Tuesday, July 31, 2007

Another Quadratic Function Challenge



The line y = k intersects the graph of y = 3x2 in points A and B. Points C and D are on the x-axis and ABCD is a rectangle. If the area of ABCD = 128/9, what is the value of k? Calculators not allowed. Show your method clearly!

The above question is designed for students in their second year of algebra or precalculus. It can also be used for practice for the Algebra 2 questions on the SAT, although it is somewhat above that level. Many students, including the more advanced, tend to struggle with problems like these because they don't have that much experience with coordinate geometry questions. These types of problems are critical for their later development. Once the students have done a few of these they do not find them so formidable. A useful pedagogical tool is to let them try it, review the method clearly, then erase the board and call on students to recall each step. Tell them you will do this, encouraging them to take good notes and pay careful attention! When using it on an assessment, make it a bonus the first time, then make it count. Some of my readers may recall a similar parabola problem a few months ago.

15 comments:

Jackie said...

Well, here's my method...

if we let the points of intersection of the parabola and the line be (a, k) and (-a, k). Then the area of the rectangle is (2a)(k).

So we have:
2ak=128/9
3a^2=k (this point also satisfies the equation y=3x^2)

Substituting into the first equation we have
2a(3a^2)=128/9
6a^3 = 128/9
a^3 = 64/27
a = 4/3

So then k = 16/3

Anyone do something different?

Dave Marain said...

Ah, jackie--
You made it seem so simple. That's because, as a math person, you've already developed a strong conceptual understanding of how the coordinates of the points link to the base and height of the rectangle. Further, the obvious substitution of 3x^2 for the height may not be so obvious to all! Now why do you think that so many of the talented students with whom I am working didn't make such an obvious connection? Perhaps, because their 'base' of experience in coordinate problems has been limited. Where would they see many problems of this type? Students need to make these connections in Algebra 2. Some texts are moving in this direction. If the skills and concepts inherent in these coordinate problems are valued enough to be assessed on state tests (not just on SATs), then they will begin to appear more in textbooks. What is assessed gets taught!
Students should also be encouraged to use symmetry wherever possible, particularly since that is a key feature of the parabola. Thus, we can work with the half of the rectangle in the first quadrant whose area is 64/9. Anytime students can avoid working with negative coordinates is beneficial IMHO!
I'm quite sure that many readers would not consider this problem to be so challenging. Well, if you're in the classroom in September, give it to your precalculus students and let me know the results. Will they find it trivial?
When all is said and done, the kinds of problems I am publishing can be viewed purely as extra challenges or they can constitute an implied curriculum when one considers the body of concepts, skills, strategies, algorithms, and reasoning needed to solve them. My goal is and has always been to challenge all of our students by asking harder questions on a regular basis, not just as an extra credit problem for the honors classes. I wonder how these problems are really being viewed by those educators, curriculum specialists and publishers who happen to read them? I already know how my regular readers view them and that's very gratifying...

Jackie said...

Dave,

Thanks. As I was working the problem, I thought about issues that students might have. My list was similar to yours (symmetry, substitution, trying to use a distance formula for the length of the base...).

I do not think this would be an easy problem for my students. I'll be sure to throw it in my pre-calc class this fall.

Sadly, I don't see many of these types of problems in either our Honors Geometry textbook or our Alg II text. My views on their inclusion in the curriculum are, I think, similar to yours. I'd love to hear opinions of anyone else out there.

It never occurred to me to use only the area of 1/2 the rectangle - thanks!

Mike said...

Dave,

I'll be starting a new career as a high school math teacher this fall, teaching geometry and algebra 2. I find your problems to be excellent. Thank you for writing them! I will do my best to make use of them, and similar problems.

Mike

Mike said...

... oh, and about the problem: I saw the symmetry and deduced that the right half of the rectangle had area of 64/9. From there I figured that xy = 64/9, which implies that y = 64/9x, and that y = 3x^2. Then I set 64/9x equal to 3x^2 and solved for x, then y.

Dave Marain said...

thanks, mike--
Comments like these are so encouraging! As you can imagine, I'm not pulling these out of my hat - they require considerable thought and reworking before I publish them. It's particularly nice to know that they may be of some benefit to math teachers starting their career. I would love to hear students' reactions in the fall!

Totally_clueless said...

The meaning of co-ordinates is a pretty basic concept, and I find it disturbing that talented students are not making the connection. I find repeated drills of similar problems with variations to be invaluable in ensuring that students get the basic concept (experience only from working with my child, not with other students. The Alg 1,2 and Geometry textbooks used here seem to have enough problems in the exercises that doing most/all of them was quite good in getting the concept across, but this may be a biased viewpoint due to a small sample space of students)

There is also the obvious generalization to the calculus students: The area between the line and the parabola (below the line) is (say) 16. Find the value of k.

TC

Dave Marain said...

tc--
I completely agree. This problem should not have caused much difficulty. The issue was experience. Every student should have had experience with these problems, not just those students whose textbooks include practice with these. Some texts do more than others but the curriculum and standards should be the same no matter what, right?!? If we agree that all students from Algebra 2 on should be able to handle this, then the problem is not the students or the teachers. The problem is lack of consistency in curriculum from district to district, state to state. That's been my concern from the beginning, but let's not go there at this moment...

The Calc application is nice. Thanks.

Jackie said...

I was thinking about this problem (driving home from a family picnic) and how I could extend it - here's my thought:

Change the function to y = 3(x-4)^2, leave the rest the same and ask for the x coordinates of point B (or A or...).

Any thoughts on this extension?

Dave Marain said...

Nice jackie!
You're getting too much like me, though, thinking of extensions while driving!!
Do you think many students would simplify your problem by translating your parabola 4 units to the left, solving the y = 3x^2 problem, then adding 4 to the x-coordinates producing 8/3 and 16/3.
These problems are almost identical to my post of 2-16-07. There's a link to this at the bottom of the body of the current post. Revisit it - you'll see the similarity. The difference is that the current problems are less abstract. That's why I wrote this. I felt the original was a bit intimidating.

Jackie said...

Ahh - thanks. As a new reader, I haven't yet gone back to read prior posts. Perhaps that will be my "break" this week as I seriously begin planning for next(this?) year.

I'm not sure how many would tackle the problem by performing the translation first. I think this would be "easier" than thinking of the points as (a + 4, k) and (-a + 4, k) using notation from my original comment.

As for getting to be too much like you - I consider that a compliment! My husband has stopped asking what I'm thinking about while we're driving :)

jonathan said...

I'll share this with colleagues, but... This year I am teaching Algebra I, Geometry, and my elective (combinatorics, aka counting, in the fall), and no advanced algebra or pre-calculus. A lot of stuff I can't use myself. Oh well.

Miss Kopels said...

I like Jackie's method, but my students and I developed a different strategy. We knew the height of the rectangle was k, and that y=3x^2, so we could solve for the roots x in terms of k to get our width.

so when y = k, y = 3x^2 (substitute)
k = 3x^2
k/3 = x^2
x = sqrt(k/3) and -sqrt(k/3)

Then our length is 2 * sqrt(k/3)...
so we have

2k*sqrt(k/3)=128/9 --> pull in 2k

sqrt(4k^3/3) = 2^7/3^2 ---> square both sides

4k^3/3 = 2^14/3^4 ----> get k by itself

k^3 = (2^14/3^4) * (3/4) ----> reduce

k^3 = 2^12/3^3 --------> take cubed root

k = 2^4/3 = 16/3


Requires strong knowledge of how to work with powers and roots

Miss Kopels said...

I like Jackie's method, but my students and I developed a different strategy. We knew the height of the rectangle was k, and that y=3x^2, so we could solve for the roots x in terms of k to get our width.

so when y = k, y = 3x^2 (substitute)
k = 3x^2
k/3 = x^2
x = sqrt(k/3) and -sqrt(k/3)

Then our length is 2 * sqrt(k/3)...
so we have

2k*sqrt(k/3)=128/9 --> pull in 2k

sqrt(4k^3/3) = 2^7/3^2 ---> square both sides

4k^3/3 = 2^14/3^4 ----> get k by itself

k^3 = (2^14/3^4) * (3/4) ----> reduce

k^3 = 2^12/3^3 --------> take cubed root

k = 2^4/3 = 16/3


Requires strong knowledge of how to work with powers and roots

Dave Marain said...

Miss Kopels,
Congratulate your students for me. The exponent manipulations were awesome and a great example of what happens when calculators are taken away! Yi