# Lesson 14

Graphs That Represent Situations

## 14.1: A Jumping Frog (5 minutes)

### Warm-up

In this warm-up, students begin to apply their new understandings about graphs to reason about quadratic functions contextually. They evaluate a simple quadratic function, find its maximum, and interpret these values in context. The input values to be evaluated produce an output of 0, reminding students of the meaning of the zeros of a function and their connection to the horizontal intercepts ($$x$$-intercepts) of the graph.

To find the maximum value of the function, students could graph the function, but applying what they learned about the connection between the horizontal intercepts and the vertex (without graphing) would be more efficient. Identify students who make this connection and ask them to share their thinking in the discussion.

### Student Facing

The height in inches of a frog's jump is modeled by the equation $$h(t) = 60t-75t^2$$ where the time, $$t$$,  after it jumped is measured in seconds.

1. Find $$h(0)$$ and $$h(0.8)$$. What do these values mean in terms of the frog’s jump?
2. How much time after it jumped did the frog reach the maximum height? Explain how you know.

### Student Response

Student responses to this activity are available at one of our IM Certified Partners

### Activity Synthesis

Select students to share their responses and explanations. Even though students are not asked to graph the function, make sure that they begin to connect the quadratic expression that defines the function to the features of the graph representing that function.

Ask students: “If we graph the equation, what would the graph look like? Where would the intercepts be? Would the graph open up or down? Where would the vertex be?”

Highlight the following points:

• The graph intersects the horizontal axis at $$t=0$$ and $$t=0.8$$, because $$h(0)$$ and $$h(0.8)$$ both have a value of 0.
• $$h(0.4)$$ is the maximum height of the frog since $$t=0.4$$ is halfway between the horizontal intercepts. If we were to graph it, $$(0.4,h(0.4))$$ would be the vertex.
• The graph opens downward because the coefficient of the squared term is a negative number (-75).

## 14.2: A Catapulted Pumpkin (15 minutes)

### Activity

Earlier in this unit, students read off of a diagram the position of an object dropped from the top of a building and then observed that the data is modeled by a quadratic function. This lesson continues to develop this idea with an additional layer of complexity, appropriate for this stage in the unit.

### Launch

Arrange students in groups of 2. Display the equation $$d = 10 + 406t- 16t^2$$ for all to see. Remind students that earlier in the unit, we saw this equation used to model the height of a cannonball that is shot up in the air as a function of time in seconds. The height was measured in feet. Ask students to recall what each term in the equation represents and briefly discuss their thinking.

Next, give students a few minutes of quiet think time to read the first question (all four parts) to themselves and think about their responses. Ask them to share their thoughts with a partner before proceeding to the graphing questions. Clarify if needed that “horizontal intercept” and “vertical intercept” are a more general way to refer to $$x$$- and $$y$$-intercepts when the equation that defines a function uses variables other than $$x$$ and $$y$$.

Provide access to devices that can run Desmos or other graphing technology. If needed, remind students how to use the available graphing technology to identify the coordinates of points on a graph.

Speaking, Reading: MLR5 Co-Craft Questions. Locate and display a video or image of a pumpkin that is catapulted up in the air. If unable to find an image or video, display only the task statement without the questions that follow. Before students record possible mathematical questions that could be asked about the situation, invite them to share any questions they may have about the context. Invite students to compare the mathematical questions they have before revealing the remainder of the task. Listen for and amplify any questions involving interpretations of features of the quadratic graph.
Design Principle(s): Maximize meta-awareness; Support sense-making
Representation: Access for Perception. Invite students to follow along as you read the task statement and the questions from the first problem aloud. Students who both listen to and read the information will benefit from extra processing time. Pause for think time and then give students 2–3 minutes to discuss their initial thoughts with a partner before moving on.
Supports accessibility for: Language; Conceptual processing

### Student Facing

The equation $$h = 2 + 23.7t - 4.9t^2$$ represents the height of a pumpkin that is catapulted up in the air as a function of time, $$t$$, in seconds. The height is measured in meters above ground. The pumpkin is shot up at a vertical velocity of 23.7 meters per second.

1. Without writing anything down, consider these questions:
• What do you think the 2 in the equation tells us in this situation? What about the $$\text-4.9t^2$$?
• If we graph the equation, will the graph open upward or downward? Why?
• Where do you think the vertical intercept would be?
• What about the horizontal intercepts?
2. Graph the equation using graphing technology.
3. Identify the vertical and horizontal intercepts, and the vertex of the graph. Explain what each point means in this situation.

### Student Response

Student responses to this activity are available at one of our IM Certified Partners

### Student Facing

#### Are you ready for more?

What approximate vertical velocity would this pumpkin need for it stay in the air for about 10 seconds? (Assume that it is still shot from 2 meters in the air and that the effect of gravity pulling it down is the same.)

### Student Response

Student responses to this activity are available at one of our IM Certified Partners

### Activity Synthesis

Invite students to share their graph and interpretations of the features of the graph. Discuss with students:

• “The graph shows two horizontal intercepts, one with a positive $$t$$-coordinate and the other with a negative $$t$$-coordinate. How do we make sense of the negative $$t$$-coordinate in this situation?” (Only the positive one has any meaning in this situation, since negative value of time—the time before the baseball was hit—is not relevant in this model.)
• “How would you estimate the coordinates of the vertex without using a graph? Suppose you know both of the horizontal intercepts.” (We can find the halfway point of the two horizontal intercepts and evaluate the function at that value of $$t$$.)

## 14.3: Flight of Two Baseballs (15 minutes)

### Activity

This activity prompts students to analyze two quadratic functions—one represented by a graph, and the other by an equation—and solve a problem. Graphs can make it easier to compare functions, but because students are asked not to graph the second function, they need to rely on their knowledge of the connections between equations and their graphs to make the comparison. To answer the questions, students have to interpret various pieces of the given information (the graph, the numbers in the equation, the descriptions in the questions, etc.), allowing them to reason quantitatively and abstractly (MP2).

As students work, look for students who reason in the ways outlined in the Activity Synthesis so they can share their thinking later.

### Launch

Give students a moment to read the task statement. Make sure students understand what it means for a baseball to “stay in flight.”

Consider keeping students in groups of 2 and and asking them to think quietly before discussing their thinking with a partner.

### Student Facing

Here is a graph that represents the height of a baseball, $$h$$, in feet as a function of time, $$t$$, in seconds after it was hit by Player A.

The function $$g$$ defined by $$g(t) = (\text-16t-1)(t-4)$$ also represents the height in feet of a baseball $$t$$ seconds after it was hit by Player B. Without graphing function $$g$$, answer the following questions and explain or show how you know.

1. Which player’s baseball stayed in flight longer?
2. Which player’s baseball reached a greater maximum height?
3. How can you find the height at which each baseball was hit?

### Student Response

Student responses to this activity are available at one of our IM Certified Partners

### Anticipated Misconceptions

Some students may find it challenging to find the zeros of function $$g$$ because of the $$\text-16$$ in $$\text-16t-1$$. Support these students reasoning about the equation $$\text-16t-1=0$$ by asking whether this equation will have a positive or negative solution and whether it will have a solution close to 0 or not. Another challenge students may face is estimating the location of the vertex. Point out the graph of $$h$$ and ask them how the vertex relates to the horizontal intercepts. It is not expected students will compute the precise location by averaging $$\text-\frac{1}{16}$$ and 4. Encourage them to use the number between 0 and 4 as a good estimate of the $$t$$-coordinate of the vertex. Encourage struggling students to use technology to graph $$g$$ to check their thinking.

### Activity Synthesis

Invite students to share their responses and reasoning. If not mentioned in students’ explanations, highlight the following points:

• To determine which baseball stayed in flight longer, we can compare where the graphs intersect the horizontal axis or the zeros of the functions, which tell us when the baseball hits the ground.
• The graph of function $$h$$ intersects the horizontal axis around 5.
• Function $$g$$ is given in factored form, so the zeros will be a very small negative number (about $$\text- \frac{1}{16}$$) and the other will be 4.
• The maximum height of Player A’s baseball (function $$h$$) is shown in the graph. For Player B (function $$g$$), we can approximate the halfway point between the horizontal intercepts, which is around $$t=2$$, and then find $$g(2)$$.
• The baseball was hit at $$t=0$$, so to find the height we would look at the vertical intercept ($$y$$-intercept).
• For function $$h$$, this point is shown on the graph.
• For function $$g$$, it is not immediately apparent because the equation is in factored form. If it was in standard form, the constant term is the vertical intercept. We can either expand the $$(\text-16t-1)(t-4)$$ and write an equivalent expression in standard form, or find $$g(0)$$.
Conversing: MLR2 Collect and Display. As students discuss their explanations with a partner, capture student language that reflects a variety of ways to describe the different representations of quadratic functions. Write the students’ words on a visual display of the graph. Emphasize connections between terms such as “constant term” and “$$y$$-intercept”, between “starting height” and “$$t=0$$”, and between “$$x$$-intercepts” and “when the ball hits the ground”. Remind students to borrow language from the display as needed.
Design Principle(s): Optimize output (for explanation); Maximize meta-awareness

## 14.4: Info Gap: Rocket Math (20 minutes)

### Optional activity

This optional activity gives students an additional opportunity to apply what they learned about key characteristics of a quadratic function to solve contextual problems. Unlike in previous activities where an equation or a graph that represents the function was given up front, here students are presented only with a familiar situation and need to consider what relevant details are missing.

The info gap structure requires students to make sense of problems by determining what information is necessary, and then to ask for information they need to solve it. This may take several rounds of discussion if their first requests do not yield the information they need (MP1). It also allows them to refine the language they use and ask increasingly more precise questions until they get the information they need (MP6).

Here is the text of the cards for reference and planning:

### Launch

Tell students they will continue to work with interpreting graphs and equations of a function in terms of a situation involving toy rockets.

Explain the info gap structure, and consider demonstrating the protocol if students are unfamiliar with it.

Arrange students in groups of 2. In each group, distribute a problem card to one student and a data card to the other student. After reviewing students’ work on the first problem, give them the cards for a second problem and instruct partners to switch roles.

Conversing: This activity uses MLR4 Information Gap to give students a purpose for discussing information necessary to solve contextual problems using what they learned about key characteristics of a quadratic function. Display questions or question starters for students who need a starting point such as: “Can you tell me . . . (specific piece of information)”, and “Why do you need to know . . . (that piece of information)?"
Design Principle(s): Cultivate Conversation
Engagement: Develop Effort and Persistence. Display or provide students with a physical copy of the written directions. Check for understanding by inviting students to rephrase directions in their own words. Keep the display of directions visible throughout the activity.
Supports accessibility for: Memory; Organization

### Student Facing

Your teacher will give you either a problem card or a data card. Do not show or read your card to your partner.

If your teacher gives you the data card:

2. Ask your partner “What specific information do you need?” and wait for your partner to ask for information. Only give information that is on your card. (Do not figure out anything for your partner!)
3. Before telling your partner the information, ask “Why do you need to know (that piece of information)?”
4. Read the problem card, and solve the problem independently.
5. Share the data card, and discuss your reasoning.

If your teacher gives you the problem card:

3. Explain to your partner how you are using the information to solve the problem.
4. When you have enough information, share the problem card with your partner, and solve the problem independently.

### Student Response

Student responses to this activity are available at one of our IM Certified Partners

### Activity Synthesis

After students have completed their work, share the correct answers and ask students to discuss the process of solving the problems. Here are some questions for discussion:

• “How did you determine when the Kiran’s rocket reached its highest point?” (The rocket launched at 0 seconds and landed at 1.6 seconds. Those times tell us the $$x$$-intercepts of the graph. The rocket will reach its maximum height at the vertex of the graph, which is halfway between the $$x$$-intercepts.)
• “How did you determine how high the rocket went?” (The maximum height is at $$x=0.8$$. If we replace $$x$$ in the expression $$\text-16x(x-1.6)$$ with 0.8, then $$\text-16(0.8)(0.8-1.6)=10.24$$.)
• “How do you know that the function that models the flight of Kiran’s toy rocket is quadratic?” (The equation that models the flight of the rocket has an $$x^2$$ term when you write it in standard form.)
• “What does the $$\text-16x^2$$ in the equation represent?” (The $$\text-16x^2$$ represents the influence of gravity.)
• “If we graphed the equation, what would the point $$(0.8, 10.24)$$ represent?” (The vertex of the graph.)

Highlight for students how the information in the problem related to the vertex and intercepts of the graph and the parts of the equation.

## Lesson Synthesis

### Lesson Synthesis

The examples students saw in this lesson were related to the motion of projectiles. Emphasize that the graphs that represent quadratic functions that model other kinds of situations can also be interpreted in context. For example, in an earlier lesson we saw a quadratic function defined by $$r(d) = d(600-75d)$$. It described the revenue in thousands of dollars collected from ticket sales as a function of the price of one ticket.

• “What can you tell about the ticket sales and revenue from the equation?” (If the ticket is sold for 0 dollars, or if $$d=0$$, the revenue will be 0.)
• “In what form is the quadratic expression defining $$r$$ written?” (factored form) “If we rewrite it in standard form, what new information would we gain?” (In standard form, the equation is $$r(d)=60d-75d^2$$. It gives us the vertical intercept, but that is not new information. We saw that one of the horizontal intercepts is at $$(0,0)$$, so the vertical intercept is known.)
• “Here is the graph representing the same function. What information does the horizontal intercepts, the vertical intercept, and the vertex of the graph give us?” (The horizontal intercepts tell us that no revenue will be collected if the tickets are sold at \$0 and \$8. The vertical intercept gives the same information: that if the ticket is free, there will be no revenue. The vertex tells us that the greatest revenue will be collected when the ticket is sold at \$4 each, and that amount is \$1,200.)

## 14.5: Cool-down - Beach Ball Trajectory (5 minutes)

### Cool-Down

Cool-downs for this lesson are available at one of our IM Certified Partners

## Student Lesson Summary

### Student Facing

Let’s say a tennis ball is hit straight up in the air, and its height in feet above the ground is modeled by the equation $$f(t) = 4 + 12t - 16t^2$$. Here is a graph that represents the function, from the time the tennis ball was hit until the time it reached the ground.

In the graph, we can see some information we already know, and some new information:

• The 4 in the equation means the graph of the function intersects the vertical axis at 4. It shows that the tennis ball was 4 feet off the ground at $$t=0$$, when it was hit.
• The horizontal intercept is $$(1,0)$$. It tells us that the tennis ball hits the ground 1 second after it was hit.
• The vertex of the graph is at approximately $$(0.4, 6.3)$$. This means that about 0.4 second after the ball was hit, it reached the maximum height of about 6.3 feet.

The equation can be written in factored form as $$f(t) = (\text-16t-4)(t-1)$$. From this form, we can see that the zeros of the function are $$t = 1$$ and $$t= \text- \frac{1}{4}$$. The negative zero, $$\text- \frac{1}{4}$$, is not meaningful in this situation, because the time before the ball was hit is irrelevant.