Tag: Eureka Math Algebra 1 Module 4

Engage NY Eureka Math Algebra 1 Module 4 Lesson 17 Answer Key

Eureka Math Algebra 1 Module 4 Lesson 17 Example Answer Key

Example
A high school baseball player throws a ball straight up into the air for his math class. The math class was able to determine that the relationship between the height of the ball and the time since it was thrown could be modeled by the function h(t) = – 16t² + 96t + 6, where t represents the time (in seconds) since the ball was thrown, and h represents the height (in feet) of the ball above the ground.
a. What do you notice about the equation, just as it is, that will help us in creating our graph?
Answer:
The leading coefficient is negative, so we know the graph opens down. h(0) = 6, so the point (0,6), which is the y – intercept, is on the graph.

b. Can we factor to find the zeros of the function? If not, solve h(t) = 0 by completing the square.
Answer:
This function is not factorable, so we complete the square to find the zeros to be (6.06,0) and ( – 0.06,0).

c. What is the vertex of the function? What method did you use to find the vertex?
Answer:
Since we already completed the square (and the zeros are irrational and more difficult to work with), we can easily find the vertex using the completed – square form, h(t) = – 16(t – 3)² + 150, which means the vertex is (3,150).

d. Now plot the graph of h(t) = – 16t² + 96t + 6, and label the key features on the graph.

Answer:

Eureka Math Algebra 1 Module 4 Lesson 17 Exercise Answer Key

Opening Exercise
A high school baseball player throws a ball straight up into the air for his math class. The math class was able to determine that the relationship between the height of the ball and the time since it was thrown could be modeled by the function h(t) = – 16t² + 96t + 6, where t represents the time (in seconds) since the ball was thrown, and h represents the height (in feet) of the ball above the ground.
a. What does the domain of the function represent in this context?
Answer:
The time (number of seconds) since the ball was thrown

b. What does the range of this function represent?
Answer:
The height (in feet) of the ball above the ground

c. At what height does the ball get thrown?
Answer:
The initial height of the ball is when t is 0 sec. (i.e., h(0)), which is the y – intercept. The initial height is 6 ft.

d. After how many seconds does the ball hit the ground?
Answer:
The ball’s height is 0 when h(t) = 0. We can solve using any method. Since this does not appear to be easily factorable, and the size of the numbers might be cumbersome in the quadratic formula, let’s solve by completing the square.
– 16t² + 96t + 6 = 0
– 16(t²–6t) = – 6
– 16(t² – 6t + 9) = – 6 – 144
From here, we see the completed – square form: h(t) = – 16(t – 3)² + 150.
– 16(t – 3)² = – 150
(t – 3)² = \(\frac{150}{16}\)
t – 3 = ±\(\frac{\sqrt{150}}{4}\)
t = 3 ±\(\frac{\sqrt{150}}{4}\)
t ≈ 6.0618 or – 0.0618
For this context, the ball hits the ground at approximately 6.1 seconds.

e. What is the maximum height that the ball reaches while in the air? How long will the ball take to reach its maximum height?
Answer:
Completing the square (and using the work from the previous question), we get h(t) = – 16(t – 3)² + 150, so the vertex is (3,150), meaning that the maximum height is 150 ft., and it will reach that height in 3 sec.

f. What feature(s) of this quadratic function are visible since it is presented in the standard form, f(x) = ax² + bx + c?
Answer:
We can see the initial position, or height of the ball, or the height when t = 0, in the constant term. We can also see the leading coefficient, which tells us about the end behavior and whether the graph is wider or narrower than the graph of f(x) = x².

g. What feature(s) of this quadratic function are visible when it is rewritten in vertex form, f(x) = a(x – h)² + k?
Answer:
We can only see the coordinates of the vertex and know that x = h is the equation of the axis of symmetry. We can still see the leading coefficient in this form, which tells us about the end behavior and whether the graph is wider or narrower than the graph of f(x) = x².

A general strategy for graphing a quadratic function from the standard form:
Answer:
→ Look for hints in the function’s equation for general shape, direction, and y – intercept.
→ Solve f(x) = 0 to find the x – intercepts by factoring, completing the square, or using the quadratic formula.
→ Find the vertex by completing the square or using symmetry. Find the axis of symmetry and the x – coordinate of the vertex using \(\frac{ – b}{2a}\) and the y – coordinate of the vertex by finding f(\(\frac{ – b}{2a}\)).
→ Plot the points you know (at least three are required for a unique quadratic function), sketch the graph of the curve that connects them, and identify the key features of the graph.

Exercises
Exercise 1.
Graph the function n(x) = x² – 6x + 5, and identify the key features.

Answer:

x – intercepts: (5, 0),(1, 0)
y – intercept: (0, 5)
Vertex: (3, – 4)

Exercise 2.
Graph the function f(x) = \(\frac{1}{2}\) x² + 5x + 6, and identify the key features.

Answer:

x – intercepts: ( – 5 + \(\sqrt{13}\),0),( – 5 – \(\sqrt{13}\),0)
y – intercept: (0, 6)
Vertex: ( – 5, – 6.5)

Exercise 3.
Paige wants to start a summer lawn – mowing business. She comes up with the following profit function that relates the total profit to the rate she charges for a lawn – mowing job:
P(x) = – x² + 40x – 100.
Both profit and her rate are measured in dollars. Graph the function in order to answer the following questions.
a. Graph P.

Answer:

x – intercepts: (20 + 10\(\sqrt{3}\),0),(20 – 10\(\sqrt{3}\))
y – intercept: (0, – 100)
Vertex: (20, 300)

b. According to the function, what is her initial cost (e.g., maintaining the mower, buying gas, advertising)? Explain your answer in the context of this problem.
Answer:
When Paige has not mown any lawns or charged anything to cut grass, her profit would be – 100. A negative profit means that Paige is spending $100 to run her business.

c. Between what two prices does she have to charge to make a profit?
Answer:
Using completing the square, we find the intercepts at (20 + 10\(\sqrt{3}\)) and (20 – 10\(\sqrt{3}\)). However, since this question is about money, we approximate and find that her rates should be between $2.68 and $37.32.

d. If she wants to make a $275 profit this summer, is this the right business choice?
Answer:
Yes. Looking at the graph, the vertex, (20, 300), is the maximum profit. So, if Paige charges $20 for each lawn she mows, she can make a $300 profit, which is $25 more than she wants.

Exercise 4.
A student throws a bag of chips to her friend. Unfortunately, her friend does not catch the chips, and the bag hits the ground. The distance from the ground (height) for the bag of chips is modeled by the function
h(t) = – 16t² + 32t + 4, where h is the height (distance from the ground in feet) of the chips, and t is the number of seconds the chips are in the air.
a. Graph h.

Answer:

– intercepts: (1 + \(\frac{\sqrt{5}}{2}\),0),(1 – \(\frac{\sqrt{5}}{2}\),0)
h – intercept: (0, 4)
Vertex: (1, 20)

b. From what height are the chips being thrown? Tell how you know.
Answer:
4 ft. This is the initial height, or when t = 0.

c. What is the maximum height the bag of chips reaches while airborne? Tell how you know.
Answer:
From the graph, the vertex is (1,20), which means that at 1 second, the bag is 20 ft. above the ground for this problem. Since this is the vertex of the graph, and the leading coefficient of the quadratic function is negative, the graph opens down (as t → ±∞, h(t) → – ∞), and the vertex is the maximum of the function. This means 20 ft. is the maximum height of the thrown bag.

d. How many seconds after the bag was thrown did it hit the ground?
Answer:
By completing the square, we find that t = 1 ±\(\frac{\sqrt{5}}{2}\). Since this is time in seconds, we need a positive value, 1 + \(\frac{\sqrt{5}}{2}\), which is about 2.12 sec.

e. What is the average rate of change of height for the interval from 0 to 1/2 second? What does that number represent in terms of the context?
Answer:
\(\frac{\left[f\left(\frac{1}{2}\right) – f(0)\right]}{\frac{1}{2} – 0}\) = \(\frac{[16 – 4]}{\frac{1}{2}}\) = 24. The rate of change for the interval from 0 to \(\frac{1}{2}\) sec. is 24 ft/s, which represents the average speed of the bag of chips from 0 to \(\frac{1}{2}\) sec.

f. Based on your answer to part (e), what is the average rate of change for the interval from 1.5 to 2 sec.?
Answer:
The average rate of change for the interval from 1.5 to 2 sec. will be the same as it is from 0 to \(\frac{1}{2}\) except that it will be negative: – 24 ft/s.

Exercise 5.
Notice how the profit and height functions both have negative leading coefficients. Explain why this is.
Answer:
The nature of both of these contexts is that they have continually changing rates, and both require the graph to open down since each would have a maximum. Problems that involve projectile motion have maxima because an object can only go so high before gravity pulls it back down. Profits also tend to increase as prices increase only to a point before sales drop off, and profits begin to fall.

Eureka Math Algebra 1 Module 4 Lesson 17 Problem Set Answer Key

Question 1.
Graph f(x) = x² – 2x – 15, and identify its key features.

Answer:

x – intercepts: ( – 3, 0) (5, 0)
y – intercept: (0, – 15)
Vertex: (1, – 16)
End behavior: As x → ±∞, y → ∞.

Question 2.
Graph f(x) = – x² + 2x + 15, and identify its key features.

Answer:

x – intercepts: ( – 3, 0) (5, 0)
y – intercept: (0, 15)
Vertex: (1, 16)
End behavior: As x → ±∞, y → – ∞.

Question 3.
Did you recognize the numbers in the first two problems? The equation in the second problem is the product of – 1 and the first equation. What effect did multiplying the equation by – 1 have on the graph?
Answer:
The graph gets reflected across the x – axis. The x – intercepts remain the same. The y – intercept becomes the opposite of the original y – intercept. The end behavior of the graph reversed. The vertex became the maximum instead of the minimum.

Question 4.
Giselle wants to run a tutoring program over the summer. She comes up with the following profit function:
P(x) = – 2x² + 100x – 25,
Answer:
where x represents the price of the program. Between what two prices should she charge to make a profit? How much should she charge her students if she wants to make the most profit?
Using the quadratic formula, the two roots are 25 ±\(\frac{35 \sqrt{2}}{2}\), which is about $0.25 and $50. If Giselle charges between $0.25 and $50, she can expect to make a profit. If she charges $25, she will make a maximum profit of $1,225.

Question 5.
Doug wants to start a physical therapy practice. His financial advisor comes up with the following profit function for his business:
P(x) = – \(\frac{1}{2}\) x² + 150x – 10000
where x represents the amount, in dollars, that he charges his clients. How much will it cost for him to start the business? What should he charge his clients to make the most profit?
Answer:
The formula suggests it would cost him $10,000 to start his business. He should charge $150 to make the most profit.

Eureka Math Algebra 1 Module 4 Lesson 17 Exit Ticket Answer Key

Question 1.
Graph g(x) = x² + 10x – 7, and identify the key features (e.g., vertex, axis of symmetry, x – and y – intercepts).

Answer:

Engage NY Eureka Math Algebra 1 Module 4 Lesson 24 Answer Key

Eureka Math Algebra 1 Module 4 Lesson 24 Example Answer Key

Example 1.
Use the points (0, 4), (1, 9), and (-3, 1) to write the equation for the quadratic function whose graph contains the three points.
Answer:
Demonstrate for students how, if we know the y-intercept and two other points for a quadratic function, we can form a system of linear equations to determine the standard form of the quadratic function defined by those points. Use the example with the blue points above: (0, 4), (1, 9), and (-3, 1).
Notice that we have the y-intercept, which allows us to find the value of c quickly and first. After that, we can substitute the other two coordinates into the equation, giving us two linear equations to solve simultaneously.
Using (0, 4)
f(x) = ax² + bx + c
4 = a(0)² + b(0) + c
4 = c

Using (1, 9)
f(x) = ax² + bx + 4
9 = a(1)² + b(1) + 4
9 = a + b + 4
a + b = 5

Using (-3, 1)
f(x) = ax² + bx + 4
1 = a(-3)² + b(-3) + 4
1 = 9a-3b + 4
9a-3b = -3
→ Since c = 4, the resulting system has two variables: . Use substitution or elimination to determine that a = 1 and b = 4.
→Substitute a = 1, b = 4, and c = 4 into standard form: f(x) = x² + 4x + 4 is the quadratic function that contains the given points.

Demonstrate that the graph of the function we just found does, in fact, pass through all three points by showing the graph on the board or screen.
→ Notice that in the graph below, we have included the two different fourth points from the Opening Exercise, (-1, 1) and (2, 5). Clearly (-1, 1) is on the graph of the function, but (2, 5) is not.

Eureka Math Algebra 1 Module 4 Lesson 24 Exercise Answer Key

Opening Exercise
Draw as many quadratic graphs as possible through the following two points on the graph. Check with your neighbors for ideas. These points are (0, 4) and (1, 9).

Answer:
After a few minutes, gather the class together, and have students share some of their graphs. You might have three or four students come to the board and sketch one of their graphs, each in a different color. There are an infinite number of solutions. Make sure that some of the sketches have one of the points as a vertex and that some open up and some down.
Now, introduce a third point and ask students to repeat the exercise. Now the points are (0, 4), (1, 9), and (-3, 1).

Ultimately, students should conclude that only one quadratic graph can pass through all three points simultaneously. Therefore, it requires no less than three points to determine a quadratic function.

Students may be curious about what happens if a fourth point is introduced. Add a fourth point in two different places, and have them study the possibilities. Try adding a point in another color that is on the quadratic graph, (-1, 1), and then add one that is not, (2, 5).


Explain that a fourth point, in this case (2, 5), may either belong to the quadratic graph (see: Fourth Point #1 graph) or not (see: Fourth Point #2 graph), but the function has already been determined by the first three (blue) points.

Exercise 1.
Write in standard form the quadratic function defined by the points (0, 5), (5, 0), and (3, -4).
Answer:
Using (0, 5)
f(x) = ax² + bx + c
5 = a(0)² + b(0) + c
5 = c

Using (5, 0)
f(x) = ax² + bx + 5
0 = a(5)² + b(5) + 5
0 = 25a + 5b + 5
25a + 5b = -5
5a + b = -1

Using (3, -4)
f(x) = ax² + bx + 5
-4 = a(3)² + b(3) + 5
-4 = 9a + 3b + 5
9a + 3b = -9
3a + b = -3

Since c = 5, the resulting system has two variables: .
Use substitution or elimination, and find that a = 1 and b = -6.
Substitute a = 1, b = -6, and c = 5 into standard form: f(x) = x²-6x + 5 is the quadratic function that contains the given points.

Exercise 2.
Louis dropped a watermelon from the roof of a tall building. As it was falling, Amanda and Martin were on the ground with a stopwatch. As Amanda called the seconds, Martin recorded the floor the watermelon was passing. They then measured the number of feet per floor and put the collected data into this table. Write a quadratic function to model the following table of data relating the height of the watermelon (distance in feet from the ground) to the number of seconds that had passed.

a. How do we know this data will be represented by a quadratic function?
Answer:
The relationship between height and time for all free-falling objects is represented by a quadratic equation. Also, we can see mathematically that the function values have a first difference of -16, -48, -80, and -112. The second differences are constant at -32.

b. Do we need to use all five data points to write the equation?
Answer:
No, only three are needed.

c. Are there any points that are particularly useful? Does it matter which we use? Write the quadratic function that models the data.
Answer:
(0, 300) is useful because it is the y-intercept. We will need to use (0, 300), but the other two can be selected based on efficiency (the least messy or smallest numbers).
Use (0, 300)
f(t) = at² + bt + c
300 = a(0)² + b(0) + c
300 = c

Use (1, 284)
f(t) = at² + bt + c
284 = a(1)² + b(1) + 300
-16 = a + b

Use (2, 236)
f(t) = at² + bt + c
236 = a(2)² + b(2) + 300
-64 = 4a + 2b

Since c = 300, the resulting system has two variables: .
Use substitution or elimination and find that a = -16 and b = 0.
Substitute a = -16, b = 0, and c = 300 into standard form: f(t) = -16t² + 300.

d. How does this equation for the function match up with what you learned about physics in Lesson 23? Is there a more efficient way to find this equation?
Answer:
It matches perfectly. This equation shows that the initial position (height) of the object is 300 ft. and that the initial velocity is 0. It correctly uses -16 as the leading coefficient. We could have written the equation directly from the information provided since we already know the initial height and velocity.

e. Can you use your quadratic function to predict at what time, t, the watermelon will hit the ground (i.e., f(t) = 0)?
Answer:
Yes.
f(t) = -16t² + 300
0 = -16t² + 300
-300 = -16t²
18.75 = t²
±4.33 ≈ t
So, the watermelon hit the ground after about 4.33 sec.

Eureka Math Algebra 1 Module 4 Lesson 24 Problem Set Answer Key

Question 1.
Write a quadratic function to fit the following points, and state the x-values for both roots. Then, sketch the graph to show that the equation includes the three points.

Answer:

Using the three points:
Use (0, 4)
f(x) = ax² + bx + c
4 = a(0)² + b(0) + c
4 = c

Use (-2, 0)
f(x) = ax² + bx + c
0 = a(-2)² + b(-2) + 4
-4 = 4a-2b

Use (1, 3)
f(x) = ax² + bx + c
3 = a(1)² + b(1) + 4
-1 = a + b

Since c = 4, the resulting system has two variables: .
Use substitution or elimination and find that a = -1 and b = 0.
Substitute a = -1, b = 0, and c = 4 into standard form: f(x) = -x² + 4.

Question 2.
Write a quadratic function to fit the following points: (0, 0.175), (20, 3.575), (30, 4.675).
Answer:
Use (0, 0.175)
f(x) = ax² + bx + c
0.175 = a(0)² + b(0) + c
0.175 = c

Use (20, 3.575)
f(x) = ax² + bx + c
3.575 = a(20)² + b(20) + 0.175
3.4 = 400a + 20b

Use (30, 4.675)
f(x) = ax² + bx + c
4.675 = a(30)² + b(30) + 0.175
4.5 = 900a + 30b
Since c = 0.175, the resulting system has two variables: .
Use substitution or elimination and find that a = -0.002 and b = 0.21.
Substitute a = -0.002, b = 0.21, and c = 0.175 into standard form: f(x) = -0.002x² + 0.21x + 0.175.

Eureka Math Algebra 1 Module 4 Lesson 24 Exit Ticket Answer Key

Question 1.
Write a quadratic function from the following table of data.

Answer:
Using the three points:
Use (0, 4.7)
f(x) = ax² + bx + c
4.7 = a(0)² + b(0) + c
4.7 = c

Use (100, 8.7)
f(x) = ax² + bx + c
8.7 = a(100)² + b(100) + 4.7
4 = 10, 000a + 100b

Use (200, 10.7)
f(x) = ax² + bx + c
10.7 = a(200)² + b(200) + 4.7
6 = 40, 000a + 200b

Since c = 4.7, the resulting system has two variables: .
Use substitution or elimination and find that a = \(\frac{-1}{10, 000}\) = -0.0001 and b = \(\frac{1}{20}\) = 0.05.
Substitute a = -0.0001, b = 0.05, and c = 4.7 into standard form: f(x) = -0.0001x² + 0.05x + 4.7.

Engage NY Eureka Math Algebra 1 Module 4 Lesson 22 Answer Key

Eureka Math Algebra 1 Module 4 Lesson 22 Exploratory Challenge Answer Key

Exploratory Challenges 1–3
Solve each problem, and show or explain how you found your answers.
Question 1.
Xavier and Sherleese each threw a baseball straight up into the air. The relationship between the height (distance from the ground in feet) of Sherleese’s ball with respect to the time since it was thrown, in seconds, is given by the function:
S(t) = – 16t² + 79t + 6.
The graph of the height as a function of time of Xavier’s ball is represented below.

Xavier claims that his ball went higher than Sherleese’s. Sherleese disagrees. Answer the questions below, and support your answers mathematically by comparing the features found in the equation to those in the graph.

a. Who is right?
Answer:
Sherleese is right. Xavier’s baseball went 75 ft. at its highest point (based on the maximum point of the graph), which was lower than Sherleese’s. The maximum height of Sherleese’s ball is approximately 103.5 ft. This could be determined by completing the square, S(t) = – 16(t – \(\frac{79}{32}\))² + 103.5, using the vertex formula to find the vertex, or using a table to approximate.

b. For how long was each baseball airborne?
Answer:
Sherleese’s baseball was airborne for approximately 5 sec.; the zeros of S are (5.01, 0) and ( – 0.07, 0). Since the ball lands on the ground at about 5 sec., that is approximately how long it was in flight. Xavier’s baseball was airborne for approximately 4.3 sec.

c. Construct a graph of Sherleese’s throw as a function of time (t) on the same set of axes as the graph of Xavier’s, and use the graph to support your answers to parts (a) and (b).

Answer:
In the graph below, you can see that Sherleese’s baseball clearly has a higher vertex. You can also read the t – axis to find the length of time each baseball was airborne (the distance between t = 0 and the positive t – intercept).

Question 2.
In science class, students constructed pendulums of various lengths and then recorded the time required for the pendulum to complete one full oscillation (out and back). The results are displayed in the table shown below.
a. Jack looks at the first three rows of the table and says that a linear function should be used to model the data. Based on the data, do you agree with him? Justify your reasoning.

Answer:
No, I do not agree with Jack. Looking at the first three rows, the data does appear to be linear because the average rate of change is the same for the intervals [5, 10] and [10, 15]. However, after that, the average rate of change does not remain constant; it decreases for each successive interval. This indicates that a linear model is not appropriate for the data.

b. Create a scatterplot of length versus oscillation time.

Answer:

c. Based on the scatterplot, what sort of function might be used to model the data?
Answer:
It appears that the data could be modeled using a square root function or possibly a cube root function.

d. Mr. Williams, the science teacher, tells the students that the oscillation time for a pendulum can be found using the formula T = 2π\(\sqrt{\frac{L}{9.8}}\) where L is the length of the pendulum, in meters, and T is the oscillation time, in seconds. Does this formula support the results from the table?
Answer:
Yes. When values for L from the table are substituted into the formula, the values for T are approximately equal to the values in the table. For example, for a length of 25 cm, the oscillation time is approximately 1.0035 sec. (as shown in the calculation below), which is very close to the value recorded in the table (1.00).
L = 0.25→ T = 2π\(\sqrt{\frac{0.25}{9.8}}\)≈1.0035

c. Looking at the table of values, what effect does quadrupling the length of the pendulum have on the oscillation time? Use the formula from part (d) to demonstrate why this is the case.
Answer:
Quadrupling the length of the pendulum, doubles the oscillation time.
T = 2π\(\sqrt{\frac{4L}{9.8}}\) = 2π\(\sqrt{4 \cdot \frac{L}{9.8}}\) = 2π∙\(\sqrt{4}\)∙\(\sqrt{\frac{L}{9.8}}\) = 2(2π\(\sqrt{\frac{L}{9.8}}\))

Question 3.
The growth of a Great Dane puppy can be represented by the graph below, where y represents the shoulder height (in inches) and x represents the puppy’s age (in months).

The growth of a lion cub can be modeled by the function represented in the table below.

a. Which animal has the greater shoulder height at birth?
Answer:
The Great Dane has the taller shoulder height at birth, which is around 12 in. The lion cub has a shoulder height of only 8 in. at birth.

b. Which animal will have the greater shoulder height at 3 years of age (the age each animal is considered full – grown)?
Answer:
Estimating from the graph, the lion’s shoulder height at 36 months appears to be close to 25 in. Looking at the table of values, the average rate of change for each interval is decreasing. On the interval [24, 32], the average rate of change is 1.452. So on the interval [32, 36], the average rate of change should be less than 0.7. Therefore, I think the Great Dane’s height at 36 months will be closer to 4 in. than to 25 in., and the lion will have the greater shoulder height at 3 years of age.

c. If you were told that the domain for these functions is the set of all real numbers, would you agree? Why or why not?
Answer:
There are physical limitations on time and height, so the domain must be greater than or equal to zero but must not exceed the finite limit of their life spans, and the range must both be greater than or equal to the finite heights recorded at birth, and less than or equal to their maximum possible heights. Both the domain and range may have other limitations as well. For example, neither animal will grow continuously for his lifetime; rather, both are likely to shrink if they live to an old age.

Eureka Math Algebra 1 Module 4 Lesson 22 Exercise Answer Key

Opening Exercise
Populate the table on the right with values from the graph.

Answer:

Briefly discuss ways to recognize key features in both representations of this function.
→ What is the vertex for the function? Find it and circle it in both the table and the graph.
( – 2, – 1)
→ What is the y – intercept for the function? Find it and circle it in both the table and the graph.
(0, 3)
→ What are the x – intercepts for the function? Find them and circle them in both the table and the graph.
( – 3, 0) and ( – 1, 0)

→ What components of the equation for a function give us clues for identifying the key features of a graph?
Given a quadratic function in the form f(x) = ax² + bx + c, the y – intercept is represented by the constant, c; the vertex (h, k) can be seen in the completed – square form, f(x) = a(x – h)² + k; and the zeros of the function are found most readily in the factored form, f(x) = a(x – m)(x – n).

→ How can the key features of the graph of a quadratic function give us clues about how to write the function the graph represents?
Given a graph of a quadratic function, (h, k) represents the vertex (i.e., maximum or minimum point). These values, h and k, can be substituted into the vertex form. Then, substituting any other ordered pair for (x, y), which represents a point on the quadratic curve (e.g., the y – intercept), will allow you to solve for the leading coefficient, a, of the vertex form of a quadratic function.

→ When both x – intercepts are visible, we can write the equation of the graph in factored form using the coordinates of any other point to determine the leading coefficient. And when the y – intercept and only one x – intercept are visible, we can most easily write the function by using standard form. In this case, the y – intercept tells us the value of the constant term, c, and we can use two other points to substitute for x and y into the form f(x) = ax² + bx + c to determine the specific values for a and b.

Eureka Math Algebra 1 Module 4 Lesson 22 Problem Set Answer Key

Question 1.
One type of rectangle has lengths that are always two inches more than their widths. The function f describes the relationship between the width of this rectangle in x inches and its area, f(x), in square inches and is represented by the table below.

A second type of rectangle has lengths that are always one half of their widths. The function g(x) = \(\frac{1}{2}\) x² describes the relationship between the width given in x inches and the area, g(x), given in square inches of such a rectangle.
a. Use g to determine the area of a rectangle of the second type if the width is 20 inches.
Answer:
g(x) = \(\frac{1}{2}\) x²
g(20) = \(\frac{1}{2}\) (20)²
g(20) = 200
The area of the rectangle is 200 in².

b. Why is (0, 0) contained in the graphs of both functions? Explain the meaning of (0, 0) in terms of the situations that the functions describe.
Answer:
For the first type of rectangle (with length two inches more than its width), a width of 0 in. means that the rectangle’s length is 2 in. However, if the width is 0 in., its area is 0 in², and there is no rectangle. For the second type of rectangle (represented by g), a width of 0 in. means that the rectangle’s area is 0 in². Although the point (0, 0) is contained in the graphs of both functions, it does not represent any rectangle width area pair because rectangles by definition have positive widths and lengths.

c. Determine which function has a greater average rate of change on the interval 0 ≤ x ≤ 3.
Answer:

On the interval 0 ≤ x ≤ 3, f has a greater average rate of change than g, as seen in the graphs of the functions above. The average rate of change for f is (\(\frac{f(3) – f(0)}{3 – 0}\) = 5, while the average rate of change for g is \(\frac{g(3) – g(0)}{3 – 0}\) = \(\frac{3}{2}\).

d. Interpret your answer to part (c) in terms of the situation being described.
Answer:
The area of rectangles, on average, on the interval 0 ≤ x ≤ 3 is growing a little more than 3 times faster for rectangles described by the function f than those described by the function g of a certain width.

e. Which type of rectangle has a greater area when the width is 5 inches? By how much?
Answer:
According to the table, the first rectangle will have an area of 35 in² if its width is 5 in. Using the formula for g, g(5) = 12.5, the second rectangle will have an area of 12.5 in² if its width is 5 in. The first type of rectangle has an area that is 22.5 in² greater than the second type of rectangle when their widths are 5 in.

f. Will the first type of rectangle always have a greater area than the second type of rectangle when widths are the same? Explain how you know.
Answer:
If we consider rectangles with width c, where c > 0, then c + 2 > \(\frac{1}{2}\) c, so (c + 2) ⋅ c > ( \(\frac{1}{2}\) c) ⋅ c. This means that the areas of the first type of rectangle will always be greater than the areas of the second type of rectangle when they have the same width.

Question 2.
The function given by the equation y = \(\sqrt{x}\) gives the edge length, y units, of a square with area x square units. Similarly, the graph below describes the length of a leg, y units, of an isosceles right triangle whose area is x square units.

a. What is the length of a leg of an isosceles right triangle with an area of 12 square units?
Answer:
The length of a leg of an isosceles triangle with an area of 12 square units is approximately 4.9 units.

b. Graph the function that represents a square with area x square units using the same graph that was given. Which function has a greater average rate of change on the interval 0 ≤ x ≤ 3?
Answer:

The graphs of both functions intersect at the origin. According to the provided graph, when the area of an isosceles triangle is 3 square units, the length of a leg is approximately 2.4 units. Thus, the average rate of change is approximately 0.8. Using the equation y = \(\sqrt{x}\), if x = 3, y ≈ 1.7. This means that when the area of a square is 3 square units, the length of a leg is approximately 1.7 units. Thus, the average rate of change is approximately 0.6. The average rate of change of the legs of an isosceles triangle is greater than the average rate of change of the sides of a square on the interval 0 ≤ x ≤ 3.

c. Interpret your answer to part (b) in terms of the situation being described.
Answer:
Over the interval 0 ≤ x ≤ 3, the average rate of change for the lengths of the legs of an isosceles triangle per a given area is approximately \(\frac{4}{3}\) greater than the average rate of change for the sides of the square. In other words, for every unit of change in area, the legs grow (on average) by 0.2 units more over the interval
0 ≤ x ≤ 3.

d. Which will have a greater value: the edge length of a square with area 16 square units or the length of a leg of an isosceles right triangle with an area of 16 square units? Approximately by how much?
Answer:
According to the graph, the length of the leg of an isosceles right triangle with an area of 16 square units is approximately 5.7 units. Using the equation y = \(\sqrt{x}\), the length of the sides of a square with an area of 16 square units is 4 units. The leg of the isosceles right triangle is approximately 1.7 units greater than the side lengths of the square when their areas are 16 square units.

Question 3.
A portion of a graph of a cube root function, f, and select values of a square root function, g, are given below. The domain of g is ≥ 0 .

Fill in each blank with one of the following: >, <, or = .
a. f(2) _______ g(2)
Answer:
f(2) = 0 and g(2) is between 3.5 and 4.
f(2) < g(2)

b. y – intercept of f ______ y – intercept of g
Answer:
The y – intercept of f is between – 2 and – 3. The y – intercept of g is 3.
y – intercept of f < y – intercept of g

c. Average rate of change of f on interval [0, 16] ______ Average rate of change of g on interval [0, 16]
Answer:
f(x) = 2\(\sqrt [ 3 ]{ x – 2 }\)
Average rate of change of f on interval [0, 16] = \(\frac{f(16) – f(0)}{16 – 0} = \frac{2 \sqrt[3]{14} – 2 \sqrt[3]{ – 2}}{16}\) ≈ 0.459
Average rate of change of g on interval [0, 16] = \(\frac{g(16) – g(0)}{16 – 0}\) = \(\frac{5 – 3}{16}\) = 0.125
Average rate of change of f on interval [0, 16] > Average rate of change of g on interval [0, 16]

Eureka Math Algebra 1 Module 4 Lesson 22 Exit Ticket Answer Key

Question 1.
Two people, each in a different apartment building, have buzzers that don’t work. They both must throw their apartment keys out of the window to their guests, who will then use the keys to enter.
Tenant 1 throws the keys such that the height – time relationship can be modeled by the graph below. On the graph, time is measured in seconds, and height is measured in feet.

Tenant 2 throws the keys such that the relationship between the height of the keys (in feet) and the time that has passed (in seconds) can be modeled by h(t) = – 16t² + 18t + 9.
a. Whose window is higher? Explain how you know.
Answer:
The window for Tenant 1 is higher (15 ft.) than that of Tenant 2 (9 ft.), which can be seen by comparing the values of the y – intercepts. You can see this in the graph below, showing both functions on the same coordinate plane.

b. Compare the motion of Tenant 1’s keys to that of Tenant 2’s keys.
Answer:
Tenant 2’s keys reach a maximum height at the vertex and then fall back toward the ground. (See the graph on the previous page.) The vertex for the graph of h can be found by completing the square:
h(t) = – 16(t² – (\(\frac{18}{16}\))t + ) + 9
= – 16(t² – (\(\frac{9}{8}\))t + (\(\frac{9}{16}\))² ) + 9 + 16(\(\frac{9}{16}\))²
= – 16(t – \(\frac{9}{16}\))² + \(\frac{225}{16}\).
So, the keys will reach a height of \(\frac{225}{16}\) ft., or about 14 ft., before beginning the descent.
By comparison, Tenant 1’s keys’ motion is free falling. No linear term for Tenant 1 means an initial velocity of 0 ft/sec initial velocity; this is a quadratic graph whose axis of symmetry is the y – axis.

c. In this context, what would be a sensible domain for these functions?
Answer:
Both domains would be positive. For Tenant 2, the zeros are ( – 0.375, 0) and (1.5, 0), so the domain is [0, 1.5]. For Tenant 1, the domain is [0, 1] since the keys would be on the ground at about 1 second.

Engage NY Eureka Math Algebra 1 Module 4 Lesson 23 Answer Key

Eureka Math Algebra 1 Module 4 Lesson 23 Exercise Answer Key

Mathematical Modeling Exercise 1
Use the information in the Opening to answer the following questions.
Chris stands on the edge of a building at a height of 60 ft. and throws a ball upward with an initial velocity of 68 ft/s. The ball eventually falls all the way to the ground. What is the maximum height reached by the ball? After how many seconds will the ball reach its maximum height? How long will it take the ball to reach the ground?
a. What units will we be using to solve this problem?
Answer:
Feet for height, seconds for time, and feet per second for velocity

b. What information from the contextual description do we need to use in the function equation?
Answer:
Gravity = – 32 ft/s²
Initial Velocity (v₀) = 68 ft/s
Initial Height (h₀) = 60 ft.
So, the function is h(t) = – 16t² + 68t + 60.

c. What is the maximum point reached by the ball? After how many seconds will it reach that height? Show your reasoning.
Answer:
The maximum function value is at the vertex. To find this value, we first notice that this function is factorable and is not particularly friendly for completing the square. So, we will rewrite the function in factored form, f(t) = – 4(4t² – 17t – 15)  f(t) = – 4(4t + 3)(t – 5).
Second, we find the zeros, or the t – intercepts, of the function by equating the function to zero (zero height).
So, – 4(4t + 3)(t – 5) = 0.
Then, (4t + 3) = 0 or (t – 5) = 0.
t – intercepts are t = – \(\frac{3}{4}\) or t = 5.
Then, using the concept of symmetry, we find the midpoint of the segment connecting the two t – intercepts (find the average of the t – coordinates): t = \(\frac{ – \frac{3}{4} + 5}{2}\) = 2 \(\frac{1}{8}\), or 2.125.
Now, we find h(2.125) in the original function, h(t) = – 16t² + 68t + 60, h(2.125) = 132.25.
Therefore, the ball reached its maximum height of 132.25 ft. after 2.125 sec.
(Note that students may try to complete the square for this function. The calculations are very messy, but the results will be the same:
h(t) = – 16(t – \(\frac{17}{8}\))² + \(\frac{529}{4}\) = – 16(t – 2.125)² + 132.25. Students may also opt to use the vertex formula.)

d. How long will it take the ball to land on the ground after being thrown? Show your work.
Answer:
We factored in the previous question and found the zeros of this function to be t = – \(\frac{3}{4}\) and t = 5. The ball begins its flight at 0 sec. and ends at 5 sec. Therefore, it will be in flight for 5 sec.

e. Graph the function of the height, h, of the ball in feet to the time, t, in seconds. Include and label key features of the graph such as the vertex, axis of symmetry, and t – and y – intercepts.
Answer:
The graph should include identification of the intercepts, vertex, and axis of symmetry.
Vertex: (2.125,132.25)
t – intercepts: ( – 0.75,0) and (5,0)
y – intercept: (0,60) (See the graph of y = h(t).)

Mathematical Modeling Exercise 2
Read the following information about Business Applications:
Many business contexts can be modeled with quadratic functions. This is because the expressions representing price (price per item), the cost (cost per item), and the quantity (number of items sold) are typically linear. The product of any two of those linear expressions will produce a quadratic expression that can be used as a model for the business context. The variables used in business applications are not as traditionally accepted as variables that are used in physics applications, but there are some obvious reasons to use c for cost, p for price, and q for quantity (all lowercase letters). For total production cost we often use C for the variable, R for total revenue, and P for total profit (all uppercase letters). You have seen these formulas in previous lessons, but we will review them here since we use them in the next two lessons.

Business Application Vocabulary
UNIT PRICE (PRICE PER UNIT): The price per item a business sets to sell its product, sometimes represented as a linear expression.
QUANTITY: The number of items sold, sometimes represented as a linear expression.
REVENUE: The total income based on sales (but without considering the cost of doing business).
UNIT COST (COST PER UNIT) OR PRODUCTION COST: The cost of producing one item, sometimes represented as a linear expression.
PROFIT: The amount of money a business makes on the sale of its product. Profit is determined by taking the total revenue (the quantity sold multiplied by the price per unit) and subtracting the total cost to produce the items (the quantity sold multiplied by the production cost per unit): Profit = Total Revenue – Total Production Cost.

The following business formulas will be used in this lesson:
Total Production Costs = (cost per unit)(quantity of items sold)
Total Revenue = (price per unit)(quantity of items sold)
Profit = Total Revenue – Total Production Costs

Now answer the questions related to the following business problem:
A theater decided to sell special event tickets at $10 per ticket to benefit a local charity. The theater can seat up to 1,000 people, and the manager of the theater expects to be able to sell all 1,000 seats for the event. To maximize the revenue for this event, a research company volunteered to do a survey to find out whether the price of the ticket could be increased without losing revenue. The results showed that for each $1 increase in ticket price, 20 fewer tickets will be sold.
a. Let x represent the number of $1.00 price – per – ticket increases. Write an expression to represent the expected price for each ticket.
Answer:
Let x represent the number of $1 increases. If each ticket is $10, plus a possible price increase in $1 increments, the price per ticket will be 10 + 1x.

b. Use the survey results to write an expression representing the possible number of tickets sold.
Answer:
Since 20 fewer seats will be sold for each $1 increase in the ticket price, 20x represents the number of seats fewer than 1,000 that will be sold.
1000 – 20x represents the expected number of tickets sold at this higher price.

c. Using x as the number of $1 – ticket price increases and the expression representing price per ticket, write the function, R, to represent the total revenue in terms of the number of $1 – ticket price increases.
Answer:
Total Revenue = (price per ticket)(number of tickets)
R(x) = (10 + x)(1000 – 20x) = 10 000 + 1000x – 200x – 20x² = – 20x² + 800x + 10 000

d. How many $1 – ticket price increases will produce the maximum revenue? (In other words, what value for x produces the maximum R value?)
Answer:
We need to find the vertex of the revenue equation. The equation is originally in factored form, so we can just go back to that form, or we can complete the square (which seems to be pretty efficient).
1. By completing the square: R(x) = – 20(x² – 40x + ) + 10 000
– 20(x² – 40x + 400) + 10 000 + 8000
– 20(x – 20)² + 18 000, so the vertex is (20,18 000).

2. Using the factors, set the equation equal to zero to find the zeros of the function:
R(x) = (1000 – 20x)(10 + x) = 0
x = – 10 and 50 are the zeros, and the vertex will be on the axis of symmetry (at the midpoint between – 10 and 50), which is x = 20.
Finally, we reach the conclusion that after 20 price increases, the theater will maximize its revenue.

e. What is the price of the ticket that will provide the maximum revenue?
Answer:
Price per ticket expression is 10 + x. For x = 20, we have 10 + 20 = 30. The price per ticket that will provide the maximum revenue is $30.

f. What is the maximum revenue?
Answer:
According to the R – value at the vertex, the maximum revenue will be $18,000.

g. How many tickets will the theater sell to reach the maximum revenue?
Answer:
With the maximum revenue of $18,000 at $30/ticket, the theater is selling 600 tickets.

h. How much more will the theater make for the charity by using the results of the survey to price the tickets than they would had they sold the tickets for their original $10 price?
Answer:
At $10 per ticket, the theater would have brought in $10,000 after selling all 1,000 seats. The theater will make an additional $8,000 by using the survey results to price their tickets.

Exercise 1.
Two rock climbers try an experiment while scaling a steep rock face. They each carry rocks of similar size and shape up a rock face. One climbs to a point 400 ft. above the ground, and the other climbs to a place below her at 300 ft. above the ground. The higher climber drops her rock, and 1 second later the lower climber drops his. Note that the climbers are not vertically positioned. No climber is injured in this experiment.
a. Define the variables in this situation, and write the two functions that can be used to model the relationship between the heights, h₁ and h₂, of the rocks, in feet, after t seconds.
Answer:
h₁ represents the height of the rock dropped by the higher climber,
h₂ represents the height of the rock dropped by the lower climber,
t represents the number of seconds passed since the higher climber dropped her rock,
t – 1 represents the number of seconds since the lower climber dropped his rock.
h₁ (t) = – 16t² + 400 and h₂ (t) = – 16(t – 1)² + 300

b. Assuming the rocks fall to the ground without hitting anything on the way, which of the two rocks will reach the ground last? Show your work, and explain how you know your answer is correct.
Answer:
We are looking for the zeros in this case. Setting each function equal to zero we get:
h₁ (t) = – 16t² + 400 = 0
– 16(t² – 25) = 0
– 16(t + 5)(t – 5) = 0
t = – 5 or 5
For this context, it will take 5 seconds for the higher climber’s rock to hit the ground.
h₂ (t) = – 16(t – 1)² + 300 = 0
The standard form for this equation is h₂ (t) = – 16t² + 32t + 284, which is not factorable. We will solve from the completed – square form.
– 16(t – 1)² + 300 = 0
– 16(t – 1)² = – 300
(t – 1)² = \(\frac{300}{16}\) [square root both sides]
t – 1 = ±\(\frac{\sqrt{300}}{4}\)
t = 1±\(\frac{\sqrt{300}}{4}\) ≈ – 3.3 or 5.3
For this context, it will take approximately 5.3 seconds for the lower climber’s rock to hit the ground.

The rock dropped from the higher position will hit the ground approximately 0.3 seconds before the rock dropped from the lower position. (Notice that the first function equation is easy to factor, but the other is not. Students may try to factor but may use the completed – square form to solve or may opt to use the quadratic formula on the second one.)

c. Graph the two functions on the same coordinate plane, and identify the key features that show that your answer to part (b) is correct. Explain how the graphs show that the two rocks hit the ground at different times.
Answer:
The two times are indicated on the x – axis as the x – intercepts. The red graph shows the height – to – time relationship for the rock dropped from 300 ft., and the blue graph shows the same for the rock dropped from 400 ft. For the red graph, h(t) = 0 when t = 5.3, and for the blue graph, h(t) = 0 when t = 5.

d. Does the graph show how far apart the rocks were when they landed? Explain.
Answer:
No, the graph only shows the height of the rocks with respect to time. Horizontal position and movement are not indicated in the function or the graph.

Exercise 2.
Amazing Photography Studio takes school pictures and charges $20 for each class picture. The company sells an average of 12 class pictures in each classroom. They would like to have a special sale that will help them sell more pictures and actually increase their revenue. They hired a business analyst to determine how to do that. The analyst determined that for every reduction of $2 in the cost of the class picture, there would be an additional 5 pictures sold per classroom.
a. Write a function to represent the revenue for each classroom for the special sale.
Answer:
Let x represent the number of $2 reductions in price.
Then the price expression would be 20 – 2x.
The quantity expression would be 12 + 5x.
So, the revenue is R(x) = (20 – 2x)(12 + 5x) = 240 + 100x – 24x – 10x² = – 10x² + 76x + 240.

b. What should the special sale price be?
Answer:
Find the vertex for R(x) = – 10x² + 76x + 240.
– 10(x² – 7.6x + ____) + 240
– 10(x² – 7.6x + 3.8²) + 240 + 3.8² (10) = – 10(x – 3.8)² + 240 + 144.40
So, R(x) = – 10(x – 3.8)² + 384.4, and the studio should reduce the price by between three or four
$2 – increments.
If we check the revenue amount for 3 reductions, the price is $20 – $2(3) = $14. The quantity will be
12 + 5(3) = 27 pictures per classroom, so the revenue would be $378.
Now check 4 reductions: The price is $20 – $2(4) = $12. The quantity will be 12 + 5(4) = 32 pictures per classroom, and the revenue for 4 reductions would be $384.
The special sale price should be $12 since the revenue was greater than when the price was $14.

c. How much more will the studio make than they would have without the sale?
Answer:
The revenue for each class will be $384 during the sale. Without the sale, they would make $20 per picture for 12 pictures, or $240. They will increase their revenue by $144 per classroom.

Eureka Math Algebra 1 Module 4 Lesson 23 Problem Set Answer Key

Question 1.
Dave throws a ball upward with an initial velocity of 32 ft/s. The ball initially leaves his hand 5 ft. above the ground and eventually falls back to the ground. In parts (a)–(d), you will answer the following questions: What is the maximum height reached by the ball? After how many seconds will the ball reach its maximum height? How long will it take the ball to reach the ground?
a. What units will we be using to solve this problem?
Answer:
Height is measured in feet, time is measured in seconds, and velocity is measured in feet per second.

b. What information from the contextual description do we need to use to write the formula for the function h of the height of the ball versus time? Write the formula for height of the ball in feet, h(t), where t stands for seconds.
Answer:
Gravity: – 32 ft/s²
Initial height (h₀): 5 ft. above the ground
Initial velocity (v₀): 32 ft/s
Function: h(t) = – 16t² + 32t + 5

c. What is the maximum point reached by the ball? After how many seconds will it reach that height? Show your reasoning.
Answer:
Complete the square:
h(t) = – 16t² + 32t + 5
h(t) = – 16(t² – 2t + ▭) + 5 + ▭
h(t) = – 16(t² – 2t + 1) + 5 + 16 Completing the square
h(t) = – 16(t – 1)² + 21 The vertex (maximum height) is 21 ft. and is reached at 1 sec.

d. How long will it take for the ball to land on the ground after being thrown? Show your work.
Answer:
The ball will land at a time t when h(t) = 0, that is, when 0 = – 16t² + 32t + 5:
0 = – 16(t² – 2t – \(\frac{5}{16}\))
t = \(\frac{ – b \pm \sqrt{b^{2} – 4 a c}}{2 a}\)
t = \(\frac{ – ( – 2) \pm \sqrt{( – 2)^{2} – 4(1)\left( – \frac{5}{16}\right)}}{2(1)}\)
t = \(\frac{2 \pm \sqrt{4 + \frac{20}{16}}}{2}\)
t = \(\frac{2 \pm \sqrt{\frac{21}{4}}}{2}\)
t = 1±\(\frac{\sqrt{21}}{4}\)
t ≈ 2.146 and t ≈ – 0.146

The negative value does not make sense in the context of the problem, so the ball reaches the ground in approximately 2.1 sec.

e. Graph the function of the height of the ball in feet to the time in seconds. Include and label key features of the graph such as the vertex, axis of symmetry, and t – and y – intercepts.
Answer:

Question 2.
Katrina developed an app that she sells for $5 per download. She has free space on a website that will let her sell 500 downloads. According to some research she did, for each $1 increase in download price, 10 fewer apps are sold. Determine the price that will maximize her profit.
Answer:
Profit equals total revenue minus total production costs. Since the website that Katrina is using allows up to 500 downloads for free, there is no production cost involved, so the total revenue is the total profit.
Let x represent the number of $1 increases to the cost of a download.
Cost per download: 5 + 1x
Apps sold: 500 – 10x
Revenue = (unit price)(quantity sold)
R(x) = (5 + 1x)(500 – 10x)
0 = (5 + 1x)(500 – 10x)
0 = 5 + x or 0 = 500 – 10x
x = – 5 or x = 50
The average of the zeros represents the axis of symmetry.
\(\frac{ – 5 + 50}{2}\) = 22.5 Katrina should raise the cost by $22.50 to earn the greatest revenue.
R(x) = (5 + x)(500 – 10x)
R(x) = – 10x² + 450x + 2500 Written in standard form
R(22.5) = – 10(22.5)² + 450(22.5) + 2500
R(22.5) = 7562.50
Katrina will maximize her profit if she increases the price per download by $22.50 to $27.50 per download. Her total revenue (and profit) would be $7,562.50.

Question 3.
Edward is drawing rectangles such that the sum of the length and width is always six inches.
a. Draw one of Edward’s rectangles, and label the length and width.
Answer:

b. Fill in the following table with four different possible lengths and widths.

Answer:

c. Let x be the width. Write an expression to represent the length of one of Edward’s rectangles.
Answer:
If x represents the width, then the length of the rectangle would be 6 – x.

d. Write an equation that gives the area, y, in terms of the width, x.
Answer:
Area = length × width
y = x(6 – x)

e. For what width and length will the rectangle have maximum area?
Answer:
y = x(6 – x)
y = – x² + 6x
y = – 1(x² – 6x + ▭) + ▭
y = – 1(x² – 6x + 9) + 9 By completing the square
y = – 1(x – 3)² + 9 The vertex is (3,9).
The rectangle with the maximum area has a width of 3 in. (also length 3 in.) and an area of 9 in².

f. Are you surprised by the answer to part (e)? What special name is given for the rectangle in your answer to part (e)?
Answer:
Responses will vary. The special rectangle in part (e) is a square.

Eureka Math Algebra 1 Module 4 Lesson 23 Exit Ticket Answer Key

Question 1.
What is the relevance of the vertex in physics and business applications?
Answer:
By finding the vertex, we know the highest or lowest value for the function and also the x – value that gives that minimum or maximum. In physics, that could mean the highest point for an object in motion. In business, that could mean the minimum cost or the maximum profit or revenue.

Engage NY Eureka Math Algebra 1 Module 4 Lesson 20 Answer Key

Eureka Math Algebra 1 Module 4 Lesson 20 Example Answer Key

Example

Answer:
→ Is it possible to shrink or stretch the graph of a function? If so, how might that happen?
→ Yes, since we discovered that adding or subtracting a value to the parts of a parent function shifts its graph horizontally or vertically, it is possible that multiplying or dividing will shrink or stretch a function. Note that students may respond with comments about the points of the graph being pushed together or spread apart.

Eureka Math Algebra 1 Module 4 Lesson 20 Exploratory Challenge Answer Key

Exploratory Challenge
Complete the following to review Module 3 concepts:
a. Consider the function f(x) = |x|. Complete the table of values for f(x). Then, graph the equation y = f(x) on the coordinate plane provided for part (b).

Answer:

b. Complete the following table of values for each transformation of the function f. Then, graph the equations y = g(x), y = h(x), y = j(x), and y = k(x) on the same coordinate plane as the graph of y = f(x). Label each graph.

Answer:

c. Describe how the graph of y = kf(x) relates to the graph of y = f(x) for each case.
i. k > 1
Answer:
The graph is stretched vertically by a factor equal to k.

ii. 0 < k < 1
Answer:
The graph is shrunk vertically by a factor equal to k.

iii. k = – 1
ans;
The graph is reflected across the x – axis.

iv. – 1 < k < 0
Answer:
The graph is reflected across the x – axis and shrunk vertically by a factor equal to |k|.

v. k < – 1
Answer:
The graph is reflected across the x – axis and stretched vertically by a factor equal to |k|.

d. Describe the transformation of the graph of f that results in the graphs of g, h, and k given the following formulas for each function. Then, graph each function and label each graph.
f(x) = x³
g(x) = 2x³
h(x) = 0.5x³
k(x) = – 3x³

Answer:

The graph of g shows a vertically stretched graph of f with a scale factor of 2. The graph of h is a vertically shrunk, or compressed, graph of f with a scale factor of 0.5. The graph of k shows a vertically stretched graph of f with a scale factor of 3 and is reflected across the x – axis.

e. Consider the function f(x) = \(\sqrt [ 3 ]{ x }\). Complete the table of values; then graph the equation y = f(x).

Answer:

f. Complete the following table of values, rounding each value to the nearest hundredth. Graph the equations y = g(x), y = h(x), and y = j(x) on the same coordinate plane as your graph of y = f(x) above. Label each graph.

Answer:

Describe the transformations of the graph of f that result in the graphs of g, h, and j.
When the x – values of f are multiplied by 2, the graph is shrunk horizontally by a factor of 0.5. When the
x – values of f are multiplied by 0.5, the graph is stretched horizontally by a factor of 2. When the x – values of f are multiplied by – 2, the graph is shrunk horizontally by a factor of 0.5 and is reflected about the
y – axis.

f. Describe how the graph of y = f(\(\frac{1}{k}\) x) relates to the graph of y = f(x) for each case.
i. k > 1
Answer:
The graph is stretched horizontally by a factor equal to k.

ii. 0 < k < 1
Answer:
The graph is shrunk horizontally by a factor equal to k.

iii. k = – 1
Answer:
The graph is reflected across the y – axis.

iv. – 1 < k < 0
Answer:
The graph is shrunk horizontally by a factor equal to |k| and is reflected across the y – axis.

v. k < – 1
Answer:
The graph is stretched horizontally by a factor equal to |k| and is reflected across the y – axis.

Eureka Math Algebra 1 Module 4 Lesson 20 Exercise Answer Key

Opening Exercise
The graph of a quadratic function defined by f(x) = x² has been translated 5 units to the left and 3 units up. What is the formula for the function, g, depicted by the translated graph?
Answer:
g(x) = (x + 5)² + 3

Sketch the graph of the equation y = g(x).

Answer:

Exercise 1.
For each set of graphs below, answer the following questions:
What are the parent functions?
How does the transformed graph relate to the graph of the parent function?
Write the formula for the function depicted by the transformed graph.
a.

Answer:
The parent function (in red) is f(x) = \(\sqrt{x}\). The graph in blue is a vertical scaling of the graph of f with a scale factor of 3. The function depicted by the blue graph is g(x) = 3\(\sqrt{x}\). The other graph (in pink) is a vertical scaling of the graph of f with a scale factor of 0.5. The function depicted by the graph is h(x) = 0.5\(\sqrt{x}\).

b.

Answer:
The parent function (in red) is f(x) = x². The graph in blue is a horizontal scaling of the graph of f with a scale factor of 4. The function depicted by the blue graph is g(x) = (\(\frac{1}{4}\) x)². The other graph (in pink) is a vertical scaling of the graph of f with a scale factor of 2 and is reflected over the x – axis. The function depicted by the graph is h(x) = – 2x².

Exercise 2.
Graph each set of functions in the same coordinate plane. Do not use a graphing calculator.
a. f(x) = |x|
g(x) = 4|x|
h(x) = |2x|
k(x) = – 2|2x|

Answer:

y = f(x) in red
y = g(x) in purple
y = h(x) in pink
y = k(x) in green

b. g(x) = \(\sqrt [ 3 ]{ x }\)
p(x) = 2\(\sqrt [ 3 ]{ x }\)
q(x) = – 2\(\sqrt [ 3 ]{ 2x }\)

Answer:

y = g(x) in red
y = p(x) in purple
y = q(x) in pink

Eureka Math Algebra 1 Module 4 Lesson 20 Problem Set Answer Key

Question 1.
Graph the functions in the same coordinate plane. Do not use a graphing calculator.
f(x) = |x|
g(x) = 2|x|
h(x) = |3x|
k(x) = – 3|3x|
Answer:

Question 2.
Explain how the graphs of functions j(x) = 3|x| and h(x) = |3x| are related.
Answer:
Each of these transformations of the absolute value functions creates the same graph.

Question 3.
Explain how the graphs of functions q(x) = – 3|x| and r(x) = | – 3x| are related.
Answer:
The two graphs have the same scaling factor of 3, but they are reflections of each other across the x – axis. Multiplying an absolute value by a negative number will reflect it across the x – axis. However, multiplying by a negative number INSIDE the absolute value has the same effect as multiplying by a positive number on the outside.

Question 4.
Write a function, g, in terms of another function, f, such that the graph of g is a vertical shrink of the graph f by a factor of 0.75.
Answer:
g(x) = 0.75 f(x)

Question 5.
A teacher wants the students to write a function based on the parent function f(x) = \(\sqrt [ 3 ]{ x }\). The graph of f is stretched vertically by a factor of 4 and shrunk horizontally by a factor of \(\frac{1}{3}\). Mike wrote g(x) = 4\(\sqrt [ 3 ]{ 3x }\) as the new function, while Lucy wrote h(x) = 3\(\sqrt [ 3 ]{ 4x }\). Which one is correct? Justify your answer.
Answer:
Mike is correct. A vertical stretch by a factor of 4 means multiplying f(x) by 4, and a horizontal shrink by a factor of \(\frac{1}{3}\) means that the x – values of f(x) must be multiplied by 3.

Question 6.
Study the graphs of two different functions below. Which is a parent function? What is the constant value(s) multiplied to the parent function to arrive at the transformed graph? Now write the function defined by the transformed graph.

Answer:
The parent function is f(x) = \(\sqrt{x}\). The graph of y = g(x) is the graph of y = f(x) reflected across the x – axis. The function depicted by the transformed graph is g(x) = – \(\sqrt{x}\).

Eureka Math Algebra 1 Module 4 Lesson 20 Exit Ticket Answer Key

Question 1.
How would the graph of f(x) = \(\sqrt{x}\) be affected if it were changed to g(x) = – 2\(\sqrt{x}\)?
Answer:
The graph of f would be stretched vertically by a factor of 2 and reflected across the x – axis.

Question 2.
Sketch and label the graphs of both f and g on the grid below.

Answer:

Engage NY Eureka Math Algebra 1 Module 4 Lesson 21 Answer Key

Eureka Math Algebra 1 Module 4 Lesson 21 Example Answer Key

Example 1: Quadratic Expression Representing a Function
a. A quadratic function is defined by g(x) = 2x² + 12x + 1. Write this in the completed – square (vertex) form and show all the steps.
Answer:
g(x) = 2x² + 12x + 1
Step 1 = (2x² + 12x) + 1 Gather variable terms.
Step 2 = 2(x² + 6x) + 1 Factor out the GCF.
Step 3 = 2(x² + 6x + 9) + 1 – 18 Complete the square and balance the equality.
Step 4 = 2(x + 3)² – 17 Factor the perfect square.
g(x) = 2(x + 3)² – 17

b. Where is the vertex of the graph of this function located?
Answer:
The vertex is at ( – 3, – 17).

c. Look at the completed – square form of the function. Can you name the parent function? How do you know?
Answer:
The parent function is f(x) = x². The function is quadratic.

d. What transformations have been applied to the parent function to arrive at function g? Be specific.
Answer:
The parent function f(x) = x² is translated 3 units to the left, stretched vertically by a factor of 2, and translated 17 units down.

e. How does the completed – square form relate to the quadratic parent function f(x) = x²?
Answer:
The completed – square form can be understood through a series of transformations of the quadratic parent function, f.

Example 2.
The graph of a quadratic function f(x) = x² has been translated 3 units to the right, vertically stretched by a factor of 4, and moved 2 units up. Write the formula for the function that defines the transformed graph.
Answer:
g(x) = 4(x – 3)² + 2

Eureka Math Algebra 1 Module 4 Lesson 21 Exercise Answer Key

Exercises
Exercise 1.
Without using a graphing calculator, sketch the graph of the following quadratic functions on the same coordinate plane, using transformations of the graph of the parent function f(x) = x².
a. g(x) = – 2(x – 3)² + 4
b. h(x) = – 3(x + 5)² + 1
c. k(x) = 2(x + 4)² – 3
d. p(x) = x² – 2x
e. t(x) = x² – 2x + 3

Answer:
Note: By completing the square, we have p(x) = (x – 1)² – 1 and t(x) = (x – 1)² + 2.

Exercise 2.
Write a formula for the function that defines the described transformations of the graph of the quadratic parent function f(x) = x².
a. 3 units shift to the right
b. Vertical shrink by a factor of 0.5
c. Reflection across the x – axis
d. 4 units shift up
Then, graph both the parent and the transformed functions on the same coordinate plane.

Answer:
g(x) = – 0.5(x – 3)² + 4

Exercise 3.
Describe the transformation of the quadratic parent function f(x) = x² that results in the quadratic function g(x) = 2x² + 4x + 1.
Answer:
First, rewrite g(x) into the completed – square form.
g(x) = 2x² + 4x + 1
= (2x² + 4x) + 1
= 2(x² + 2x) + 1
= 2(x² + 2x + 1) + 1 – 2
= 2(x² + 2x + 1) – 1
g(x) = 2(x + 1)² – 1
This means that the graph of f is translated 1 unit to the left, vertically stretched by a factor of 2, and translated 1 unit down.

Exercise 4.
Sketch the graphs of the following functions based on the graph of the function f(x) = x². If necessary, rewrite some of the functions in the vertex (completed – square) form. Label your graphs.
a. g(x) = – (x – 4)² + 3
b. h(x) = 3(x – 2)² – 1
c. k(x) = 2x² + 8x
d. p(x) = x² + 6x + 5

Answer:
c. k(x) = 2(x + 2)² – 8
d. p(x) = (x + 3)² – 4

Eureka Math Algebra 1 Module 4 Lesson 21 Problem Set Answer Key

Question 1.
Write the function g(x) = – 2x² – 20x – 53 in completed – square form. Describe the transformations of the graph of the parent function f(x) = x² that result in the graph of g.
Answer:
g(x) = – 2x² – 20x – 53
= ( – 2x² – 20x) – 53
= – 2(x² + 10x) – 53
= – 2(x² + 10x + 25) – 53 + 50
= – 2(x² + 10x + 25) – 3
g(x) = – 2(x + 5)² – 3
The graph of f is translated 5 units to the left, vertically stretched by a factor of 2, and translated 3 units down. The graph of f is facing up, while the graph of g is facing down because of the negative value of a.

Question 2.
Write the formula for the function whose graph is the graph of f(x) = x² translated 6.25 units to the right, vertically stretched by a factor of 8, and translated 2.5 units up.
Answer:
g(x) = 8(x – 6.25)² + 2.5

Question 3.
Without using a graphing calculator, sketch the graphs of the functions below based on transformations of the graph of the parent function f(x) = x². Use your own graph paper, and label your graphs.
a. g(x) = (x + 2)² – 4
b. h(x) = – (x – 4)² + 2
c. k(x) = 2x² – 12x + 19
d. p(x) = – 2x² – 4x – 5
e. q(x) = 3x² + 6x
Answer:
c. k(x) = 2(x – 3)² + 1
d. p(x) = – 2(x + 1)² – 3
e. q(x) = 3(x + 1)² – 3

Eureka Math Algebra 1 Module 4 Lesson 21 Exit Ticket Answer Key

Question 1.
Describe in words the transformations of the graph of the parent function f(x) = x² that would result in the graph of g(x) = (x + 4)² – 5. Graph the equation y = g(x).

Answer:
The graph of g is a translation of the graph of f, 4 units to the left and 5 units down.

Engage NY Eureka Math Algebra 1 Module 4 Lesson 19 Answer Key

Eureka Math Algebra 1 Module 4 Lesson 19 Example Answer Key

Example
For each graph, answer the following:

• What is the parent function?
• How does the translated graph relate to the graph of the parent function?
• Write the formula for the function depicted by the translated graph.

a.

Answer:
y = f(x)
y = g(x)
The parent function is f(x) = x². The graph is shifted 4 units to the right. The function defined by the translated graph is g(x) = (x – 4)².

b.

Answer:
y = f(x)
y = g(x)
The parent function is f(x) = \(\sqrt{x}\). The constant value added to f(x) is 5 because the graph is shifted 5 units up. The function defined by the translated graph is g(x) = \(\sqrt{x}\) + 5.

c.

Answer:
y = f(x)
y = g(x)
The parent function is f(x) = |x|. The constant values added to f(x) are – 3 and + 2 because the graph is shifted 3 units down and 2 units to the left. The function defined by the translated graph is g(x) = |x + 2| – 3.

Eureka Math Algebra 1 Module 4 Lesson 19 Exercise Answer Key

Opening Exercise
Graph each set of three functions in the same coordinate plane (on your graphing calculator or a piece of graph paper). Then, explain what similarities and differences you see among the graphs.
a. f(x) = x
g(x) = x + 5
h(x) = x – 6

b. f(x) = x²
g(x) = x² + 3
h(x) = x² – 7

c. f(x) = |x|
g(x) = |x + 3|
h(x) = |x – 4|
Answer:
Part (a)—The graphs are parallel lines, but they have different x – and y – intercepts.

Part (b)—The graphs look the same (because they are congruent), but they have different vertices, which in this case means different minimum values. They are related by vertical translations.

Part (c)—The overall shapes of the graphs look the same (because they are congruent), but they have different vertices. They are related by horizontal translations.

Exercises
Exercise 1.
For each of the following graphs, use the formula for the parent function f to write the formula of the translated function.
a.

Answer:
Parent Function: f(x) = |x|
Translated Functions: g(x) = |x| + 2.5,
h(x) = |x| – 4

b.

Answer:
Parent Function: f(x) = \(\sqrt [ 3 ]{ x }\)
Translated Functions: g(x) = \(\sqrt [ 3 ]{ x }\) + 1,
h(x) = \(\sqrt [ 3 ]{ x + 5 }\)

Exercise 2.
Below is a graph of a piecewise function f whose domain is – 5 ≤ x ≤ 3. Sketch the graphs of the given functions on the same coordinate plane. Label your graphs correctly.
g(x) = f(x) + 3 h(x) = f(x – 4)

Answer:

Exercise 3.
Match the correct equation and description of the function with the given graphs.


Answer:

Eureka Math Algebra 1 Module 4 Lesson 19 Problem Set Answer Key

Question 1.
Graph the functions in the same coordinate plane. Do not use a graphing calculator.
f(x) = \(\sqrt{x}\)
p(x) = 10 + \(\sqrt{x}\)
q(x) = \(\sqrt{x + 8}\)

Answer:

Question 2.
Write a function that translates the graph of the parent function f(x) = x² down 7.5 units and right 2.5 units.
Answer:
f(x) = (x – 2.5)² – 7.5

Question 3.
How would the graph of f(x) = |x| be affected if the function were transformed to f(x) = |x + 6| + 10?
Answer:
The graph would be shifted 10 units up and 6 units to the left.

Question 4.
Below is a graph of a piecewise function f whose domain is the interval – 4≤x≤2. Sketch the graph of the given functions below. Label your graphs correctly.
g(x) = f(x) – 1 h(x) = g(x – 2) [Be careful; this one might be a challenge.]

Answer:
Point out that the graph of h is related to g rather than f. Make sure students recognize that they must find the graph of g first, and then translate it to find h.

Question 5.
Study the graphs below. Identify the parent function and the transformations of that function depicted by the second graph. Then, write the formula for the transformed function.

Answer:
y = g(x)
y = f(x)
The parent function is f(x) = x², in red. The graph of the transformed function, in black, is the graph of y = f(x) shifted 3 units to the right and 5 units up. The function defined by the translated graph is g(x) = (x – 3)² + 5.

Eureka Math Algebra 1 Module 4 Lesson 19 Exit Ticket Answer Key

Question 1.
Ana sketched the graphs of f(x) = x² and g(x) = x² – 6 as shown below. Did she graph both of the functions correctly? Explain how you know.

Answer:
The function f was graphed correctly, but not g. The graph of g should have been translated 6 units below the graph of f.

Question 2.
Use transformations of the graph of f(x) = \(\sqrt{x}\) to sketch the graph of f(x) = \(\sqrt{x – 1}\) + 3.

Answer:
The graph should depict the graph of the square root function translated 1 unit right and 3 units up.

Engage NY Eureka Math Algebra 1 Module 4 Lesson 15 Answer Key

Eureka Math Algebra 1 Module 4 Lesson 15 Exercise Answer Key

Opening Exercise
Solve the following:
a. 4x² + 5x + 3 = 2x² – 3x
Answer:
Students should recognize that this is a difficult quadratic equation to solve. Accordingly, they should set it equal to zero and solve it using the quadratic formula:
2x² + 8x + 3 = 0

Check by substituting the approximated decimal value(s) into the original equation or by completing the square:
2(x² + 4x) = – 3 → 2(x² + 4x + 4) = – 3 + 8 → 2(x + 2)² = 5
(x + 2)² = \(\frac{5}{2}\) → x + 2 = ±\(\sqrt{\frac{5}{2}}\) → x = – 2 ± \(\sqrt{\frac{5}{9}}\)
(Have students use their calculators to check that these are the same decimal values as the previous solutions.)

b. c² – 14 = 5c
Answer:
Initially, students may approach this problem by using the quadratic formula. While this approach works, encourage students to look for a more efficient pathway to the solution (in this case, to solve by factoring):
c² – 5c – 14 = 0 → (c – 7)(c + 2) = 0 → c = 7 or – 2
Checks:
7² – 14 = 5(7) → 49 – 14 = 35 → 35 = 35
( – 2)² – 14 = 5( – 2) → 4 – 14 = – 10 → – 10 = – 10

Exercises
Solve Exercises 1–5 using the quadratic formula.
Exercise 1.
x² – 2x + 1 = 0
Answer:
a = 1, b = – 2, c = 1
x = \(\frac{ – ( – 2) \pm \sqrt{( – 2)^{2} – 4(1)(1)}}{2(1)}\) = \(\frac{2 \pm \sqrt{0}}{2}\) = 1

Exercise 2.
3b² + 4b + 8 = 0
Answer:
a = 3, b = 4, c = 8
b = \(\frac{ – 4 \pm \sqrt{4^{2} – 4(3)(8)}}{2(3)}\) = \(\frac{ – 4 \pm \sqrt{ – 80}}{6}\) = ???

Exercise 3.
2t² + 7t – 4 = 0
Answer:
a = 2, b = 7, c = – 4

Exercise 4.
q² – 2q – 1 = 0
Answer:
a = 1, b = – 2, c = – 1
q = \(\frac{ – ( – 2) \pm \sqrt{( – 2)^{2} – 4(1)( – 1)}}{2(1)}\) = \(\frac{2 \pm \sqrt{8}}{2}\) = \(\frac{2 \pm 2 \sqrt{2}}{2}\) = 1 ± \(\sqrt{2}\)

Exercise 5.
m² – 4 = 3
Answer:
a = 1, b = 0, c = – 7
m = \(\frac{0 \pm \sqrt{0^{2} – 4( – 7)}}{2(1)}\) = \(\frac{0 \pm \sqrt{28}}{2}\) = \(\frac{0 \pm 2 \sqrt{7}}{2}\) = 0 ± \(\sqrt{7}\) = ±\(\sqrt{7}\)

For Exercises 6–9, determine the number of real solutions for each quadratic equation without solving.
Exercise 6.
p² + 7p + 33 = 8 – 3p
Answer:
a = 1, b = 10, c = 25 → 10² – 4(1)(25) = 0 → one real solution

Exercise 7.
7x² + 2x + 5 = 0
Answer:
a = 7, b = 2, c = 5 → 2² – 4(7)(5) = – 136 → no real solutions

Exercise 8.
2y² + 10y = y² + 4y – 3
Answer:
a = 1, b = 6, c = 3 → 6² – 4(1)(3) = 24 → two real solutions

Exercise 9.
4z² + 9 = – 4z
Answer:
a = 4, b = 4, c = 9 → 4² – 4(4)(9) = – 128 → no real solutions

Exercise 10.
On the line below each graph, state whether the discriminant of each quadratic equation is positive, negative, or equal to zero. Then, identify which graph matches the discriminants below.

Answer:

Exercise 11.
Consider the quadratic function f(x) = x² – 2x – 4.
a. Use the quadratic formula to find the x – intercepts of the graph of the function.
Answer:
x = 1 + \(\sqrt{5}\) and x = 1 – \(\sqrt{5}\)

b. Use the x – intercepts to write the quadratic function in factored form.
Answer:
f(x) = (x – (1 + \(\sqrt{5}\)))(x – (1 – \(\sqrt{5}\)))

c. Show that the function from part (b) written in factored form is equivalent to the original function.
Answer:
f(x) = x² – (1 + \(\sqrt{5}\))x – (1 – \(\sqrt{5}\))x + (1 + \(\sqrt{5}\))(1 – \(\sqrt{5}\))
f(x) = x² – x – \(\sqrt{5}\) x – x + \(\sqrt{5}\) x + 1 – 5
f(x) = x² – 2x – 4

Extension: Consider the quadratic equation ax² + bx + c = 0.
a. Write the equation in factored form, a(x – m)(x – n) = 0, where m and n are the solutions to the equation.
Answer:

Eureka Math Algebra 1 Module 4 Lesson 15 Problem Set Answer Key

Without solving, determine the number of real solutions for each quadratic equation.
Question 1.
b² – 4b + 3 = 0
Answer:
a = 1, b = – 4, c = 3 → ( – 4)² – 4(1)(3) = 4 → two real solutions

Question 2.
2n² + 7 = – 4n + 5
Answer:
a = 2, b = 4, c = 2 → (4)² – 4(2)(2) = 0 → one real solution

Question 3.
x – 3x² = 5 + 2x – x²
Answer:
a = – 2, b = – 1, c = – 5 → ( – 1)² – 4( – 2)( – 5) = – 39 → no real solutions

Question 4.
4q + 7 = q² – 5q + 1
Answer:
a = – 1, b = 9, c = 6 → (9)² – 4( – 1)(6) = 105 → two real solutions

Based on the graph of each quadratic function, y = f(x), determine the number of real solutions for each corresponding quadratic equation, f(x) = 0.
Question 5.

Answer:
One real solution

Question 6.

Answer:
No real solutions

Question 7.

Answer:
Two real solutions

Question 8.

Answer:
One real solution

Question 9.
Consider the quadratic function f(x) = x² – 7.
a. Find the x – intercepts of the graph of the function.
Answer:
x = ±\(\sqrt{7}\)

b. Use the x – intercepts to write the quadratic function in factored form.
Answer:
f(x) = (x – \(\sqrt{7}\))(x + \(\sqrt{7}\))

c. Show that the function from part (b) written in factored form is equivalent to the original function.
Answer:
f(x) = (x – \(\sqrt{7}\))(x + \(\sqrt{7}\)) = x² + \(\sqrt{7}\) x – \(\sqrt{7}\) x – 7 = x² – 7

Question 10.
Consider the quadratic function f(x) = – 2x² + x + 5.
a. Find the x – intercepts of the graph of the function.
Answer:
x = \(\frac{1 \pm \sqrt{41}}{4}\)

b. Use the x – intercepts to write the quadratic function in factored form.
Answer:
f(x) = – 2(x – \(\frac{1 + \sqrt{41}}{4}\)) (x – \(\frac{1 – \sqrt{41}}{4}\))

c. Show that the function from part (b) written in factored form is equivalent to the original function.
Answer:
f(x) = – 2(x – \(\frac{1 + \sqrt{41}}{4}\))(x – \(\frac{1 – \sqrt{41}}{4}\)) = – 2(x² – \(\frac{1 + \sqrt{41}}{4}\) x – \(\frac{1 – \sqrt{41}}{4}\)x – \(\frac{5}{2}\)) = – 2(x² – \(\frac{1}{2}\) x – \(\frac{5}{2}\))
= – 2x² + x + 5

Eureka Math Algebra 1 Module 4 Lesson 15 Exit Ticket Answer Key

Question 1.
Solve the following equation using the quadratic formula: 3x² + 6x + 8 = 6.
Answer:
3x² + 6x + 2 = 0 →

Question 2.
Is the quadratic formula the most efficient way to solve this equation? Why or why not?
Answer:
This is a personal preference. Some may consider the quadratic formula to be more efficient, while others may prefer completing the square. After the leading coefficient is factored out, the linear term coefficient is still even, making this a good candidate for completing the square.

Question 3.
How many zeros does the function f(x) = 3x² + 6x + 2 have? Explain how you know.
Answer:
Since the discriminant of the original equation is positive, 12, and yields two real solutions, the function must have two zeros. OR After solving the equation 3x² + 6x + 2 = 0, I found that there were two irrational solutions. This means that the corresponding function has two zeros.