Analysis
There is background knowledge required for students to work on the problem suites in this unit. Assuming they recognize the general form of a quadratic function as ax 2 + bx + c, students must, at the lowest level, be able to solve equations by using tables and/or graphs on a graphing calculator. At a higher level, students should be able to solve quadratic functions by algebraic methods including square roots, factoring, completing the square or using the Quadratic Formula. Students should also be able to find the vertex (coordinates of the maximum or minimum point) by using a graphing calculator or algebraically from any form of the quadratic function.
We spent considerable time in our seminar categorizing problems in a problem suite according to similarities and differences. The first order of business is to define a problem territory. My problem territory is Quadratic Functions, which I am breaking down into two subgroups, namely Projectile Motion and Geometry. The second order of business is to designate the dimensions that I use for grouping and categorizing the problem suites that I assembled. In this section, I will describe the dimensions in detail using examples. Many more word problems can be found in Appendix B, broken down according to the dimensions I describe.
Problem Suite A: Projectile Motion
The projectile motion problems in my problem suite come from the equation (which is derived from the laws of physics)
h(t) = h 0 + v 0 t + ½at 2
where h(t) describes the vertical height of an object with respect to time, t (seconds), and
h 0= initial height
v 0= initial upward velocity
a = acceleration due to gravity (a = -32 ft/s or -9.8 m/s).
Dimension 1A: Write the equation
The problem suite begins with students practicing writing projectile motion equations. I would expect students to extract the initial height and initial upward velocity from the information given in the word problem and substitute these values for h 0 and v 0 , respectively, in the equation given above. They also need to select the appropriate value for a, depending on the units (feet or meters) used in the problem.
Dimension 2A: Evaluate the equation
The simplest question to ask students is to find the height of an object at a given time. In this case, the student simply substitutes the time (in seconds) in place of t in the equation. By doing the arithmetic (square the number of seconds and multiply by the a value times ½, then add the product of bt, and then add the value of c), the student is evaluating the equation to find the height.
Dimension 3A: h 0 = 0; find the time it takes an object to return to the ground
If a projectile is launched from the ground, the initial height is zero, or, in terms of the quadratic function ax 2 + bx + c, c = 0. For example, consider a soccer ball goal kick that a defender kicks from the 6-yard line at an initial upward velocity of 52 ft/s. Since the velocity is given in ft/s, the acceleration in this problem will be -32 ft/s, leading to the equation, h(t) = -16t 2 + 52t. An equation in this form will always be factorable by factoring out the variable, t, giving h(t) = t(-16t + 52). This equation can be factored further by factoring out a common factor of -4, giving h(t) = -4t(4t - 13). A quadratic equation in this form can be solved for x-intercepts ("zeroes") or coordinates of the vertex, as described below. The Quadratic Formula will yield the same result, but the factored format leads to solutions quickly, as demonstrated in this section and the next.
To find the time it takes for the ball to return to the ground, first students must set the function equal to zero because the height of the ball on the ground is zero. Next, they need to find the x-intercepts, also known as the roots or the zeroes of the equation. Mathematically, when they find the roots of an equation where h 0 = 0, they will find two of them. One of the roots in this case will always be zero because the object is on the ground at the start. So, it's the other root that answers the question of when the object returns to the ground.
Continuing with the example started above, solving the equation -4t(4t - 13) = 0 can be done by setting each of the two factors equal to zero. So, -4t = 0 when t = 0 and 4t - 13 = 0 when t = 13/4. Therefore, the soccer ball will return to the ground after 13/4 = 3.25 seconds in the air. Again, the Quadratic Formula will work to find the "zeroes." And, it's always a good idea to confirm the answers by checking them against a table or graph on the graphing calculator.
Dimension 4A: Find the time it takes an object to reach its maximum height
Here, students must recognize that this question is asking for the x-value (time) that would give the maximum y-value. In other words, they are looking for the x-coordinate of the vertex. Since we can rewrite quadratic functions in vertex form by "completing the square," we know that every quadratic function is a parabola with a vertical line of symmetry that passes through the vertex. Because of that symmetry, two points on the parabola having the same y-value (as in the "zeros") must be reflections of each other across the line of symmetry. Therefore, the line of symmetry must be halfway between them. There are several ways for students to find the coordinates of the vertex point, but I will continue with the soccer example that is already in factored form. We found that the x-intercepts are 0 and 3.25 seconds. To find the line of symmetry, we find the average of 0 and 3.25, which is 1.625. Since the vertex is the only point on the parabola with the maximum y-value, it must be on the line of symmetry. So for this example, the time it takes the soccer ball to reach its maximum height will be 1.625 seconds.
Dimension 5A: Find the maximum height reached by an object
Once you know the time it takes an object to reach its maximum height, what you really know is the x-coordinate of the vertex. So, to find the maximum height, simply evaluate the quadratic function for that x-value. For some reason, my students often forget that they know how to "plug" a number (x-value) into an equation to find its corresponding y-value. Perhaps, now that I included Dimension 2A (evaluating) in this problem suite, my students will be more successful at remembering to use the x-value of the line of symmetry to find the corresponding (maximum) y-value of a function.
To complete the soccer example, the maximum height of the soccer ball can be found by evaluating h(1.625) = -4(1.625)(4-1.625 - 13) = 42.25. Therefore, the maximum height reached by the soccer ball is 42.25 feet. Again, we should verify our answers for the two coordinates of the vertex by finding them on the graphing calculator.
Often, one problem will ask students to find all of the things I separated into different dimensions: the time it takes an object to return to the ground, the time it takes to reach a maximum height, and what that maximum height is. I am choosing to keep the questions separated so that students must consider what they need to find, rather than just going through a process of finding "everything."
Dimension 6A: h 0 ¹ 0; find the max, find the time to reach max or ground
When the initial height of the object is not zero, the quadratic function in the form ax 2 + bx + c will contain all three terms with c = h 0 . Students will be asked to answer the same three questions previously discussed. The difference will probably be in the solution method. Most likely, the quadratic function cannot be factored easily and students will use the Quadratic Formula to find the x-intercepts. When h 0> 0, one of the x-intercepts will be negative. I would hold a discussion to be sure students understand why a negative time for the ball to be on the ground does not apply to these situations. Looking at a graph of the function on the calculator and seeing that the y-intercept is equal to h 0 (i.e. the graph shows the ball starting above the ground represented by the x-axis on the graph) should help them see that the graph to the left of the y-axis is excluded in this situation and the positive x-intercept represents when the ball hits the ground. This time shows up clearly on the graph, as well.
I will use another soccer example to demonstrate two other algebraic methods for finding the coordinates of the vertex. Suppose a player bumps the ball with her head. If she is standing so that her head is 5 feet above the ground when she bumps it and the ball goes straight up with an initial velocity of 12 ft/s, then the equation would be h(t) = -16t 2 + 12t + 5. The first method for finding the coordinates of the vertex is "completing the square." The steps in the process would be:
So, the original equation in the form ax 2 + bx + c has been transformed into the vertex form (x
+ h) 2 + k where ( -h , k ) represents the coordinates of the vertex. By transforming the original
equation, we can see that the vertex point (in a more simplified form) is 
.
Returning to the example, the soccer ball reaches its maximum height of 29/4 = 7.25 feet in 3/8= 0.375 seconds.
The second method for finding the coordinates of the vertex uses the Quadratic Formula. Once again, using the fact that the vertex of the parabola lies on the line of symmetry, we can find the line of symmetry from the first part of the Quadratic Formula, namely, x = (-b/2a)x. For the same soccer example, the line of symmetry occurs at x=-12 / -32 = 3/8 = 0.375 seconds. Then evaluating the equation h(0.375) = -16(0.375) 2 + 12(0.375) + 5 = 7.25 feet agrees (fortunately) with the result we got above.
Dimension 7A: Find the time(s) to reach specified height, h(t)¹ 0
If students are solving these equations using tables and graphs on a calculator, this dimension is a non-issue. They are just looking for the x-value(s) that corresponds to a different number in the y-column of the table, or a specific y-value on the graph.
This dimension does add complexity to solving quadratic functions algebraically because the quadratic expression is set equal to a number other than zero, as in ax 2+bx+c = h. However, all algebraic solution methods that we teach are based on finding the x-value(s) that make y = 0. So, students must manipulate the equation to make something equal to zero. The manipulation involves subtracting the specified height, h, from both sides of the equation. Only the c-value is changed on the left-hand side, and the resulting equation ax 2+bx+c' = 0 (c' = c - h) is still quadratic, but now the quadratic expression is set to zero. In this form we can solve it by factoring or using the Quadratic Formula to find the roots.
Example: Suppose a baseball is thrown straight up with an initial velocity of 19 m/s from a height of 2 m above the ground. When is the ball 15 m above the ground?
The equation to solve is -4.9t 2 + 19t + 2 = 15. To begin, subtract 15 from both sides of the equation giving -4.9t 2 + 19t - 13 = 0. Next, I would apply the Quadratic Formula giving x = 0.89 seconds and x = 3.0 seconds. There should be two times that a ball is at the same height-once on the way up, and once on the way down. Of course, we should confirm these times by checking a graph, table, or substituting the results into the original equation.
Dimension 8A: Find the initial upward velocity
For more practice with algebraic manipulations, as well as solidifying the projectile motion ideas, problems in this dimension give information about a certain point on the graph (time, height) and ask for the initial upward velocity. For example: If a softball player hit the ball from a height of 1.2 m above the ground and it hit the ground after 2.75 sec, what was the initial upward velocity of the ball when it was hit? This problem does not provide a lot of information outright, but we know the force of gravity, and we know that the height of the ball when it hits the ground in 2.75 sec is zero. With this added knowledge, we can write the equation 0 = ½(-9.8)(2.75) 2 + v 0(2.75) + 1.2 and solve algebraically for v 0 = 13 m/s.
Another way to ask for v 0 would be to give the time and height of the maximum and ask for the initial upward velocity. For the same softball situation, the problem would be: If a softball player hit the ball and it reached its maximum height of 9.8 m in 1.33 sec, what was the initial upward velocity of the ball when it was hit? Substituting the vertex (k,h) into the quadratic y = a(x - k) 2 + h, we get y = -4.9(x - 1.33) 2 + 9.8, which can be written in expanded form as y = -4.9x 2 + 13x + 1.1. From this we see that v 0 = 13 m/s which agrees with our answer above!
Dimension 9A: Find the initial height
Similar to Dimension 8A, we can give students enough information to solve for the initial height of an object. They would need to take the information given, add some implied information (i.e. gravity, using the correct units) and substitute into some form of the projectile motion quadratic equation. Solving for h 0 then requires applying algebraic skills.
Dimension 10A: Interpret the result/compare result to information given
So far, all of the problems in the suite have asked students to find the value of one of the variables in the word problem. In this group, students must figure out what variable they are looking for and then use the result to answer a question. For example:
A woodland jumping mouse hops along a parabolic path given by y = -0.2x 2 + 1.3x where x is the mouse's horizontal position and y is the corresponding height, both in feet. Can the mouse jump over a fence that is 2 ft high? The answer is yes. But to find the answer, students must find the maximum height the mouse can jump. Although this problem brings in horizontal distance as the x-variable, rather than time, the question still requires finding the y-value (height) of the vertex point by any method they choose. The maximum height the mouse jumps occurs at a horizontal distance of 3.25 ft and is 2.11 feet. Since the maximum height is greater than the fence height, yes, the mouse can jump over it.
Dimension 11A: Including the x and y components of velocity
If I have a very advanced group of students, or ones that solve all problems in the problem suite described so far, I would challenge them with problems that require using trigonometry to determine both the vertical and horizontal components of the initial velocity. These problems are typical of what they will see in Physics. In our curriculum they have already studied trigonometric relationships, so these problems are within their grasp. I think the greater challenge will come from the multiple steps required to answer these questions. I am including some of these problems in the Appendix, but will not include any examples here.
Problem Suite B: Geometry
Within the Geometry problem suite, students will encounter many of the same dimensions that I discussed within the Projectile Motion problem suite. They will encounter problems where c = 0 and c ¹ 0. They will find problems where they must manipulate the equation to equal zero (as described in Dimension 7A above) before applying one of the algebraic solution methods. They will be asked to find the dimensions that yield the maximum area or volume and/or what the maximum area or volume is. In some problems they will need to interpret their answer in order to answer the question. Since students already worked with these dimensions as they related to projectile motion, I am assuming they are fairly adept at solving them, and I will not repeat them here. Instead, the dimensions I will describe are concerned with how to set up the quadratic equations that need to be solved.
For each of the Geometry problems, I would strongly recommend drawing a picture to visualize the problem and labeling the dimensions given. At first students may need help labeling the dimensions in terms of only one unknown, so that they have only one variable in the equation. They will also need to know, or have available to them, basic area, surface area and volume formulas for different shapes and figures.
Dimension 1B: Find the maximum area, given the perimeter
Beginning with rectangular areas, there is a category of problems that provide a perimeter and ask students to find the maximum area that can be enclosed. For example, if you have a 500-foot roll of fencing and a large field, and you want to construct a rectangular playground, what is the largest possible area, and what are its dimensions?
I would first insist that my students draw a rectangle to represent the playground area. Next, they need to label the dimensions. The names "l" and "w" work, but that means there are two variables to solve for. The 500 ft is the perimeter and can be used to relate the length and width of the playground. In other words, 2l + 2w = 500. Solving for l (it could be w instead) and simplifying, l = 250 - w. Now, using the area formula for a rectangle, we can write A = lw = (250 - w)w, which is a quadratic function of w. Since we are looking for the maximum, we can leave it in this factored form to find the roots, w = 0 and w = 250. The maximum will occur halfway between the roots, on the line of symmetry at w = 125. So, the width of the playground area should be 125 ft, and, substituting, the length should be 250-125 = 125 ft, and its maximum area would be 125 2 = 15,625 ft 2.
After doing several problems of this type, I would hope that some students recognize that the maximum area for a given perimeter occurs when the rectangle is a square. I would review that observation during a short class discussion. It is an observation that many of my students remember from previous math classes, but it never hurts to reinforce things when they reach the same conclusion from another direction.
There are further subcategories for finding the maximum area, given the perimeter. Students may be asked to find the maximum area of a rectangular area when one side uses a physical boundary and the perimeter refers to only three sides of the rectangle. Altering the playground problem above, if one side of the playground is bordered by a school building, what would be the maximum area, and what are its dimensions? In this case, 500 = l + 2w (or 2l + w), so l = 500 - 2w. The quadratic function for area would be A = (500 - 2w) w. The zeroes would be w = 0 and w = 250, and the maximum would occur at w = 125. The dimensions do change, however. While the width of the maximum area is still 125 ft, the length would be l =500 - 2(125) =250 ft and the maximum area for the playground would be (250)(125) = 31,250 ft 2 (twice as large as the previous example!).
Another subcategory occurs when the perimeter must enclose 2 or more areas that need to be maximized. Continuing with the playground example, if the 500 ft of fencing must enclose two separate playgrounds for different age groups and both must enclose the same area, the picture would look like this:
Then P = 2l + 3w = 500 and l = 250 ñ (3/2)w. Area = (250 ñ (3/2)w)w. The zeroes are w = 0 and w= 500/3, so the maximum area will occur when w = 250/3. The maximum area for both playgrounds together would be approximately 10,417 ft 2 with dimensions of 125 ft by 250/3 ft. The area for each playground would be approximately 5,208 ft 2 with dimensions of 62.5 ft by 250/3 ft.
The final subcategory is to vary the shape of the area enclosed by a given perimeter. In other words, students may need to use the area formula for shapes other than rectangles, depending on the information given in the word problem.
Dimension 2B: Find the dimensions, given the area and perimeter
In my search through textbooks and Internet sites, I found many word problems that state the perimeter and required area for a region, and students are asked to find the dimensions that satisfy both. An example of this type would be: A student environmental group wants to build a rectangular ecology garden. The area of the garden should be 800 ft 2 to accommodate all the species of plants the group wants to grow. A construction company has donated 120 ft of fencing to enclose the garden. What should the dimensions of the garden be? In this case, P = 2l + 2w = 120, or w = 60 - l. Then A = l(60 - l) = 800. To solve, I would distribute the l, subtract 800 and rearrange the order to get -l 2 +60l - 800 = 0. There are two solutions, l = 20 and l = 40. In this example, both solutions work (the garden doesn't know which is length and which is width), and both solutions yield the same dimensions. The garden should be 20 ft by 40 ft.
Dimension 3B: Borders
Another category of area problems that results in quadratic functions involves borders. In each problem, the border will be a uniform width, x, surrounding the inner region. This dimension can be broken down into four subdivisions, two of which have a very subtle difference. Sometimes, the word problem presents the specific dimensions (as in length and width of a rectangle) of the inner area (we can calculate the area from the dimensions) and the area of the entire region after the border area has been added. Other times, we are given the specific dimensions of the outer area, and the area of the inner region. For rectangular examples of these two types, we either add 2x (x in each direction) to each of the inner dimensions, or subtract 2x from each of the outer dimensions (again, x in each direction).
Example: An elementary teacher wants to paint a 4-square court in the center of a 20 ft by 30 ft fenced area. If the teacher wants a walkway of uniform width around the court that leaves a court area of 336 ft 2, how wide is the walkway? The outer (original) area is 20 x 30 = 600 ft 2 and the inner area is 336 ft 2. Since the walkway must be the same width on all four sides of the rectangle, the inner width can be represented by 20 - 2x, and the inner length can be represented by 30 - 2x. Then the inner area will be (20 - 2x)(30 - 2x) = 336. Expanding, subtracting 336, and simplifying gives us 4x 2 - 100x + 264 = 0. The two solutions are x = 3 and x = 22. Since the walkway cannot be wider than the width, x = 22 is impossible, and the walkway must be 3 ft wide. (Check: 14x24 = 336 ft 2)
The third subdivision is very similar to the first two, except that the area of the border is given. Problems of this type require adding the border area to the inner area or subtracting the border area from the outer area when writing the representative area equation. The fourth subdivision would be for shapes that are not rectangular. The formulas would differ, but they are solved in the same manner.
Dimension 4B: Volume
The most common variety of volume problems that result in quadratic functions are those that begin with a rectangular piece of cardboard/metal. If a square is cut from each of the four corners and the sides folded up, it forms a box/tray without a lid. In some of the problems, students are given the side length of the squares cut out, while in other problems they are given the dimensions of the original material and must find the size of the square cutout.
Example: A square piece of cardboard was used to construct a tray by cutting 2-inch squares out of each corner and turning up the flaps. Find the size of the original cardboard if the resulting tray has a volume of 128 in 3. Since the original cardboard is a square of length x, the length of each side of the base of the tray after cutting out the squares can be expressed as x - 4 (2 in from each end). Then the volume formula for a "box" gives V = lwh = 2(x - 4) 2 = 128. After expanding, distributing, subtracting 128 and simplifying, we get 2x 2 - 16x - 96 = 0. The two solutions are x = 12 and x = -4. Since a length cannot be a negative number, the original length of each side of the cardboard was 12 inches. (Check: 2x8x8 = 128 in 3)
Dimension 5B: Pythagorean Theorem
I must admit that the nearly all of quadratic problems that I found that required the Pythagorean Theorem are contrived problems. However, I include them in this unit because they are good reinforcement for quadratic functions, algebraic manipulations and Pythagorean Theorem.
Example: A nature conservancy group decides to construct a raised wooden walkway through a wetland area. To enclose the most interesting part of the wetlands, the walkway will have the shape of a right triangle with one leg 700 yd longer than the other and the hypotenuse 100 yd longer than the longer leg. Find the total length of the walkway. This problem is asking students to find the perimeter of the triangle. We start by expressing the lengths of each side in terms of the length of the shortest leg, x. Then the longer leg has length x +700, and the hypotenuse has length x + 800. Applying the Pythagorean Theorem, we get x 2 + (x + 700) 2 = (x + 800) 2. After expanding, rearranging, simplifying, etc., we have the equation x 2 - 200x - 150,000 = 0 to solve. The solutions are x = 500 and x = -300. Again, since length cannot be a negative number, the length of the legs are 500 yd and 1200 yd, and the length of the hypotenuse is 1300 yd. The length of the walkway is then 500 + 1200 + 1300 = 3000 yd. (Check: 500 2 + 1200 2 = 1300 2)
Dimension 6B: Surface Area
I only found a few problems involving surface area, but they were different enough to include in this unit for a change of pace. The problems can be found in the Appendix but can be omitted because of time constraints, if necessary.
Dimension 8B: Dilations
Dilations form their own problem suite. I use area problems, described in the dimensions above, as a basis. I ask students to double or triple the area, make a prediction about the new dimensions of the figure. Then they calculate the new dimensions, and finally, compare their prediction to their calculated dimensions. Or, I ask students to double (for example) the dimensions of a figure, predict the new area, calculate the new area and compare the two. As students compare their predictions to their calculations, I expect them to reason why their predictions were correct or incorrect. Their reasoning will be a source of classroom discussion to help students internalize the effects of scale factors on area, perimeter and volume, my personal mission!
Example: A plumbing contractor realized he needed more storage space for his supplies. If he wants to double the space that he has now, a 10 ft by 12 ft shed, by adding the same amount to both the length and width, what are the new dimensions of the shed? I would expect students to predict the new space to be 20 ft x 24 ft (even though they are ignoring the condition of adding the same amount to length and width). To calculate the new dimensions, let x be the number of feet added to each dimension. Then, (10 + x)(12 + x) = 2(10-12) = 240. After expanding and manipulating, the equation to solve is x 2 + 22x - 120 = 0, yielding x » 4.5 and x » -26.5. Since length cannot be negative, the amount to add to each dimension is 4.5 ft. Thus, the new storage area would be 14.5 ft by 16.5 ft giving an area of 239.25 ft 2, essentially double the original 120 ft 2, as desired. If each of the dimensions were doubled (as in the prediction above), the new area would be 480 ft 2; four (2 2) times the original area! Furthermore, the average ratio of new to old dimensions (14.5/10 & 16.5/12) is 1.41»√2, an observation that I will be sure to point out if my students don't see it themselves. Also, a follow-up discussion on similarity with respect to multiplying versus adding to alter dimensions might be appropriate.
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