Usually habit creates the least cognitive load. Students who habitually use CAS for equations may find copying the given equation creates less cognitive load than plugging in multiple answer choices.
Obviously, there other problems for which CAS can be more useful (including grid-ins, linear systems for which you have to find x+y, etc.)
How about this one? Official Practice Test 4, Section 4, Question 25
f(x) = 2x^3 + 6x^2 + 4x
g(x) = x^2 + 3x + 2
The polynomials f(x) and g(x) are defined above. Which of the following polynomials is divisible by
2x + 3?
A) h(x)=f(x)+g(x)
B) p(x)=f(x)+3g(x)
C) r(x)=2f(x)+3g(x)
D) s(x) = 3f(x) + 2g(x)
CAS solution
Define f(x) = 2x^3 + 6x^2 + 4x
Define g(x) = x^2 + 3x + 2
f(x)+g(x) Enter
returns 2x^3+7x^2+7x+2
f(x)+3g(x) Enter
returns (2x+3)(x^2+3x+2)
Practice Test 3 Section 4 Question 33 (grid-in)
(−3x^2 + 5x − 2) − 2(x^2 − 2x − 1)
If the expression above is rewritten in the form ax^2 + bx + c, where a, b, and c are constants, what is the value of b ?
CAS Solution
(−3x^2 + 5x − 2) − 2(x^2 − 2x − 1) Enter
returns 9x - 5x^2
@Plotinus Oh yeah, that was the one I couldn’t immediately figure out without a calculator. But after realizing that f(x) = 2x g(x), it became apparent.
You can also solve that problem by plugging in the answer choices, without a CAS calculator.
Practice Test 3 Section 4 Question 22
The sum of three numbers is 855. One of the numbers, x, is 50% more than the sum of the other two numbers. What is the value of x ?
A) 570 B) 513 C) 214 D) 155
CAS Solution
solve(x+y+z=855
x=(y+z)+.5(y+z), x,y,z) Enter
returns x=513
True, but the point is that some people who can solve this problem with a CAS calculator can’t solve it nearly as quickly or possibly at all without CAS.
CAS can facilitate certain steps in more complicated problems. For example, in Official PSAT Practice Test 1 Section 4 problem 28, we have the graph of a line intersecting a parabola, and have to find the x-coordinate of the point of intersection (v,w) . We are given the vertex (4,19) and the y-intercept (0,3), of the parabola, and two other points on the line (0,-9) and (2,-1).
The first step is to find the equation of the parabola.
We can use the vertex form y=a(x-4)^2+19. Since (0.3) is a point on the parabola, we can substitute the point and solve for a with CAS:
solve(3=a*(0-4)^2+19,a) Enter
returns a=-1
Since (v,w) is on both the parabola and the line, we can set up a system of two equations in v and w:
solve(w=-(v-4)^2+19
(w+1)/(v-2)=(-1+9)/(2-0),v,w) Enter
returns v=-2 and w=-17 or v=6 and w=15
From the graph, we can see that (v,w) is the point of intersection in the first quadrant, so v=6
I feel like any student who knows how to express a quadratic function in vertex form should be able to solve 3 = a(-4)^2 + 19 in faster time than it takes to punch everything into a CAS calculator. That is just my take.
If I was really short on time, and if I was really familiar with the CAS functionality, I might use it to bypass the annoying algebra as you did in the 2nd step.
However, given all of these CAS “tricks,” I can’t say I recommend many of them to weaker (e.g. ~400-500) math students simply because I believe it is strictly better to simply learn the material and get a very solid understanding.
As another example, problem 30 on that same test can be done cheaply with a graphing calculator:
fMax(100t - (1/2)9.8t^2, t, 0, infinity) (had to look up the command name, correct me if I messed up the syntax)
which should return 510.2, round to 510.
31 can also be done a similar way, but I find it much easier to solve 0 = 100 - 9.8*t or t = 100/9.8 (s).
Practice Test 3 Section 4 Question 22 The sum of three numbers is 855. One of the numbers, x, is 50% more than the sum of the other two numbers. What is the value of x ? A) 570 B) 513 C) 214 D) 155
[/QUOTE]
CAS Solution
solve(x+y+z=855
x=(y+z)+.5(y+z), x,y,z) Enter
returns x=513<<<
Why not simply entering 855 x 3/5? And without any calculator, evaluating 57 + 38? Isn’t the SAT about reasoning and basic calculations? Either way, that problem requires a couple of seconds to solve mentally.
Miter94, there was no issue with the problem itself, except that Plotinus failed to add the four choices. Hence my reply!
Vertex and x-intercept forms of the parabola are very important for the Redesigned SAT. I would say vertex form is a 600-level skill at best for the new test. Anyone who is taking a stab at this problem needs to know vertex form. Otherwise invest the time in something more promising.
There are students who are very, very fast at punching things into a CAS but who tend to make sign and other mistakes when doing problems like this one by hand. I had more than one >600 level student make a sign mistake on this equation.
This is a field report: even students who don’t like math can be incredibly fast at punching things in. Their calculators are smoking. I would never be able to keep up with them in a calculator race, and only sometimes in a race of my pencil against their calculators, and I am pretty fast with pencil/mental math (but not as fast as @MITer94). Their calculators are extensions of their bodies. You have to see it to believe it.
This requires more and higher-level math reasoning than just translating the English into equations. I purposely translated “the first number is 50% more than the sum of the other two numbers” as “x=y+z+.5(y+z)” instead of “x=1.5(y+z)” because weaker math students understand percentage increase as an addition and not a multiplication.
If you ask a 550 M student “How much is 12% more than 30?”, he is going to do 30+.12(30). He is not going to do 1.12(30). If the student uses CAS, it does not make any difference how messy the equations are, as long as they are correct translations. Translation and data entry skills matter; algebra and numerical skills don’t matter.
The new test is “aligned with Common Core” not “Reasoning”. There seems to me to be less reasoning required than there was on the old test.
As I said before, I tell my students to solve the problems both with and without a calculator during homework practice, and then to use whatever method is fastest and easiest for them on the day of the test. Ideally, it should not be “either-or”. However, I guarantee you that at least for the old SAT, many students chose CAS on the day of the test for some problems either because it was faster and easier for them under the strain of test conditions, or because they didn’t know how to do the problem without the calculator. I don’t have much statistical data about how students will use CAS for the new test, but if the real tests are like the official practice tests or the October PSAT’s, I suspect that on test day, my students will solve some of the problems with CAS.
I recommend CAS to the following students:
This is maybe around 85% of my students.
@Zeldie I don’t think my students who qualified for NCAA or were admitted to top schools because of the CAS boost would agree with that cartoon. It’s not a joke if it gets you into an Ivy League school or a scholarship worth $250K.
That could be true but I do not think that you could muster much evidence about the CAS boost on a historical basis for the SAT Reasoning test. The opposite is eminently true for the SAT Subject test that rewarded the mastery of graphical calculator more than real knowledge.
While I find your examples fascinating, I remain convinced that most students would fare better if they did forgo jumping to a calculator in an elusive attempt to stumble on a solution for a problem they hardly understand.
For reference, both of my children took the SAT without using the calculator as they had prepared under such a condition. The oldest one researched the impact of a calculator in great depth and concluded that a well-prepared student would complete each section faster without relying on a crutch.
Again, I love your examples and I am happy to agree that it might help a few people on the new SAT. Fair enough?
Experimenting with the calculator on test day is a no-no. The calculator on test day is just for procedures you know how to do well. Technology is dangerous unless used the right way.
What research did your son do? Any references? Maybe he did not understand well what can be done with a CAS or how quickly it can be done. I agree that having a teacher who can show you how to use it properly can be an advantage because there is a learning curve.
It is impossible for me to pin down exactly how many points students were able to add with CAS on the old SAT, but I would estimate around 50-80 for bad-algebra students and around 30-40 for reasonably-good algebra students. I had a kid who scored 760 M who loved the CAS. I don’t know if he added any points because of it but it made him happier anyway. I myself far prefer pencil algebra to CAS, but I would definitely bring my calculator to the new SAT. I might not use it at all, but it might come in handy for annoying algebra or double-checking for careless errors. You never know for sure what might turn up, especially on the new test with its emphasis on Algebra 2. You can also use the history to review things later.
fmax and fmin return the X-coordinates, not the y-coordinates, of the maximum and minimum. Further, you do not need to add the interval endpoints if you are considering all real numbers. I would do these two problems (and all max/min problems) as follows:
CAS solution
Define f(t)=100t-4.9t^2
fmax(f(t),t) Enter
returns t=10.20141 — This is the answer to question 31
f(10.2) Enter
returns 510.204 – This is the answer to question 30.
It looks like projectile motion problems will be popular on the new SAT. So far I have only seen the simplified kind with purely vertical motion. A side note: Official PSAT Practice Test questions 30-31 are not very “real world”: who shoots an arrow straight up in the air?
Another standard question of this general type would be:
If we shoot a projectile straight up with a given initial velocity and from a given initial height, after how much time will it hit the ground? Supposing in the previous problem, the arrow is shot from a height of 2 meters with an initial velocity of 100 m/sec, and we want to know after how many seconds it hits the ground.
CAS Solution
Define f(t)=2+100t-4.9t^2
solve(f(t)=0,t)
returns t=-0.01998 or t=20.4281
I’d like to return to the point that it is going to be really important for the new SAT to know the vertex and x-intercept forms of the parabola.
I have seen several questions that give a messy version of a quadratic and then ask, “Which of the following is an equivalent form that contains the zeroes of the function?” or “Which of the following is an equivalent form that contains the maximum/minimum value of the function?”
These are just confusing and verbose ways to ask student to convert the quadratic into x-intercept or vertex form.
The student needs to know that “Which of the following is an equivalent form of the expression above that displays the zeroes of the function as constants or coefficients in the expression?”= bad math verbiage for “Put the equation in x-intercept form”.
Knowing that, you can use CAS to see which x-intercept form among the answer choices is equivalent to the original.
I know @Zeldie and @MITer94 are going to tell me the following question is too easy to use CAS, because it is really easy to factor x^2+4x-12. Maybe so. But some students are going to make sign mistakes. In fact, the whole problem is set up to trap students into making sign mistakes. Further, the CAS method would work even with quadratics that are much harder to factor.
PSAT October 28 Section 4 Question 9
f(x)=x^2+4(x-3)
Which of the following is an equivalent form of the expression above that displays the zeroes of the function as constants or coefficients in the expression?
A. f(x)=x^2-4x-12
B. f(x)=(x-6)(x+2)
C. f(x)=(x+6)(x-2)
D. f(x)=(x+2)^2-16
As I said before, the student should know that the question is asking for the equivalent x-intercept form, and that only B and C are in x-intercept form. At that point, this question becomes, which of the following is an equivalent expression, B or C?
I showed above that to use CAS to determine whether expressions are equivalent, you subtract the expressions and see if the difference=0.
If you know anything about how ETS/CB construct traps, you suspect the correct answer is going to come after the trap not before it, so let’s check choice C first.
CAS Solution
C: x^2+4(x-3) - (x+6)(x-2) Enter
returns 0
And if you just want to be 100% sure:
B: x^2+4(x-3) - (x-6)(x+2) Enter — copy the previous command line from the history and edit the two signs
returns 8x
PSAT Oct. 28 Section 4 Question 23
In the xy plane, if the point (a,b) lies on the parabola with equation y=x^2-1, which of the following must be equal to a^4?
A. b^2-1
B. b^2+1
C. (b-1)^2
D. (b+1)^2
CAS Solution
solve(b=a^2-1,a) Enter
returns a=-sqrt(b+1) and b≥-1 or a=sqrt(b+1) and b≥-1
sqrt(b+1)^4 Enter
returns (b+1)^2
