# Lesson 17

These materials, when encountered before Algebra 1, Unit 7, Lesson 17 support success in that lesson.

## 17.1: Area Between Triangles (10 minutes)

### Warm-up

The purpose of this warm-up is to remind students how to solve quadratic equations using a variety of methods as well as determine how reasonable the solutions are in a situation. While quadratic equations often result in 2 solutions, sometimes only 1 of the values makes sense for the context. As students work with solving quadratic equations and connecting them to real situations, they should be mindful that some solutions may not be reasonable.

Monitor for students who solve the quadratic equation by:

1. Guessing and checking values
3. Factoring the equation

### Student Facing

The area of the shaded region from the image can be represented by the expression $$\frac{1}{2}(2+2a)(2+2a) - \frac{1}{2}\boldcdot2^2$$ which can be rearranged to $$2a^2 + 4a$$. To find the value of $$a$$ when the shaded area is 30 square centimeters, Mai sets up the equation $$2a^2 + 4a = 30$$.

1. One solution to the equation is $$a = \text{-}5$$. Find another solution. Explain or show your reasoning.
2. What do the 2 solutions to the equation represent in this situation? Do the values make sense?

### Student Response

For access, consult one of our IM Certified Partners.

### Activity Synthesis

The purpose of the discussion is to share methods for solving the quadratic equation as well as determining that some solutions may not be reasonable in a situation. Select previously identified students to share their methods for solving the quadratic equation. Ask students,

• “How many solutions do we expect a quadratic equation to have? How can this be seen in the quadratic formula?” (We expect 2 solutions, which can be seen in the quadratic formula with the $$\pm$$ since this results in 2 values (unless the value under the square root is zero).)
• “Although there are 2 solutions to this quadratic equation, Mai says there is only 1 way to make the shaded shape so that is has an area of 30 square centimeters. Do you agree? Explain your reasoning.” (I agree with Mai because one of the solutions involves a negative length which does not make sense in this situation.)

## 17.2: Getting the Ball Off the Roof (15 minutes)

### Activity

In this activity, students interpret a quadratic equation in context. In the associated Algebra 1 lesson, students solve quadratic equations in context using the methods they learned throughout the unit. The work of this activity supports students to interpret the context to set up the equations so their work can focus on solving the equations.

### Student Facing

A ball is kicked off the roof of a building so that its height above the ground, given in feet, $$t$$ seconds after it is kicked is represented by the equation $$h(t) = \text{-}16t^2 + 33t + 37$$.

1. At what height is the ball when it is kicked? Explain or show your reasoning.
2. At what height is the ball 2 seconds after it is kicked? Explain or show your reasoning.
3. What does it mean for the situation when $$h(t) = 8$$?
4. What does it mean for the situation when $$t = 1.3$$?
5. Graph the function.
1. Approximate the number of seconds after the ball is kicked when it will hit the ground. Explain how you know.
2. Approximate the number of seconds after the ball is kicked when it will reach its highest point. Explain how you know.
3. Approximate the number of seconds after the ball is kicked when it will reach its starting height again. Explain how you know.
6. Write an equation that represents the exact moment when the ball hits the ground.

### Student Response

For access, consult one of our IM Certified Partners.

### Activity Synthesis

The purpose of the discussion is to focus on interpreting the quadratic function in context. Select students to share their solutions. Ask students,

• “How could you find the exact time when the ball hits the ground?” (Solve the equation from the last question using a method like completing the square.)
• “The graph also touches the $$x$$-axis near the point $$(\text{-}0.806,0)$$. What does this point mean for the situation?” (It is meaningless in this situation since the equation only holds for non-negative times after the ball is kicked. In theory, it could mean that the ball was on the ground about -0.8 seconds before being kicked, but we know the ball was on the roof before it was kicked.)

## 17.3: Kicking the Field Goal (15 minutes)

### Activity

For students who question the initial height of the ball, tell students that the height of the ball in this situation is measured as the center of the ball since it will likely be spinning around the center as it moves through the air. The length of a football is 11 inches, so it is reasonable to expect the center of the ball to be approximately 0.5 feet above the ground when it is kicked.

### Student Facing

Andre kicks a football for a field goal. The height above ground, given in feet, $$t$$ seconds after it is kicked, is represented by the equation $$g(t)=\text{-}16t^2+56t+0.5$$.

1. At what height is the ball when it is kicked? Explain or show your reasoning.
2. At what height is the ball 2 seconds after it is kicked?
3. What does it mean for the situation when $$g(t)=10$$?
4. What does it mean for the situation when $$t=1.7$$?
5. Graph the function.
1. Approximate the number of seconds after the ball is kicked when it will hit the ground. Explain how you know.
2. Approximate the number of seconds after the ball is kicked when it will reach its highest point. Explain how you know.
3. Approximate the number of seconds after the ball is kicked when it will be 10 feet above the ground for the second time. Explain how you know.
6. Write an equation that would give the exact time when the ball is 10 feet above the ground.

### Student Response

For access, consult one of our IM Certified Partners.

### Activity Synthesis

The purpose of the discussion is to interpret the function in terms of the situation. Select students to share their solutions. Ask students,

• “Andre solves the last equation and finds solutions at $$t = 0.179$$ and $$t = 3.321$$. Do both solutions make sense? Explain your reasoning.” (Yes, they both make sense. The first value is when the ball is 10 feet above the ground while it is going up, then it comes back down again and is at a height of 10 feet above the ground again later.)
• “The point $$(3.596,\text{-}5)$$ is on the graph. What does this mean in the situation? Does it make sense?” (It means that almost 3.6 seconds after the ball is kicked, it is 5 feet below the ground. It probably does not make sense for the ball to go underground.)