer number of days only; other scenarios may involve independent variables of non – integer values (e.g., gallons of gasoline purchased). In this case, a table could be used to show the fee for each day but could also show the accumulated fees for the total number of days. For example, for 15 days, the fees would be $1.00 for the first 10 plus $2.50 for the next 5, for a total of $3.50.

Question 6.
What suggestions would you make to the library about how it could better share this information with its customers? Comment on the accuracy and helpfulness of this graph.
Answer:
Rather than displaying the late fee system in a graph, a table showing the total fine for the number of days late would be clearer. If a graph is preferred, it might be better to use a discrete graph, or even a step graph, since the fees are not figured by the hour or minute but only by the full day. While the given graph shows the rate for each day, most customers would rather know, at a glance, what they owe, in total, for their overdue books.