Lesson 1
Finding Unknown Inputs
1.1: What Goes Up Must Come Down (5 minutes)
Warmup
This task reminds students that they can use a graph of a function to gain some information about the situation that the function models.
Every question can be answered, or at least estimated, by analyzing the graph. As students work, look for those who use the given equation to solve or verify the answers. For example, the first question can be answered by evaluating \(h(1)\) and the second question by evaluating \(h(8)\). Invite them to share their strategy during synthesis.
Launch
Given their earlier work on quadratic functions, students should be familiar with projectiles. If needed, give a brief orientation on the context. Tell students that there are devices that use compressed air or other means to generate a great amount of force and launch a potato or similarsized object.
(If desired and time permitting, find and show a short video clip of someone using an airpowered or a catapulttype device. Warn students that some of these devices can be dangerous and they shouldn’t try to build one without help from an adult.)
Student Facing
A mechanical device is used to launch a potato vertically into the air. The potato is launched from a platform 20 feet above the ground, with an initial vertical velocity of 92 feet per second.
The function \(h(t) = \text16t^2 + 92t + 20\) models the height of the potato over the ground, in feet, \(t\) seconds after launch.
Here is the graph representing the function.
For each question, be prepared to explain your reasoning.
 What is the height of the potato 1 second after launch?
 8 seconds after launch, will the potato still be in the air?
 Will the potato reach 120 feet? If so, when will it happen?
 When will the potato hit the ground?
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Activity Synthesis
Focus the discussion on how students used the graph to help them answer the questions. Encourage students to use precise mathematical vocabulary in their explanation. Invite students, especially those who do not rely solely on the graph, to share their responses and reasoning.
Point out that we can gather quite a bit of information about the function from the graph, but the information may not be precise.
 “Is there a way to get more exact answers rather than estimates?” (Some questions can be answered by evaluating the function. For questions that are not easy to calculate at this point, we can only estimate from the graph.)
 “Which questions in the activity could be answered by calculating?” (the first two questions, about where the potato is after some specified number of seconds) “Which questions were not as easy to figure out by calculation? (those about when the potato reaches certain heights)
 “How can we verify the time when the potato is 120 feet above the ground or when, exactly, it hits the ground?” (We can find the \(t\) values that make \(h(t)\) equal 120 by looking at the graph and then evaluating the function at those values of \(t\). To find when the potato hits the ground exactly, we would need to find the zeros of \(h\) by solving the equation \(h(t)=0\).)
Tell students that, in this unit, they will investigate how answers to these questions could be calculated rather than estimated from a graph or approximated by guessing and checking.
1.2: A Trip to the Frame Shop (20 minutes)
Activity
The purpose of this task is to motivate the need to write and solve a quadratic equation in order to solve a problem.
Students are asked to frame a picture by cutting up a rectangular piece of “framing material” into strips and arranging it around a picture such that they create a frame with a uniform thickness. The framing material has a different lengthtowidth ratio and is smaller than the picture, so students cannot simply center the framing material on the back of the picture.
Students are not expected to succeed in the task through trial and error. They are meant to struggle, just enough to want to know a way to solve the problem that is better than by guessing and checking. Some students may try writing an equation to help them figure out the right measurement for the frame, but they don’t yet have the knowledge to solve the quadratic equation. (The solutions to this equation are irrational, so it is also unlikely for students to find them by chance.)
This activity was inspired by the post “When I Got Them to Beg” on http://fawnnguyen.com/gotbeg/ by Fawn Nguyen, used with permission.
Launch
Ask students if they ever had an artwork or a picture framed at a frame shop. If students are unfamiliar with custom framing, explain that it is very expensive—often hundreds of dollars for one piece. The framing materials, which are cut to exact specifications, can be costly. The time and labor needed to properly frame a picture further push up the cost.
Tell students that they are now going to frame a picture, using a sheet of paper as their framing material. The sheet is to be cut such that:
 All of it is used. (Framing material is expensive!)
 The framing material does not overlap.
 The resulting frame has uniform thickness all the way around.
Distribute scissors, along with the pictures and the “framing material” (copies of the blackline master).
Supports accessibility for: Language; Memory
Student Facing
Your teacher will give you a picture that is 7 inches by 4 inches, a piece of framing material measuring 4 inches by 2.5 inches, and a pair of scissors.
Cut the framing material to create a rectangular frame for the picture. The frame should have the same thickness all the way around and have no overlaps. All of the framing material should be used (with no leftover pieces). Framing material is very expensive!
You get 3 copies of the framing material, in case you make mistakes and need to recut.
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Student Facing
Are you ready for more?
Han says, “The perimeter of the picture is 22 inches. If I cut the framing material into 9 pieces, each one being 2.5 inches by \(\frac{4}{9}\) inch, I’ll have more than enough material to surround the picture because those pieces would mean 22.5 inches for the frame.”
Do you agree with Han? Explain your reasoning.
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Activity Synthesis
Ask a few students to show their “frames.” Consider asking questions such as:
 “How did you decide how thick the frame should be?” (I tried a couple of different thicknesses to see if they would work. I first tried \(\frac12\) inch, but that didn't give enough to frame the entire picture. Then, I tried \(\frac14\) inch, but that was too thin.)
 “How did you know what thickness would be too large or too small?” (It is hard to tell, but I know that the framing needs to have a linear measurement of at least 22 inches, enough for the entire perimeter of the picture, plus some more length for the corners. If the frame is too thin, there will be extra material. If it is too thick, there won't be enough to get the 22plussomeinch length.)
 “How did you determine the thickness such that no framing materials were left unused? Is it possible to determine this?” (I tried to find a way to evenly spread the area of the framing material around the picture, but I wasn't quite sure how to do that.)
 “What frame thickness did you end up with?”
 “Were all the strips or pieces the same size?”
Students are likely to share the challenges they encountered along the way. Tell students that, in this unit, they will learn strategies that are more effective than trial and error or solving problems such as this one.
Design Principle(s): Cultivate conversation
1.3: Representing the Framing Problem (10 minutes)
Activity
This activity allows students to formulate a mathematical model around the framing task they saw earlier (MP4). Students are prompted to write an equation but are not expected to solve it at this point. In writing an equation and interpreting the solution in context, students practice reasoning quantitatively and abstractly (MP2).
The prompt to write an equation is left relatively open to allow for different approaches. For instance, students may think in terms of:
 The length and width of the framed picture. The length is 7 inches plus the thickness of the frame, \(x\), on either side, which gives \(7 + 2x\). The height is 4 inches plus the same thickness, \(x\), for the top and bottom, which gives \(4 + 2x\). The total area is \(7 \boldcdot 4 + 4 \boldcdot (2.5)\) or 38 square inches, so the equation is \((7+2x)(4+2x) = 38\).
 The area of the picture plus the area of the frame. The area of the picture is 28 square inches. The area of the frame, in square inches, can be found by decomposing it into four squares (in the corners) that are \(x\) by \(x\) or \(x^2\) each, two rectangles that are \(7x\) each (top and bottom), and two rectangles that are \(4x\) each (left and right). The equation is \(4x^2 + 2(7x) + 2(4x) + 28 = 38\) or \(4x^2 + 22x + 28= 38\).
As students work, monitor for different strategies students use. Invite students with contrasting approaches to share later.
Launch
Tell students that they are now to write an equation to represent the quantities in the framing problem. Consider arranging students in groups of 2 and asking them to think quietly about the questions before conferring with their partner.
If students have trouble getting started, suggest that they start by labeling the diagram with relevant lengths. Then, if needed, use scaffolding questions such as:
 “How could we show that the frame is the same thickness all the way around?” (Use the same variable, for example \(x\), to label the thickness of the strip of framing on all four sides.)
 “What is the combined area of the picture and the framing material?” (38 square inches, because \(7 \boldcdot 4 + 4 \boldcdot (2.5)=38\))
 “How could we express the width of the framed picture?” (\(7+2x\)) “What about the height of the framed picture?” (\(4+2x\))
 “Once the framing material is cut up and arranged around the picture, how could we express its area?” (Two rectangles with area \(x(7+2x)\) and two rectangles with area \(4x\), or \(2x(7+2x) + 2(4x)\))
Design Principle: Support sensemaking
Supports accessibility for: Conceptual processing; Visualspatial processing
Student Facing
Here is a diagram that shows the picture with a frame that is the same thickness all the way around. The picture is 7 inches by 4 inches. The frame is created from 10 square inches of framing material (in the form of a rectangle measuring 4 inches by 2.5 inches).
 Write an equation to represent the relationship between the measurements of the picture and of the frame, and the area of the framed picture. Be prepared to explain what each part of your equation represents.
 What would a solution to this equation mean in this situation?
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Anticipated Misconceptions
Some students may struggle to express the overall length and width of the framed picture because of trouble combining numbers and variables. Consider drawing a segment composed of 3 pieces of length 1.5 inches, 10 inches, and 1.5 inches, and prompting students to find the length of the entire segment. Then, change each 1.5 to an \(x\) and ask for an expression for the length of the entire segment (\(10+2x\)).
If students mistake the result of adding \(x\) and \(x\) as \(x^2\), consider drawing a square with side length \(x\) and ask students to write expressions for the perimeter and area. Then, ask them to point out the difference between multiplying to find area and adding to find total length. If needed, draw a few more segments that are decomposed into parts, with each part labeled with a number or a variable expression, and ask students to write expressions for the total length.
Activity Synthesis
Select students to present their equations and reasoning. If no students bring up one of the strategies shown in the Activity Narrative, bring it up. Consider asking students whether \((7+2x)(4+2x)=38\) and \(4x^2 + 22x + 28 = 38\) (or other correct equations that come up) are equivalent and how to check if they are.
Discuss what a solution to these equations represents. Make sure students understand that a solution reveals the thickness of the frame when all of the framing material (10 square inches) is used.
Solicit some ideas on how they might go about finding the solution(s). Students may mention:
 graphing the equation \(y=(7+2x)(4+2x)\) and finding a point on the graph with a \(y\)coordinate of 38
 evaluating the expression on one side of the equation at different values of \(x\) until it has a value of 38
 using a spreadsheet to evaluate the expression at different values of \(x\) and see what values produce an output of 38
We can see that the equations \((7+2x)(4+2x)=38\) and \(4x^2 + 22x + 28 = 38\) are each composed of a quadratic expression. Tell students that these are examples of quadratic equations. They will learn several techniques for using algebra to solve equations like this.
Formally, a quadratic equation is defined as one that can be written in the form of \(ax^2+bx+c=0\) where \(a\) is not 0. This formal definition will be introduced in the next lesson, when students work with equations of that form and make sense of them in context.
Lesson Synthesis
Lesson Synthesis
To highlight reasons for writing and solving an equation, discuss questions such as:
 “In this lesson, we saw a few examples where we knew the output of a function and were interested in finding the input that produced it. In the launched potato situation, what outputs were known?” (The potato’s heights or distances from the ground.) “What inputs did we want to know?” (The times at which the potato was at certain heights.)
 “In the framing problem, what was the output that we cared about?” (The total area of the framed picture, which was 38 square inches.) “What information was the input?” (The thickness of the frame that would produce that total area.)
 “Why might it be helpful to write an equation to represent a problem such as the one about framing?” (An equation helps us see the relationship between the quantities in the problem, including those whose values are unknown. We can solve for those values.)
 “In the framing situation, if \(x\) represents the thickness of the frame, in inches, what would a solution to the equation \((2x+7)(2x+4)=50\) mean?” (The thickness of the frame that would produce a total area of 50 square inches.)
 "Equations such as \((2x+7)(2x+4)=50\) and \(4x^2+22x+28=38\) are called quadratic equations. Why do you think equations like these are described as quadratic?" (Each equation has a quadratic expression in it. Each one relates quantities in a quadratic function.)
(Quadratic equations will be defined more formally in the next lesson. It is fine and expected that students have an informal or partial understanding of what they are at this point.)
1.4: Cooldown  Interpreting a Solution (5 minutes)
CoolDown
Cooldowns for this lesson are available at one of our IM Certified Partners
Student Lesson Summary
Student Facing
The height of a softball, in feet, \(t\) seconds after someone throws it straight up, can be defined by \(f(t) = \text16t^2+32t+5\). The input of function \(f\) is time, and the output is height.
We can find the output of this function at any given input. For instance:
 At the beginning of the softball's journey, when \(t = 0\), its height is given by \(f(0)\).
 Two seconds later, when \(t=2\), its height is given by \(f(2)\).
The values of \(f(0)\) and \(f(2)\) can be found using a graph or by evaluating the expression \(\text16t^2+32t+5\) at those values of \(t\).
What if we know the output of the function and want to find the inputs? For example:

When does the softball hit the ground?
Answering this question means finding the values of \(t\) that make \(f(t)=0\), or solving \(\text 16t^2+32t+5=0\).

How long will it take the ball to reach 8 feet?
This means finding one or more values of \(t\) that make \(f(t) = 8\), or solving the equation \(\text 16t^2+32t+5=8\).
The equations \(\text 16t^2+32t+5=0\) and \(\text 16t^2+32t+5=8\) are quadratic equations. One way to solve these equations is by graphing \(y = f(t)\).
 To answer the first question, we can look for the horizontal intercepts of the graph, where the vertical coordinate is 0.
 To answer the second question, we can look for the horizontal coordinates that correspond to a vertical coordinate of 8.
We can see that there are two solutions to the equation \(\text 16t^2+32t+5=8\).
The softball has a height of 8 feet twice, when going up and when coming down, and these occur when \(t\) is about 0.1 or 1.9.
Often, when we are modeling a situation mathematically, an approximate solution is good enough. Sometimes, however, we would like to know exact solutions, and it may not be possible to find them using a graph.
In this unit, we will learn more about quadratic equations and how to solve them exactly using algebraic techniques.