## Engage NY Eureka Math Algebra 1 Module 3 End of Module Assessment Answer Key

### Eureka Math Algebra 1 Module 3 End of Module Assessment Task Answer Key

Question 1.

Given h(x) = |x + 2| – 3 and g(x) = – |x| + 4:

a. Describe how to obtain the graph of g from the graph of a(x) = |x| using transformations.

Answer:

To obtain the graph of g, reflect the graph of a about the x-axis, and translate this graph up 4 units.

b. Describe how to obtain the graph of h from the graph of a(x) = |x| using transformations.

Answer:

To obtain the graph of h, translate the graph of a 2 units to the left and 3 units down.

c. Sketch the graphs of h(x) and g(x) on the same coordinate plane.

Answer:

d. Use your graphs to estimate the solutions to the equation:

|x + 2| – 3 = – |x| + 4

Explain how you got your answer.

Answer:

Solution: x ≈ 2.5 or x ≈-4.5

The solutions are the x-coordinates of the intersection points of the graphs of g and h.

e. Were your estimations in part (d) correct? If yes, explain how you know. If not, explain why not.

Answer:

Let x = 2.5

Is |2.5 + 2|-3 = -|2.5| + 4 true?

Yes, 4.5 – 3 = -2.5 + 4 is true.

Let x = -4.5

Is |-4.5 + 2| – 3 = – |4.5| + 4 true?

Yes, 2.5 – 3 = -4.5 + 4 is true.

Yes, the estimates are correct. They each make the equation true.

Question 2.

Let f and g be the functions given by f(x) = x^{2} and g(x) = x|x|.

a. Find f(\(\frac{1}{3}\)), g(4), and g( – \(\sqrt{3}\)).

Answer:

f(\(\frac{1}{3}\)) = \(\frac{1}{9}\), g(4) = 16, g(-\(\sqrt{3}\)) = -3

b. What is the domain of f?

Answer:

D: all real numbers

c. What is the range of g?

Answer:

R: all real numbers

d. Evaluate f( – 67) + g( – 67).

Answer:

(-67)^{2} + -67|-67| = 0

e. Compare and contrast f and g. How are they alike? How are they different?

Answer:

When x is positive, both functions give the same value. But when x is negative, f gives the always-positive value of x^{2}, whereas g gives a value that is the opposite of what f gives.

f. Is there a value of x such that f(x) + g(x) = – 100? If so, find x. If not, explain why no such value exists.

Answer:

No. f and g are either both zero, giving a sum of zero; both positive, giving a positive sum; or the opposite of each other, giving a sum of zero. So, there is no way to get a negative sum.

g. Is there a value of such that (x) + g(x) = 50? If so, find x. If not, explain why no such value exists.

Answer:

Yes. If x = 5, f(x) = g(x) = 25; thus, f(x) + g(x) = 50.

Question 3.

A boy bought six guppies at the beginning of the month. One month later, the number of guppies in his tank had doubled. His guppy population continued to grow in this same manner. His sister bought some tetras at the same time. The table below shows the number of tetras, t, after n months have passed since they bought the fish.

a. Create a function g to model the growth of the boy’s guppy population, where g(n) is the number of guppies at the beginning of each month and n is the number of months that have passed since he bought the six guppies. What is a reasonable domain for g in this situation?

Answer:

g(n) = 6•2^{n} Domain: n is a whole number.

b. How many guppies will there be one year after he bought the six guppies?

Answer:

g(12) = 6•2^{12} = 24576 guppies

c. Create an equation to determine how many months it will take to reach 100 guppies.

Answer:

100 = 6•2^{n}

d. Use graphs or tables to approximate a solution to the equation from part (c). Explain how you arrived at your estimate.

Answer:

At the end of 4 months, there are 96 guppies which is not quite 100. So during the 5th month, the guppy population reaches 100.

e. Create a function, t, to model the growth of the sister’s tetra population, where t(n) is the number of tetras after n months have passed since she bought the tetras.

Answer:

t(n) = 8(n + 1), n is a whole number.

Or, t(n) = 8n + 8, n is a whole number.

f. Compare the growth of the sister’s tetra population to the growth of the guppy population. Include a comparison of the average rate of change for the functions that model each population’s growth over time.

Answer:

The guppies’ population is increasing faster than the tetras’ population. Each month, the number of guppies doubles, while the number of tetras increases by 8. The rate of change for the tetras is constant, but the rate of change for the guppies is always

g. Use graphs to estimate the number of months that will have passed when the population of guppies and tetras will be the same.

Answer:

g(2) = t(2) = 24

The guppier and tetras populations will be the same, 24, after 2 months.

h. Use graphs or tables to explain why the guppy population will eventually exceed the tetra population even though there were more tetras to start with.

Answer:

The guppy population’s growth is exponential, and the tetra population’s growth is linear. The graph in part (g) shows how the population of the guppies eventually overtakes the population of the tetras. The table below shows that by the end of the third month, there are more guppies than tetras.

i. Write the function g(n) in such a way that the percent increase in the number of guppies per month can be identified. Circle or underline the expression representing the percent increase in the number of guppies per month.

Answer:

g(n) = 6(200%)^{n}

Question 4.

Regard the solid dark equilateral triangle as Figure 0. Then, the first figure in this sequence is the one composed of three dark triangles, the second figure is the one composed of nine dark triangles, and so on.

a. How many dark triangles are in each figure? Make a table to show this data.

Answer:

b. Given the number of dark triangles in a figure, describe in words how to determine the number of dark triangles in the next figure.

Answer:

The number of triangles in a figure is 3 times the number of triangles in the previous figure.

c. Create a function that models this sequence. What is the domain of this function?

Answer:

T(n) = 3^{n}, D: n is a whole number.

d. Suppose the area of the solid dark triangle in Figure 0 is 1 square meter. The areas of one dark triangle in each figure form a sequence. Create an explicit formula that gives the area of just one of the dark triangles in the nth figure in the sequence.

Answer:

A(n) = (\(\frac{1}{4}\))^{n}

e. The sum of the areas of all the dark triangles in Figure 0 is 1 m^{2}; there is only one triangle in this case. The sum of the areas of all the dark triangles in Figure 1 is \(\frac{3}{4}\) m^{2}. What is the sum of the areas of all the dark triangles in the n^{th} figure in the sequence? Is this total area increasing or decreasing as n increases?

Answer:

I(n) = (\(\frac{3}{4}\))^{n}

The total area is decreasing as n increases.

f. Let P(n) be the sum of the perimeters of all the dark triangles in the n^{th} figure in the sequence of figures. There is a real number k so that

P(n + 1) = kP(n)

is true for each positive whole number n. What is the value of k?

Answer:

Let x represent the number of meters long of one side of the triangle in Figure 0.

P is a geometric sequence, and k is the ratio between any term and the previous term, so k= P(n+1)/P(n).

So, for example, for n = 0, k = \(\frac{(P(1))}{P(O)}=\frac{\frac{9}{2} x}{3 x}\) = \(\frac{3}{2}\) .

For n = 1, k = \(\frac{P(2)}{P(1)}\) = \(\frac{\frac{27}{4} x}{\frac{9}{2} x}\) = \(\frac{3}{2}\) .

K = \(\frac{3}{2}\)

Question 5.

The graph of a piecewise function f is shown to the right. The domain of f is – 3≤x≤3.

a. Create an algebraic representation for f. Assume that the graph of f is composed of straight line segments.

Answer:

b. Sketch the graph of y = 2f(x), and state the domain and range.

Answer:

Domain: -3 ≤ x ≤ 3

Range: -6 ≤ y ≤ 2

c. Sketch the graph of y = f(2x), and state the domain and range.

Answer:

Domain: -1.5 ≤ x ≤ 1.5

Range: -3 ≤ y ≤ 1

d. How does the range of y = f(x) compare to the range of y = kf(x), where k > 1?

Answer:

Every value in the range of y = f(x) would be multiplied by k. Since k > 1, we can represent this by multiplying the compound inequality that gives the range of

y = f(x) by k, giving -3k ≤ y ≤ k.

e. How does the domain of y = f(x) compare to the domain of y = f(kx), where k > 1?

Answer:

Every value in the domain of y = f(x) would be divided by k. Since k > 1, we can represent this by multiplying the compound inequality that gives the domain of

y = f(x) by \(\frac{1}{k}\), giving –\(\frac{3}{k}\) ≤ x ≤ \(\frac{3}{k}\).