# Eureka Math Algebra 1 Module 4 Lesson 17 Answer Key

## 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?
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.
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?
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.

### 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?
The time (number of seconds) since the ball was thrown

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

c. At what height does the ball get thrown?
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?
The ballâ€™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, letâ€™s solve by completing the square.
– 16t2 + 96t + 6 = 0
– 16(t2â€“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 = Â±$$\frac{\sqrt{150}}{4}$$
t = 3 Â±$$\frac{\sqrt{150}}{4}$$
t â‰ˆ 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?
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?
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?
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:
â†’ Look for hints in the functionâ€™s equation for general shape, direction, and y – intercept.
â†’ Solve f(x) = 0 to find the x – intercepts by factoring, completing the square, or using the quadratic formula.
â†’ 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($$\frac{ – b}{2a}$$).
â†’ 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.

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.

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.

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.
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?