Another Quadratic Function SAT Problem

Have you noticed the SAT Tips of the Week in the sidebar? These are intended both for math teachers and students.

The "new" SAT has a few 2nd year algebra questions and, typically, there is at least one 'parabola' problem usually expressed in function form. Here are two different versions - one multiple choice and one "grid-in" (student-generated response).
Can you predict which one might give most students more difficulty?
It might be interesting to list all of the skills, knowledge and concepts being tested here. Are all of these typically included in your Algebra 2 course? Do students get enough exposure to these kinds of problems?


Version I
For some constant r, the graph of the quadratic function f(x) = -x² + 2rx is a parabola with x-intercepts at P and Q and vertex V. What is the area of ΔPQV, in terms of r?

(A) r² (B) r³ (C) 2r²

(D) 2r³ (E) 4r³


Version II

For some constant r, the graph of the quadratic function f(x) = -x² + 2rx is a parabola with x-intercepts at P and Q and vertex V. If the area of ΔPQV equals 27, what is the value of r?

Answers, solutions, strategies and comments will appear below the Read more...



Answers/Solutions/Comments/...

Version I
Answer: (B) r³

Possible Solution (no frills):
Factoring, we have f(x) = -x(x-2r); x-intercepts are 0 and 2r. Therefore base of triangle has length 2r.
The x-coordinate of the vertex is r (why?), so y-coord = f(r) = -r(r-2r) = r².
Area of triangle = (1/2)(2r)(r²) = r³.

Version II
Answer: r = 3

Possible Solution:
From Version I, we obtain r³ = 27, so r = 3.

Comments

• These kinds of questions typically appear among the last 3-4 problems on a section, meaning they are of above-average difficulty. Students who are in Algebra 2 or beyond should definitely attempt it. After reviewing it, most students may conclude it's not very hard at all!
  
• Testmakers are more frequently using a parameter like 'r' to make it more difficult to merely punch it into the graphing calculator and read off the intercepts and vertex.
• This kind of question could also appear on standardized tests like the Algebra 2 End of Course Exam from Achieve/ADP or other state tests.
• Pick up a copy of 10 Real SATs from the College Board to find several other practice problems like this.
• One could modify and extend these problems in many ways. For example inscribe the parabolic region (bounded by the cruve and the x-axis) in a rectangle and determine its area, a simple variation. More interestingly is to note that the ratio of the area enclosed by the parabola to the area of the rectangle is 2:3, a famous result proved in Calculus.
• Skills, knowledge required for this question? Worth enumerating in my opinion...
  
• I also believe strongly that our students should be tackling these kinds of problems on a regular basis to deepen their understanding of the relationship among the function, the coordinates of key points and the geometry. This used to be known as Analytic Geometry.
  

...Read more

When Curves Collide: Quadratic Systems Explored...

[Another Update: Mutiple solutions to (a) and (b) are now provided in the comments (as of 10:50 PM 5-17-07).]

Target Audience: Algebra 2, Advanced Algebra, and beyond...

The previous post challenged students to consider 'basic' properties of circles and triangles. Now we will look at systems of quadratics - circles and parabolas in particular. The purpose here is to help students go beyond the standard algorithms of solving systems, by analyzing a general type of system using a parameter r. The use of parameters has become the norm on the AP Calculus Exam. Algebra students may benefit from an early introduction.


OVERVIEW FOR INSTRUCTOR
Consider the system:
x² + y² = r²
y = r² - x²

Here, r denotes a positive constant. Depending on the value of r, this system will have either 2, 3, or 4 solutions!

Simple problem of a parabola intersecting a circle, right? We will assume here that students have already solved specific cases of such systems both graphically and algebraically (substitution, etc.). They have been shown that parabolas and circles may intersect in 0, 1, 2, 3, or 4 points. As a review, begin this investigation by asking pairs of students to sketch (no equations here) graphs depicting each of these cases. That visualization is a powerful context for the algebraic solutions and may motivate the students to consider why varying the parameter r in this lesson leads to different conclusions. This type of analysis goes beyond standard problems and prepares students for the open-ended free-response types of standardized or AP questions they will later encounter in high school or college.

FOR THE STUDENT:

(a) r=1
Solve the following system first graphically, then algebraically:
x² + y² = 1
y = 1 - x²
This system demonstrates that a quadratic-quadratic system may have ________ (number) solutions.
For this system, the points of intersection are ___________________________.
For x between -1 and 1, the parabola is (above, below) the circle.

(b) r=2
Solve the following system first graphically, then algebraically:

x² + y² = 4
y = 4 - x²

This system demonstrates that a quadratic-quadratic system may have ________ (number) solutions.
For this system, the points of intersection are ___________________________
Restrict the domain to -2 ≤ x ≤ 2. For what values of x is the parabola above the circle in this system? Below the circle?

(c) What set of positive values of r have we not yet considered?
For such values of r, make a conjecture about the number of points of intersection of the system:
x² + y² = r²
y = r² - x²

Now choose a particular value of r in this set, say r = 1/2. Check the validity of your conjecture by solving the system for this particular value both graphically and analytically (algebraically).

(d) Analyze, algebraically, the following system for all positive values of r. Show carefully that your algebraic solution leads to 3 distinct cases for r:
x² + y² = r²
y = r² - x²
For each case, give the solutions (ordered pairs) in terms of the parameter r.
Explain why, for all positive values of r, there will always be at least TWO solutions of this system. That is, the possibility of zero or one solution does not exist for this system...

What goes up...Applying Quadratic Functions

This question was inspired by a released SAT question from a couple of years ago. Some of you may recall this question that appeared on the first released Sample Test for the 'new' SAT. Can you think of some reasons why this question was used by ETS as a Sample problem?

Because the height function given was not exactly in 'standard' form (using a,h, and k), even the strongest students I administered this problem to resorted to complicated algebra or used a physics formula (s = 0.5at²+...). They missed the point that when given the vertex of a parabola you're given more than just an ordered pair! I believe our students need more experience with this type of 'free-response' application. We see these in some textbooks but is enough time devoted to them or is that left to the physics teacher?

As usual, I've modified the question and developed it into an open-ended problem with several parts. Pls don't get exercised about the lack of reality of the physical model!

A model rocket is projected vertically upward from a point 877.5 ft above the ground and after 2.5 seconds reaches its maximum height of 1440 ft. We are given its height above the ground as a function of t:

h(t) = p(q-3t)² + r, where t is in sec, h is in ft; p, q, r are constants.


(a) Determine the values of p, q and r.

(b) Rewrite the given function in 'standard' form: h(t) = a(t-h)² + k.

(c) Determine, algebraically, all values of t for which h(t) = 1080. Explain, in terms of the motion of the rocket, why there are exactly 2 such values.

(d) After how many seconds did the object hit the ground? Use algebra.

(e) Verify your results by analyzing the function using graphing calculator technology.

The Genius of Archimedes: Parabolas, Tangents...

Pi day is over, but it seems fitting to continue exploring. Archimedes did more than develop an approximation procedure for pi! There are many excellent websites that explain the following in greater detail and discuss many more of Archimedes' theorems about parabolas and tangents. I attempted to draw a diagram using Draw in Word. It's crude but you'll get the idea. The object is to share this extraordinary piece of history of mathematics and have your students finish the proof that a light ray from the focus that strikes a parabolic surface is reflected in a ray that is parallel to the axis of the parabola. This is equally interesting in reverse: External light rays and other forms of electromagnetic radiation that are parallel to a parabola's axis are reflected to the focus, very useful for radar and other 'collection' devices.
Considering that Archimedes' proofs used only geometric properties makes his work even more astounding (now of course we can use coordinate geometry, calculus, etc.). This type of investigation is usually deferred to College Geometry courses, but I believe we can deliver it to motivated geometry, 2nd year algebra or precalculus students. If nothing else, it makes for a wonderful long-term project!

Ok, here goes...

In the diagram below, I've gone out of my way to make the reflecting ray NOT look parallel to the axis, even though we're trying to prove it is. This is to help students avoid assuming collinearity, when, in fact, that needs to be proved!

The two angles marked X are equal by a reflection principle (angle of incidence equals...). The two angles marked Y are equal because it can be proved that the tangent line at P is the perpendicular bisector of segment FP', where F is the focus and P' is the foot of the perpendicular from P to the directrix. I chose not to derive Archimedes' very subtle argument, but it is worth studying the proof. The proof starts by constructing the perpendicular bisector and showing that this line passes through P but no other point of the parabola, thus it is tangent. Alex Bogomolny's excellent and in-depth treatment (with java applets) of this topic (on cut-the-knot) is very worthwhile reading.

The student is being asked to prove that the reflecting ray is parallel to the axis. This is equivalent to showing that the line containing PP' and the reflecting ray are one and the same. The argument is straightforward, but students may want to continue learning more about the genius of Archimedes.

[Good luck copying this diagram (jpg). Some of you may find errors in my argument or an extremely simply argument for the parallelism, so pls share!!]

Parabolas, SATs, Quadratic Functions, Symmetry, Oh My!

The new SAT and other state math assessments are or will be including more Algebra 2 types of questions, particularly those involving quadratic functions. The following was inspired by a recent SAT math problem. As usual, my goal here is not to give conundrums and 'puzzlers'. I'll leave that to the expertise of Jonathan over at jd2718! My intent is to provide enrichment and extensions of questions that students are doing in class. More time is required for these than is normally given for an example presented by the teacher. Hopefully these can be used in the classroom.
The original question on the SAT gave a particular length for segment PQ (see below) and that may be a more reasonable start for most Algebra 2 students. The objective here is to have students apply and extend their knowledge of quadratic functions, graphs, coordinates, symmetry, etc. There are several approaches to this question. If instruction enables students to investigate this problem for 10-15 minutes, students may discover alternate methods that will deepen their understanding of the material. The teacher's role is to gauge the ability level and background of the group to determine how much structure/guidance is needed. This is not obvious at all and requires considerable pedagogical skill and experience.

Consider the graph of the quadratic function f(x) = x². Assume P, Q are points on the graph so that segment PQ is parallel to the x-axis and let the length of segment PQ be denoted by 2k.
If the graph of g(x) = b - x², intersects the graph of f(x) at P and Q, express the value of b in terms of k.

Notes:
Encourage several methods, i.e., pair students and require that they find at least two different methods. This is critical to develop that quick thinker who always has the answer before anyone else and does not want to deepen his/her insight. Many students will need to start with a numerical value for the length of segment PQ, say 4. Symmetry is a key idea in this problem, not only with respect to the y-axis, but also with respect to segment PQ! Some will see this quickly, others won't. It is our obligation to think this through in advance and be prepared to guide the investigation. Those who believe this kind of activity is a waste of precious time (so much more content could be covered) will never understand why I believe 'less is more' when it comes to learning math. Profound understanding can never be rushed. Short-cuts, IMO, are PART of a discussion, not the objective. Try it! Can you find at least THREE ways?

Another Quadratic Function Problem 2-16-07 through 2-20-07

You may want to read the comments for this post. Answers and possible solutions are discussed. There is also considerable discussion about teaching techniques for f(x-h).

The parabola problem from 2-15-07 generated some interesting discussion. I haven't had a chance to see it implemented with our Algebra 2 classes yet but I'll let you know if and when...
Today's problem is along the same lines. I'm trying to provide some problems that are exclusively high school math content for this time of year. There are dozens of outstanding problem-solving sites for MathCounts and similar middle-school competitions but there appears to be a dearth of secondary math problem-solving sites (or I haven't found them yet!). Again, how might one use the problem below? As a bonus or an extended in-class activity or a performance assessment or ??? How many would regard this question as suitable only for honors or accelerated students? My take is that if students are exposed to higher levels of thinking and know they are expected to learn how to do these and held accountable on an assessment, they will adjust. Not all will experience equal success but that's ok too! Many should be able to do part (a) or are my expectations way too high?

(a) Consider the quadratic function f(x) = 4(x+4)².
The graph intersects the line y = k, k>0, in 2 distinct points B and C.
The rectangle whose base is on the x-axis and 2 of whose vertices are B and C
has area 64. Determine the value of k. Show method clearly.

(b) Now let's generalize the result of (a).
Consider the quadratic function f(x) = a(x-h)², a>0.
The line y = k, k>0, intersects the graph of f in two distinct points B and C. The rectangle whose base is on the x-axis and two of whose vertices are B and C has area R.
(i) Explain graphically (not algebraically) why the area, R, of this rectangle is independent of h.
(ii) Express k in terms of a and R. Check that your formula for k gives the value you obtained from part (a).

Algebra 2 Challenge for 2-15-07

The following is an open-ended problem for Algebra 2 students...
Enjoy it but I'd really like to hear how you might implement this in the classroom. Part of homework? A bonus? An open-ended activity in class? Students working independently or in pairs? Part of an assessment? At what point would you use this? At the end of the chapter on quadratic functions?

(a) Consider the function f(x) = 6x - x².
If P and Q are the points of intersection of the graph of f with the x-axis and R is a point on the portion of the graph above the x-axis, what is the maximum area of triangle PQR?


(b) Consider the quadratic function whose x-intercepts are the nonzero numbers p and q, p > q, and whose y-intercept is -pq.

(i) Explain carefully why the graph of this function has a maximum point no matter what the signs of p and q are.
(ii) Write an expression for the y-coordinate of the vertex of the graph of this function in terms of p and q (simplified).
Note: This appears to be a standard problem using the formula -b/2a, but there are other approaches and the result may surprise you.

Challenge Problem for 2-7-07

Day 8 - still awaiting a response from the National Math Panel...

Pls read the comments re the problems from 2-6-07. There are hints for #1 and a good discussion about #3.

Something different today. This is another one of those weekly online challenges I gave last year just before the AMC Contest. Students found it difficult but those who persisted got it. Perhaps that's the best reason to give these challenges -- to teach persistence, a quality that distinguishes some of the best researchers and problem solvers from the rest. Remember to click on the image to magnify it if it's too small.