Lesson 6
Building Quadratic Functions to Describe Situations (Part 2)
6.1: Sky Bound (5 minutes)
Activity
In this warmup, students considers what happens if an object is launched up in the air unaffected by gravity. The work here serves two purposes. It reminds students that an object that travels at a constant speed can be described with a linear function. It also familiarizes students with a projectile context used in the next activity, in which students will investigate a quadratic function that more realistically models the movement of a projectile—with gravity in play.
Students who use a spreadsheet to complete the table practice choosing tools strategically (MP5).
Launch
Ask a student to read the opening paragraph of the activity aloud. To help students visualize the situation described, consider sketching a picture of a cannon pointing straight up, 10 feet above ground. Ask students to consider what a speed of 406 feet per second means in more concrete terms. How fast is that?
Students may be more familiar with miles per hour. Tell students that the speed of 406 feet per second is about 277 miles per hour.
Consider arranging students in groups of 2 so they can divide up the calculations needed to complete the table. Provide access to calculators, if requested.
Student Facing
A cannon is 10 feet off the ground. It launches a cannonball straight up with a velocity of 406 feet per second.
Imagine that there is no gravity and that the cannonball continues to travel upward with the same velocity.
 Complete the table with the heights of the cannonball at different times.
seconds 0 1 2 3 4 5 \(t\) distance above ground (feet) 10  Write an equation to model the distance in feet, \(d\), of the ball \(t\) seconds after it was fired from the cannon if there was no gravity.
Student Response
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Activity Synthesis
Ask students how the values in the table are changing and what equation would describe the height of the cannonball if there were no gravity. Even without graphing, students should notice that the height of the cannonball over time is a linear function given the repeated addition of 406 feet every time \(t\) increases by 1.
Tell students that, in the next activity, they will look at some actual heights of the cannonball.
6.2: Tracking a Cannonball (15 minutes)
Activity
Prior to this course, students learned that an object traveling at a constant speed can be described with a linear function whose graph is a straight line. Here they see a model that accounts for the fact that an object that is launched straight up at a constant speed does not keep going at the same rate when the influence of gravity is taken into account. Adding a quadratic term to a linear function has an effect of “bending” the graph, as the output values are no longer changing at a constant rate.
If students are unsure how to write an equation to represent the values in the table, ask them to compare how the actual heights of the cannonball at each second (when \(t\) is 1, 2, 3, etc.) differ from those in the nogravity case (as shown in the table in the warmup). Finding the differences between the two outputs (16, 64, 144, . . .) at the same input values should help students think of the numbers and the functions they saw in the previous lesson.
To generalize the relationship between time and distance, students reason repeatedly with numerical values and look for regularity (MP8). If students opt to use spreadsheet or graphing technology, they practice choosing appropriate tools strategically (MP5).
Launch
Arrange students in groups of 2. Give students a minute of quiet time to think about the first question, and then time to share their observations with their partner. Tell students that they will need to reference their work in the warmup.
Some students may choose to use a spreadsheet tool to extend the pattern, and subsequently to use graphing technology to plot the data. Make these tools accessible, in case requested.
Student Facing
Earlier, you completed a table that represents the height of a cannonball, in feet, as a function of time, in seconds, if there was no gravity.
 This table shows the actual heights of the ball at different times.
seconds 0 1 2 3 4 5 distance above ground (feet) 10 400 758 1,084 1,378 1,640 Compare the values in this table with those in the table you completed earlier. Make at least 2 observations.

 Plot the two sets of data you have on the same coordinate plane.
 How are the two graphs alike? How are they different?
 Write an equation to model the actual distance \(d\), in feet, of the ball \(t\) seconds after it was fired from the cannon. If you get stuck, consider the differences in distances and the effects of gravity from a previous lesson.
Student Response
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Anticipated Misconceptions
When comparing the tables, some students may make observations that lack the detail needed to write an equation for the actual height. Prompt them rewrite the outputs for the actual height in terms of the hypothetical height (\(400 = 416  16\), \(758 = 822  64\), \(1,\!084 = 1,\!228  144\), and so on). Show them values of \(16t^2\) from a previous lesson to help them see and extend the pattern to write the equation.
Activity Synthesis
Invite students to share their observations about the two tables (the one from the warmup and the one here) and how the two graphs compare. Highlight responses that suggest that the values in the second table account for the effect of gravity.
Help students see how the output for each \(t\) value varies across the two tables. When \(t\) is 1, the output in feet in the second table is 16 less than in the first table. When \(t\) is 2, there is a difference of 64 feet. When \(t\) is 3, that difference is 144 feet, and so on. The values 16, 64, 144, . . . correspond to the expression \(16t^2\) that we saw in the previous lesson (the distance fallen in feet as a function of time in seconds), so we can represent the values in the second table with the equation \(d=10+ 406t 16t^2\). Ask students:
 “What do the 10, \(406t\), and \(\text16t^2\) mean in this situation?” (The 10 is the vertical position of the cannonball before it was launched: 10 feet above ground. In \(406t\), the 406 tells us the vertical velocity at which it was shot up. The \(\text16t^2\) accounts for the effect of gravity on the height of the cannonball after it was shot up.)
 “Why do you think the graph that represents \(d= 10+ 406t\) changes from a straight line to a curve when \(\text16t^2\) is added to the equation?” (Before that term was added the height increased by 406 feet every second. Adding \(\text16t^2\) decreases how much the cannonball travels up by some amount, but that amount gets larger each successive second. Eventually the cannonball stops increasing in height and starts to fall.)
Design Principle(s): Support sensemaking
Supports accessibility for: Visualspatial processing
6.3: Graphing Another Cannonball (15 minutes)
Activity
In this activity, students explore another model of a projectile motion. They graph and interpret a quadratic function in context and begin considering a reasonable domain for the function. Along the way, they practice reasoning concretely and abstractly (MP2). By the end of the lesson, they relate the vertex of the graph to the maximum height of the cannonball and the positive zero of the function to the time when the cannonball hits the ground.
Launch
Provide access to devices that can run Desmos or other graphing technology. If needed, demonstrate how to adjust the graphing boundaries of the graphing tool.
Depending on the graphing tool available and their facility with it, students may approach the estimations in the third question in different ways (including by eyeballing). If desired, demonstrate how to use the graphing tool to trace the graph and identify the coordinates of any point on it (which may include values that are precise or values rounded to a specified decimal place). Or, first observe how students go about estimating and give additional guidance as needed.
To support students with the last question, ask students: “Is the equation a good model for predicting the height of the cannonball 10 seconds after it is fired? What about 1 minute after it is fired?”
Student Facing
The function defined by \(d=50+ 312t 16t^2\) gives the height in feet of a cannonball \(t\) seconds after the ball leaves the cannon.
 What do the terms 50, \(312t\), and \(\text{}16t^2\) tell us about the cannonball?
 Use graphing technology to graph the function. Adjust the graphing window to the following boundaries: \(0 < x < 25\) and \(0 < y < 2,\!000\).

Observe the graph and:
 Describe the shape of the graph. What does it tell us about the movement of the cannonball?
 Estimate the maximum height the ball reaches. When does this happen?
 Estimate when the ball hits the ground.
 What domain is appropriate for this function? Explain your reasoning.
Student Response
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Student Facing
Are you ready for more?
If the cannonball were fired at 800 feet per second, would it reach a mile in height? Explain your reasoning.
Student Response
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Activity Synthesis
Invite students to share their observations and interpretations of the graph. Highlight the following points:
 The graph representing a quadratic function is a parabola, which is a special kind of U shape. We will learn more about the geometry of this shape in a later course, but for now, notice that there is a point when the graph changes direction, from going up as \(x\) increases to going down (or changes from going down to going up). We call this point the vertex of the graph.
 In the previous activity, we plotted a limited set of points, so we could not tell where the vertex of the graph was. In this graph, we are able to identify the vertex. In this situation, the vertex tells us the maximum height that the cannonball reaches and the number of seconds after launch that it took before it starts to fall.
 In this graph, we can also see that the height of the cannonball is 0 when \(t\) is a little less than 20. That point, the horizontal intercept, relates to the zero of the function, or an input value that produces 0 for the output. In this situation, the zero tells us when the cannonball hits the ground.
 Even though we can continue the graph beyond \(t\) of 20, in this situation any output values beyond that point would not have any meaning. After the cannonball hits the ground, the function is no longer appropriate for modeling the movement of the cannonball. Likewise, the function is not appropriate before \(t=0\) or before the cannon is fired. In this situation, a domain between 0 and just below 20 seconds is appropriate.
Supports accessibility for: Conceptual processing; Language
Lesson Synthesis
Lesson Synthesis
To reinforce the connections between the parameters of a quadratic expression and the situation it describes, ask students:
 “So far, we’ve seen different expressions that represent vertical distances. Here are three expressions that all represent distance, in feet, as a function of time, in seconds, for an object that is dropped or launched. What does each of them tell us? Draw a diagram to illustrate the distances, if helpful.”
 \(16t^2\) (the distance an object travels \(t\) seconds after being dropped)
 \(40016t^2\) (the height of an object that is dropped from a height of 400 feet)
 \(50 + 100t  16t^2\) (the height of an object that is shot up from 50 feet above the ground at a vertical speed of 100 feet per second, \(t\) seconds after being launched)
 “If each expression defines a function, what does the zero of that function tell us?”
 \(16t^2\) (The zero is the time when the object has traveled a distance of 0 feet. This happens at \(t=0\), before the object is dropped.)
 \(40016t^2\) (The zero is the time when the height of the object is 0 feet, which is when it hits the ground.)
 \(50 + 100t  16t^2\) (The zero is the time when the height of the object is 0 feet, which is also when it hits the ground.)
Explain to students that the models seen here are simplified models and they ignore other factors such as air resistance, so the models that scientists use to study physical phenomena are likely to be more complex than what they’ve seen here.
If time permits, consider addressing a common misconception: that a graph of a quadratic function that represents distancetime relationship shows the physical trajectory of the object. Ask students to draw a sketch of what a bystander would see if they are facing the cannon as the ball is being launched. If needed revisit the Geogebra applet “Galileo and Gravity” (from the previous lesson) to further emphasize this difference.
Clarify that the graph represents the height of the object as a function of time, not the path that the object travels. In the examples given here, the object just goes straight up and straight down.
6.4: Cooldown  Rocket in the Air (5 minutes)
CoolDown
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Student Lesson Summary
Student Facing
In this lesson, we looked at the height of objects that are launched upward and then come back down because of gravity.
An object is thrown upward from a height of 5 feet with a velocity of 60 feet per second. Its height \(h(t)\) in feet after \(t\) seconds is modeled by the function \(h(t) = 5 + 60t  16t^2\).
 The linear expression \(5 + 60t\) represents the height the object would have at time \(t\) if there were no gravity. The object would keep going up at the same speed at which it was thrown. The graph would be a line with a slope of 60 which relates to the constant speed of 60 feet per second.
 The expression \(\text16t^2\) represents the effect of gravity, which eventually causes the object to slow down, stop, and start falling back again.
Notice the graph intersects the vertical axis at 5, which means the object was thrown into the air from 5 feet off the ground. The graph indicates that the object reaches its peak height of about 60 feet after a little less than 2 seconds. That peak is the point on the graph where the function reaches a maximum value. At that point, the curve changes direction, and the output of the function changes from increasing to decreasing. We call that point the vertex of the graph.
Here is the graph of \(h\).
The graph representing any quadratic function is a special kind of “U” shape called a parabola. You will learn more about the geometry of parabolas in a future course. Every parabola has a vertex, because there is a point where it changes direction—from increasing to decreasing, or the other way around.
The object hits the ground a little before 4 seconds. That time corresponds to the horizontal intercept of the graph. An input value that produces an output of 0 is called a zero of the function. A zero of the function \(h\) is approximately 3.8, because \(h(3.8) \approx 0\).
In this situation, input values less than 0 seconds or more than about 3.8 seconds would not be meaningful, so an appropriate domain for this function would include all values of \(t\) between 0 and about 3.8.