## 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) = 2x^{2} + 12x + 1. Write this in the completed – square (vertex) form and show all the steps.

Answer:

g(x) = 2x^{2} + 12x + 1

Step 1 = (2x^{2} + 12x) + 1 Gather variable terms.

Step 2 = 2(x^{2} + 6x) + 1 Factor out the GCF.

Step 3 = 2(x^{2} + 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) = x^{2}. 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) = x^{2} 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) = x^{2}?

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) = x^{2} 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) = x^{2}.

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) = x^{2} – 2x

e. t(x) = x^{2} – 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) = x^{2}.

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) = x^{2} that results in the quadratic function g(x) = 2x^{2} + 4x + 1.

Answer:

First, rewrite g(x) into the completed – square form.

g(x) = 2x^{2} + 4x + 1

= (2x^{2} + 4x) + 1

= 2(x^{2} + 2x) + 1

= 2(x^{2} + 2x + 1) + 1 – 2

= 2(x^{2} + 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) = x^{2}. 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) = 2x^{2} + 8x

d. p(x) = x^{2} + 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) = – 2x^{2} – 20x – 53 in completed – square form. Describe the transformations of the graph of the parent function f(x) = x^{2} that result in the graph of g.

Answer:

g(x) = – 2x^{2} – 20x – 53

= ( – 2x^{2} – 20x) – 53

= – 2(x^{2} + 10x) – 53

= – 2(x^{2} + 10x + 25) – 53 + 50

= – 2(x^{2} + 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) = x^{2} 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) = x^{2}. 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) = 2x^{2} – 12x + 19

d. p(x) = – 2x^{2} – 4x – 5

e. q(x) = 3x^{2} + 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) = x^{2} 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.