# Eureka Math Algebra 1 Module 4 Lesson 16 Answer Key

## Engage NY Eureka Math Algebra 1 Module 4 Lesson 16 Answer Key

### Eureka Math Algebra 1 Module 4 Lesson 16 Exploratory Challenge Answer Key

Exploratory Challenge
Caitlin has 60 feet of material that can be used to make a fence. Using this material, she wants to create a rectangular pen for her dogs to play in. What dimensions will maximize the area of the pen?
a. Let w be the width of the rectangular pen in feet. Write an expression that represents the length when the width is w feet.
$$\frac{60 – 2w}{2}$$ or 30 – w

b. Define a function that describes the area, A, in terms of the width, w.
A(w) = w(30 – w) or A(w) = – w2 + 30w

c. Rewrite A(w) in vertex form.
A(w) = – w2 + 30w
= – (w2 – 30w)
= – (w2 – 30w + 225) + 225
= – (w – 15)2 + 225

d. What are the coordinates of the vertex? Interpret the vertex in terms of the problem.
The vertex is located at (15, 225). Since the leading coefficient is negative, the function has a maximum.
The maximum value of the function is 225, which occurs when w = 15. For this problem, this means that the maximum area is 225 square feet, which happens when the width is 15 feet.

e. What dimensions maximize the area of the pen? Do you think this is a surprising answer?
The pen has the greatest area when the length and width are both 15 feet. Students may or may not be surprised to note that this occurs when the rectangle is a 15 × 15 square.

### Eureka Math Algebra 1 Module 4 Lesson 16 Exercise Answer Key

Opening Exercise
Graph the equations y = x2, y = (x – 2)2, and y = (x + 2)2 on the interval – 3 ≤ x ≤ 3.
Have students graph the equations y = x2, y = (x – 2)2, and y = (x + 2)2 on the interval – 3≤x≤3. Consider having students graph the equations using the graphing calculator or graph paper.
→ How are the graphs of these equations similar? How are they different?
→ The graphs look similar in that they have the same shape. Point out that the graphs are all translations of each other. They have different vertices: y = x2 has its vertex at (0, 0), while y = (x – 2)2 has its vertex at (2, 0), and y = (x + 2)2 has its vertex at ( – 2, 0).

→ Now consider the graph of y = (x – 5)2. Where would you expect this graph to be in relation to the other three?
The graph of y = (x – 5)2 is 5 units to the right of the graph of y = x2, 3 units to the right of
y = (x – 2)2, and 7 units to the right of y = (x + 2)2.

Exercises

Exercise 1.
Without graphing, state the vertex for each of the following quadratic equations.
a. y = (x – 5)2 + 3
(5, 3)

b. y = x2 – 2.5
(0, – 2.5)

c. y = (x + 4)2
( – 4, 0)

Exercise 2.
Write a quadratic equation whose graph will have the given vertex.
a. (1.9, – 4)
y = (x – 1.9)2 – 4

b. (0, 100)
y = x2 + 100

c. ( – 2, $$\frac{3}{2}$$)
y = (x + 2)2 + $$\frac{3}{2}$$

### Eureka Math Algebra 1 Module 4 Lesson 16 Problem Set Answer Key

Question 1.
Find the vertex of the graphs of the following quadratic equations.
a. y = 2(x – 5)2 + 3.5
(5, 3.5)

b. y = – (x + 1)2 – 8
( – 1, – 8)

Question 2.
Write a quadratic equation to represent a function with the following vertex. Use a leading coefficient other than 1.
a. (100, 200)
y = – 2(x – 100)2 + 200

b. ( – $$\frac{3}{4}$$, – 6)
y = 4(x + $$\frac{3}{4}$$)2 – 6

Question 3.
Use vocabulary from this lesson (i.e., stretch, shrink, opens up, and opens down) to compare and contrast the graphs of the quadratic equations y = x2 + 1 and y = – 2x2 + 1.
The quadratic equations share a vertex at (0, 1), but the graph for the equation y = – 2x2 + 1 opens down and has a vertical stretch, while the graph of the equation y = x2 + 1 opens up.

### Eureka Math Algebra 1 Module 4 Lesson 16 Exit Ticket Answer Key

Question 1.
Compare the graphs of the function, f(x) = – 2(x + 3)2 + 2 and g(x) = 5(x + 3)2 + 2. What do the graphs have in common? How are they different?