June 2014 Common Core Algebra 1 Regents Exam, Part 2

Update: I now have a Common Core Regents Review books available on Amazon.

Today was the first ever New York State Common Core Algebra 1 Regents. No one knew what to expect. Sure, math is math, and Algebra is Algebra. What questions could they ask, right? Well, it’s not just a matter of knowing the material. Some of this was covered nearly a year ago and not revisited. Not everything in the course scaffolds into new topics; not every new topic spirals back into the old.

And then there’s the question of presentation. You can do practice problems until the kids’ pencils are worn to nubs, but if the test problems are suddenly presented in a different -- particularly in an odd – way, a young teen might freeze up and yield the opportunity to work it out.

A lot of the test came down to vocabulary, and not necessarily math vocabulary, and reading comprehension. If you could figure out what they were asking, you could figure out what the answer might be. Or should I say “is”. It should be “is”, but who can be sure?

Once again, I’ll be reviewing the test. I’m starting with the open-ended. We’ll spiral back to the multiple-choice in the coming days. Part 1 is shorter than the older test and Part 2 makes up for it. Big Time.

Note: I won’t even pretend to guess at how many points you’ll get for writing what, other than to say if it’s perfect, you’ll get full credit. But who can be sure what “perfect” means?

Algebra 1 (Common Core), Part 2

25. Draw the graph of y = SQRT(x) – 1.

If you put this in your calculator, you had to be sure to close the parentheses after the x. Otherwise, the “- 1” would be part of the expression beneath the radical.

The trick to remember here is that the domain is x > 0. You can’t use negative numbers. The y-intercept is (0, -1). You should have, at the least, plotted the points (0, -1), (1, 0), (4, 1) and (9,2) before drawing a curve through them. There should be an arrow on the right side of the curve because it continues to the right. There is no arrow on the left because the line starts with (0, -1).

26. The breakdown of a sample of a chemical compound is represented by the function p(t) = 300(0.5)^t, where p(t) represents the number of milligrams of the substance and t represents the time, in years. In the function p(t), explain what 0.5 and 300 represent.

I don’t know just how specific an answer they are looking for here.

• 0.5 is the rate of decay of the substance. It is the base in the exponential function.
• 300 is the initial amount of the substance. It is the y-intercept of the function and the co-efficient of the base.

27. Given 2x + ax – 7 > -12, determine the largest integer value of a when x = -1.

Confused? Don’t feel bad. Make sure you use x = -1 and not a = -1.
Plug is -1 for x and simplify the inequality before you do anything else.
2(-1) + a(-1) – 7 > -12
-2 – a – 7 > -12
-a – 9 > -12
-a > -3
a < 3
Remember to flip the inequality symbol when you divide by -1. If a < 3, then the largest integer value of a will be 2.

28. The vertex of the parabola represented by f(x) = x² - 4x + 3 has coordinates (2, -1). Find the coordinates of the vertex of the parabola defined by g(x) = f(x -2). Explain how you arrived at your answer.

The notation for this is confusing. And when my students see this, I know that they’ll want to solve something because of the equal sign, but it’s a definition, not an equation.

Others will look at this and think it’s a recursive function because we just reviewed those a few days ago. Sigh.

For every value of x, g(x) will have the same value that the f() function had when x was 2 less than it is now. So the entire parabola will shift two places to the right. That means that the coordinates of the vertex with be (4, -1).

There are more complicated ways of achieving the same result, which, for 2 miserable points, I hope that they aren’t looking for.

29. On the set of axes below, draw the graph of the equation y = (-3/4)x + 3. Is the point (3, 2) a solution to the equation? Explain your answer based on the graph drawn.

This seems to be the easiest, most straightforward question, so far. Okay, so it’s a graph. Do the graph. You have a calculator to help you, if you need it. The y-intercept is (0, 3). The slope is -3/4 – down 3, 4 to the right, make another point, down 3, 4 to the right, make another point, … when you’re at the end of the graph, go back up the other direction. LABEL THE LINE

(3, 2) is not a solution. How do you show this using the graph? Put the point on the graph at (3, 2). Label it (3, 2). Respond: (3, 2) is not on the line so it is not a solution to the equation.

Do NOT plug (3, 2) into the equation to check. That’s not what they asked for, so they won’t give you points for it.

30. The function f has a domain of {1, 3, 5, 7} and a range of {2, 4, 6}. Could f be represented by {(1, 2), (3, 4), (5, 6), (7, 2)}? Justify your answer.

This one led to a bit of a discussion in the Math Department. One side was quite sure of their superiority of knowledge, and the other side still wasn’t satisfied with the explanation. To put it plainly: I think I know what they are asking, but I’m not entirely sure. And I’ve learned in the past, you can’t always go for what you think they want – sometimes, you have to go with what they ask.

The argument boils down to semantics, really. Or maybe it’s syntax. I don’t know. I’m not an English teacher. However, I have a problem with the word “could”. Seriously.

Is this question asking if the relation they gave fits the domain and range of f? If so, the answer is YES. Or is this question asking if the relation is the ONLY POSSIBLE FUNCTION f? If that’s the case, it’s NO. We don’t know how f is defined. There is no mapping function. It could be that this relation represents f, but it might not be. Is that what it’s asking? Literally, yes, that is what it says, word for word. And yet I’m still not sure if that’s what they mean, and I’m not sure that my students will catch that meaning as well. Nuance? I don’t know. Maybe I’m overthinking it.

Another way for me to put it is like this: Could A represent B if A is only a subset of B?

Unfortunately, not all my students are native speakers, so I hope there isn’t a problem.

One thing I know: “Yes” or “No” without a good explanation will be worth nothing.

UPDATE: I spoke with a teacher who has been to training on how to grade these exams, and he had an answer key with sample responses and their point values. Basically, the answer is YES for reasons given above. When I explained my concerns about the wording, he thought I was splitting hairs. To be honest, I agree with that. That said, the Regents has been know to split hairs in the past.

31. Factor the expression x⁴ + 6x² - 7 completely.

They changed it up a bit. Usually, a “factor completely” question has a Greatest Common Factor (GCF) component to it.

x⁴ + 6x² - 7 factors into (x² + 7) (x² - 1). If you think that this seems a little simplistic for “factor completely” instead of “factor into two binomials”, you are not wrong.

That’s because using the Difference of Squares Rule (x² - 1) can be factored into (x – 1)(x + 1), making the final answer:

(x² + 7) (x – 1)(x + 1) Note that (x² + 7) has no real roots and cannot be factored further.

32. Robin collected data on the number of hours she watched television on Sunday through Thursday nights for a period of 3 weeks. The data are shown in the table below. … Using an appropriate scale on the number line below, construct a box plot for the 15 values.

Note: A picture of the table will be added later.

Put the 15 data values in order. The appropriate scale would be start at 1 and increment by .5.

The data are: 1, 1.5, 1.5, 2, 2, 2.5, 2.5, 3, 3, 3, 3.5, 4, 4, 4.5, 5.
Note: if you don’t have 15 values, you left something out. Also, your calculator will do all this for you -- but copy it ALL down on your paper anyway!

Your five-number summary is as follows: Min: 1, Q1: 2 (4th value), Median: 3 (8th value), Q3: 4 (12th value), Max: 5. Number the scale from 1 to 5, counting by .5. Plot these five points. Draw a box using Q1 and Q3, with a vertical line through the median. Draw whiskers from Q1 to min and Q3 to max.

Done.

And that will do it for Part 2, which is much longer than the Integrated Algebra Part 2.