## Engage NY Eureka Math Algebra 2 Module 1 Lesson 35 Answer Key

### Eureka Math Algebra 2 Module 1 Lesson 35 Example Answer Key

Example: Dilation at the Origin

Let f(x) = x^{2} and let k = 2. Write a formula for the function g that results from dilating f at the origin by a factor of \(\frac{1}{2}\).

Answer:

The new function will have equation g(x) = 2f (\(\frac{1}{2}\)x). Since f(x) = x^{2}, the new function will have equation g(x) = 2(\(\frac{1}{2}\)x)^{2}. That is, g(x) = \(\frac{1}{2}\)x^{2}.

What would the results be for k = 3, 4, or 5? What about k = \(\frac{1}{2}\)?

Answer:

For k = 3, g(x) = \(\frac{1}{3}\)x^{2}.

For k = 4, g(x) = \(\frac{1}{4}\)x^{2}.

For k = 5, g(x) = \(\frac{1}{5}\)x^{2}.

For k = \(\frac{1}{2}\), g(x) = 2x^{2}.

### Eureka Math Algebra 2 Module 1 Lesson 35 Exercise Answer Key

Exercise 1.

Write equations for two parabolas that are congruent to the parabola given by y = x^{2}, and explain how you determined your equations.

Answer:

(Student answers will vary.) The parabolas given by y = (x – 2)^{2} and y = x^{2} – 3 are congruent to the parabola given by y = x^{2}. The first parabola is translated horizontally to the right by two units and the second parabola is translated down by 3 units, so they each are congruent to the original parabola.

Exercise 2.

Sketch the graph of y = x^{2} and the two parabolas you created on the same coordinate axes.

Answer:

Exercise 3.

Write the equation of two parabolas that are NOT congruent to y = x^{2}. Explain how you determined your equations.

Answer:

(Student answers will vary.) By our work in the previous lesson, we know that the equation any parabola can be written in the form y = \(\frac{1}{2p}\)(x – h)^{2} + k, and that two parabolas are congruent if and only if their equations have the same value of |P|. Then the parabolas y = 2x^{2} and y = x^{2} are both not congruent to the parabola given by y = x^{2}.

Exercise 4.

Sketch the graph of y = x^{2} and the two non-congruent parabolas you created on the same coordinate axes.

Answer:

Exercise 5.

What does it mean for two triangles to be similar? How do we use geometric transformation to determine if two triangles are similar?

Answer:

Two triangles are similar if we can transform one onto the other by a sequence of rotations, reflections, translations and dilations.

Exercise 6.

What would it mean for two parabolas to be similar? How could we use geometric transformation to determine if two parabolas are similar?

Answer:

Two para bolas should be similar if we can transform one onto the other by a sequence of rotations, reflections, translations and dilations.

Exercise 7.

Use your work in Exercises 1 – 6 to make a conjecture: Are all parabolas similar? Explain your reasoning.

Answer:

(Student answers will vary.) It seems that any pair of parabolas should be similar because we can line up the vertices through a sequence of rotations, reflections and translations, then we should be able to dilate the width of one parabola to match the other.

Exercise 8.

The parabola at right is the graph of which equation?

a. Label a point (x, y) on the graph of P.

Answer:

b. What does the definition of a parabola tell us about the distance between the point (x, y) and the directrix L, and the distance between the point (x, y) and the focus F?

Answer:

Let (x, y) be any point on the graph of P. Then, these distances are equal because P = {(x, y) I (x, y) is equidistant from F and L}.

c. Create an equation that relates these two distances.

Answer:

Distance from (x, y) to F: \(\sqrt{(x-2)^{2}+(y-0)^{2}}\)

Distance from (x, y) to L: x + 2

Therefore, any point on the parabola has coordinates (x, y) that satisfy \(\sqrt{(x-2)^{2}+(y-0)^{2}}\) = x + 2.

d. Solve this equation for x.

Answer:

The equation can be solved as follows.

\(\sqrt{(x-2)^{2}+(y-0)^{2}}\) = x + 2

(x – 2)^{2} + y^{2} = (x + 2)^{2}

x^{2} – 4x + 4 + y^{2} = x^{2} + 4x + 4

y^{2} = 8x

x = \(\frac{1}{8}\)y2

Thus,

P = {(x, y) | x = \(\frac{1}{8}\)y^{2}}.

e. Find two points on the parabola P, and show that they satisfy the equation found in part (d).

Answer:

By observation, (2, 4) and (2, – 4) are points on the graph of P. Both points satisfy the equation that defines

(2, 4): \(\frac{1}{8}\)(4)^{2} = \(\frac{16}{8}\) = 2

(2, – 4): \(\frac{1}{8}\)(-4)^{2} = \(\frac{16}{8}\) = 2

Discussion

Do you think that all parabolas are similar? Explain why you think so.

Answer:

Yes, they all have the same basic shape.

What could we do to show that two parabolas are similar? How might you show this?

Answer:

Since every parabola can be transformed into a congruent parabola by applying one or more rigid transformations, perhaps similar parabolas can be created by applying a dilation which is a non-rigid transformation.

Exercises 9 – 12

Use the graphs below to answer Exercises 9 and 10.

Exercise 9.

Suppose the unnamed red graph on the left coordinate plane is the graph of a function g. Describe g as a vertical scaling of the graph of y = f(x); that is, find a value of k so that g(x) = kf(x). What is the value of k? Explain how you determined your answer.

Answer:

The graph of g is a vertical scaling of the graph off by a factor of 2. Thus, g (x) = 2f(x). By comparing points on the graph off to points on the graph of g, you can see that the y-values on g are all twice the y-values on f.

Exercise 10.

Suppose the unnamed red graph on the right coordinate plane is the graph of a function h. Describe h as a vertical scaling of the graph of y = f(x); that is, find a value of k so that h(x) = kf(x). Explain how you determined your answer.

Answer:

The graph of h is a vertical scaling of the graph of f by a factor of \(\frac{1}{2}\). Thus, h(x) = \(\frac{1}{2}\)f(x). By comparing points on the graph off to points on the graph of h, you can see that the y-values on h are all half of the y-values on f.

Use the graphs below to answer Exercises 11 – 12.

Exercise 11.

Suppose the unnamed function graphed in red on the left coordinate plane is g. Describe g as a horizontal scaling of the graph of y = f(x). What is the value of the scale factor k? Explain how you determined your answer.

Answer:

The graph of g is a horizontal scaling of the graph off by a factor of 2. Thus, g(x) = f(\(\frac{1}{2}\)x). By comparing points on the graph off to points on the graph of g, you can see that for the same y-values, the x-values on g are all twice the x-values on f.

Exercise 12.

Suppose the unnamed function graphed in red on the right coordinate plane is h. Describe h as a horizontal scaling of the graph of y = f(x). What is the value of the scale factor k? Explain how you determined your answer.

Answer:

The graph of h is a horizontal scaling of the graph of f by a factor of \(\frac{1}{2}\). Thus, h(x) = f(2x). By comparing points on the graph off to points on the graph of h, you can see that for the same y-values, the x-values on h are all half\(\frac{1}{2}\)of the x-values on f.

### Eureka Math Algebra 2 Module 1 Lesson 35 Problem Set Answer Key

Question 1.

Let (x) = \(\sqrt{4-x^{2}}\). The graph off is shown below. On the same axes, graph the function g, where g(x) = f (\(\frac{1}{2}\)x). Then, graph the function h, where h(x) = 2g(x).

Answer:

Question 2.

Let f(x) = – |x| + 1. The graph off is shown below. On the same axes, graph the function g, where g(x) =f(\(\frac{1}{3}\)x). Then, graph the function h, where h(x) = 3g(x).

Answer:

Question 3.

Based on your work in Problems 1 and 2, describe the resulting function when the original function is transformed with a horizontal and then a vertical scaling by the same factor, k.

Answer:

The resulting function is scaled by a factor of k in both directions. It is a dilation about the origin of the original figure and is similar to it.

Question 4.

Let f(x) = x^{2}.

a. What are the focus and directrix of the parabola that is the graph of the function f(x) = x^{2}?

Answer:

Since \(\frac{1}{2 p}\) = 1, we know p = and that is the distance between the focus and the directrix. The point (0, 0) is the vertex of the parabola and the midpoint of the segment connecting the focus and the directrix. Since the distance between the focus and vertex is \(\frac{1}{2}\)p = \(\frac{1}{4}\), which is the same as the distance between the vertex and directrix; therefore, the focus has coordinates (0, \(\frac{1}{4}\)) and the directrx is y = –\(\frac{1}{4}\).

b. Describe the sequence of transformations that would take the graph of f to each parabola described below.

i. Focus: (o, –\(\frac{1}{4}\)), directrix: y = \(\frac{1}{4}\)

Answer:

This parabola is a reflection of the graph off across the x-axis.

ii. Focus: (\(\frac{1}{4}\), 0) directrix: x = –\(\frac{1}{4}\)

Answer:

This parabola is a 90Â° clockwise rotation of the graph of f.

iii. Focus: (0, 0), directrix: y = –\(\frac{1}{2}\)

Answer:

This parabola is a vertical translation of the graph of f down \(\frac{1}{4}\) unit.

iv. Focus: (o, \(\frac{1}{4}\)), directrix: y = –\(\frac{3}{4}\)

Answer:

This parabola is a vertical scaling of the graph of f by a factor of \(\frac{1}{2}\) and a vertical translation of the resulting image down \(\frac{1}{4}\) unit.

v. Focus: (0, 3), directrix: y = – 1

Answer:

This parabola is a vertical scaling of the graph of f by a factor of \(\frac{1}{8}\) and a vertical translation of the resulting image up 1 unit.

c. Which parabolas are similar to the parabola that is the graph of f? Which are congruent to the parabola that is the graph of f?

Answer:

All of the parabolas are similar. We have proven that all parabolas are similar. The congruent parabolas are (i), (ii), and (iii). These parabolas are the result of a rigid transformation of the original parabola that is the graph off. They have the same distance between the focus and directrix line as the original parabola.

Question 5.

Derive the analytic equation for each parabola described in Problem 4(b) by applying your knowledge of transformations.

Answer:

i. y = – x^{2}

ii. x = y^{2}

iii y = x^{2} – \(\frac{1}{4}\)

iv. y = \(\frac{1}{2}\)x^{2} – \(\frac{1}{4}\)

v. y = \(\frac{1}{8}\)x^{2} + 1

Question 6.

Are all parabolas the graph of a function of x in the xy-plane? If so, explain why, and if not, provide an example (by giving a directrix and focus) of a parabola that is not.

Answer:

No, they are not. Examples include the graph of the equation x = y^{2}, or a list stating a directrix and focus. For example, students may give the example of a directrix given by x = – 2 and focus (2, 0), or an even more interesting example, such as a directrix given by y = x with focus (1, – 1). Any line and any point not on that line define a parabola.

Question 7.

Are the following parabolas congruent? Explain your reasoning.

Answer:

They are not congruent, but they are similar. I can see that the parabola on the left appears to contain the point (1, 1), while the parabola on the right appears to contain the point (1, \(\frac{1}{2}\)). This implies that the graph of the parabola on the right is a dilation of the graph of the parabola on the left, so they are not congruent.

Question 8.

Are the following parabolas congruent? Explain your reasoning.

Answer:

They are congruent. Both graphs contain the points (0, 0), (1, 1), and (2, 4) that satisfy the equation y = x^{2}. The scales are different on these graphs, making them appear non-congruent.

Question 9.

Write the equation of a parabola congruent to y = 2x^{2} that contains the point (1, – 2). Describe the transformations that would take this parabola to your new parabola.

Answer:

There are many solutions. Two possible solutions:

Reflect the graph about the x-axis to get y = – 2x^{2}.

OR

Translate the graph down four units to get y = 2x^{2} – 4.

Question 10.

Write the equation of a parabola similar to y = 2x^{2} that does NOT contain the point (0, 0) but does contain the point (1, 1).

Answer:

Since all parabolas are similar, as established in the lesson, any parabola that passes through (1,1) and not (0, 0) is a valid response. One solution is y = (x – 1)^{2} + 1. This parabola is congruent to y = x^{2} and, therefore, similar to the original parabola, but the graph has been translated horizontally and vertically to contain the point (1, 1) but not the point (0, 0).

### Eureka Math Algebra 2 Module 1 Lesson 35 Exit Ticket Answer Key

Question 1.

Describe the sequence of transformations that would transform the parabola P_{x} into the similar parabola P_{y}.

Answer:

Vertical scaling by a factor of \(\frac{1}{2}\), vertical translation up 3 units, and a 900 rotation clockwise about the origin

Question 2.

Are the two parabolas defined below similar or congruent or both?

Parabola 1: The parabola with a focus of (0, 2) and a directrix line of y = – 4

Parabola 2: The parabola that Is the graph of the equation y = \(\frac{1}{6}\)x^{2}

Answer:

They are similar but not congruent because the distance between the focus and the directrix on Parabola 1 is 6 units, but on Parabola 2, it is only 3 units. Alternatively, students may describe that you cannot apply a series of rigid transformations that will map Parabola 1 onto Parabola 2. However, by using a dilation and a series of rigid transformations, the two parabolas can be shown to be similar since ALL parabolas are similar.