What follows is a portion of the Common Core Integrate Algebra exam. Other parts will be posted on other days. Illustrations will be added at a later time when they become available.

### June 2016 Algebra Regents, Part II

**25. ***Given that f(x) = 2x + 1, find g(x) if g(x) = 2[f(x)] ^{2} - 1. *

Take the expression for *f(x)* and square it. Then double that trinomial. Then subtract 1 from the expression.

Multiplying (2x + 1)(2x + 1) gives you 4x^{2} + 4x + 1.

Double it to: 8x^{2} + 8x + 2

Subtract 1: 8x^{2} + 8x + 1, which is the final answer.

**26. ***Determine is the product of 3*SQRT(2) and 8*SQRT(18) is rational or irrational. Explain your answer.*

You can put it in your calculator or do it by hand. Either way, you will get 24 * 6 = 144, which is a rational number.

144 can be written as a fraction.

144 is a Whole number and all whole numbers are Rational.

There are many similar things you can right as an explanation.

**27. ***On the set of axes below, draw the graph of y = x ^{2} - 4x - 1.
State the equation of the axis of symmetry. (image omitted)*

Put the equation into your graphing calculator and not the points in the table of values. Plot the points and draw the graph.

By looking at your graph, you will be able to see that the Axis of Symmetry is x = 2

Note that they asked for an equation for the axis of symmetry. If you just state "2" without the "x = ", you will not get the point!

**28. ***Amy solved the equation 2x ^{2} + 5x - 42 = 0. She stated that the solutions to the equation were 7/2 and -6. Do you agree with Amy's solution? Explain why or why not.*

First thing to notice: they didn't ask you to solve it like Amy did. Just whether or not you agree and could back it up. Second: don't forget to answer the question: "Yes, I agree with Amy because ..." (if you disagreed, I have some bad news for you.)

The algebraic solution will be added later. If you did it algebraically, you should have agreed with Amy.

If you put the equation y = 2x^{2} + 5x - 42 into the graphing calculator, you will see that -6 is one solution and the other solution is between 3 and 4. You can use the function of the calculator that finds the zeroes to show that the other point is 7/2 (or 3.5).

Likewise, you can check her work, showing that 2(-6)^{2} + 5(-6) - 42 = 0 and 2(7/2)^{2} + 5(7/2) - 42 = 0. I would add a note that says that quadratic equations can have up to 2 solutions, just to exclude the possibility that there are other answers besides these two.

**29. ***Sue and Kathy were doing their algebra homework. They were asked to write the equation of the line that passes through the points (-3, 4) and (6, 1). Sue wrote y - 4 = -1/3(x + 3) and Kathy wrote y = -1/3(x) + 3. Justify why both students are correct.*

Sue wrote her answer in **point-slope form** and Kathy wrote her answer in **slope-intercept form**. However, you still have to prove that they both have the correct answer. You can calculate that the slope is (4 - 1) / (-3 - 6) = 3 / (-9) = -1/3, which is in both equations.

Sue used the slope -3 and the point (-3, 4) to write her equation.

If you distribute the (-1/3) on the right side, and then add 4 to both sides to isolate the y, you will get Kathy's equation.

If this didn't occur to you, but you checked both points in both equations and found out that they all worked, you should get full credit.

**30. ***During a recent snowstorm in Red Hook, NY, Jaime noted that there were 4 inches of snow on the ground at 3:00 pm and there were 6 inches of snow on the ground at 7:00 pm.
*

*If she were to graph these data, what does the shape of the line connecting these two points represent in the context of this problem?*

Note that they never asked you to graph the data or find the slope, only to explain the slope.

The slope of the line would be the number of inches that fell every hour.

If you wanted to know how much that was: difference in snow divided by the difference in time = 2 inches / 4 hours = 1/2 inch/hour.

**31. ***The formula for the sum of the degree measures of the interior angles of a polygon is S = 180(n - 2). Solve for n, the number of sides of the polygon, in terms of S.*

Isolate *n*, and move everything else to the other side using *Inverse Operations*.

Divide by 180: S / 180 = n - 2.

Add 2 to both sides: (S / 180) + 2 = n.

**32. ***In the diagram below, f(x) = x ^{3} + 2x^{2} is graphed. Also graphed is g(x), the result of a translation of f(x).
*

Determine an equation of g(x). Explain your reasoning.

It doesn't matter if you never graphed a **cubic function** all year or if you don't remember what one looks like. All you need to remember is that this is an example of a **translation**. How do the points on the two lines differ?

All the points on g(x) are 4 places lower. That means that all the y-values are 4 less for g(x) than for f(x). So g(x) = f(x) - 4.

Therefore g(x) = x^{3} + 2x^{2} - 4

Is there another way to do this? Sure. You could have, for example, put some of the points into lists in the graphing calculator and did a **Cubic regression** to get the equation of the line. This is a totally valid approach as long as you explain what you did and where you're answer came from.

**END OF PART II**

How did you do? Any questions?

My original scratch writings are shown below.

## 38 comments:

Thank you, Love your blog!

thanks so much

do you have part 4

thanks so much

do you have part 4

You're the best!

Thanks so much! Do you have part 1?

Thanks so much! Do you have part 1?

Hello, for number 25, I am 100% positive there was no exponent on the question. I took it just this morning and I am sure that there wasn't a square.

ya there was hun

Yes there was. I took it yesterday and there was definitely a square

Please post a more clear pic of your scrap paper cause it comes out distorted! Thanks

Does anybody know the equation for the airplane cruising question?

sdfg: I only have my notes here, but I worked out the problem with the test in front of me.

There was definitely an exponent in g(x). Sorry.

Unknown: By the time the ban is lifted, I'll have better images.

I said that this posting was "quick and dirty" because people want to see answers "NOW!" and typing takes time and graphics take even more time.

I'm only one blogger.

Is there something in particular you wanted to know?

I can't post the exact answer to the airplane problem until later, but ...

part a was a slope question, using the two sets of values for time and distance

part b was a y = mx + b question, using the slope you just found + the miles traveled when it started cruising.

part c: use x = 120 and solve for y.

Are you allowed to say what the numbers were in the airplane problem? I feel like one was 192 (after 32 mins) and the other was in the 700s, though I can't remember what it is. I want to try to solve the problem again to see if I get the same thing I got on the test.

Is the equation to the airplane question y=9.5x-112? I checked it with the points that should be on the line for just cruising and it worked out fine. Shouldn't the y-intercept be -112 because the equation is just for the time when the airplane was cruising? Doesn't that mean you would have to subtract the distance it wasn't in cruising?

Did we have to state parameters for the equation as well?

So, for y=mx+b you said "part b was a y = mx + b question, using the slope you just found + the miles traveled when it started cruising." So, it would be y=9.5x+192. However, that would not fit your standards for part c, "part c: use x = 120 and solve for y". This is because I have heard part c was something else than what comes out. So what is it, y=9.5x+192 or y=9.5x-112. HELP PLZ.

It is y= 9.5x-112, 192 is not the y-intercept it is the first point on the line for y=9.5x-112

Hi. I am looking to see the common core geometry regents and algebra 2 regents problems from this year, June 2016. I remember that you posted the answers to last year's common core geometry - hoping you will do the same this year. Will you?

I had commented on your blog last year.

Rose C - retired math teacher but still interested.

There was a lot of discussion, debate, and arguing about the airplane question in my room, but also absolutely nothing we could do about it.

The state's answer for the airplane equation was y = 9.5x

No y-intercept.

To make matters more confusing, you needed a slightly different equation for the third part of the problem.

No, none of the teachers at my table or in my room thought that the wording was clear and unambiguous. Most of us would have gotten it wrong based on the third part of the question.

Rose C:

I plan to post some in the coming week. I have been "asked" not to just yet.

Likewise, I haven't been able to get copies yet.

Could you let us know what you had to do to solve the third part of the problem with this "slightly different equation?"

You needed to account for the time you weren't at cruising altitude. If you just plugged in 120 into the equation in the middle of the page, you didn't get the correct answer.

What was the answer to the third part?

They were asked to write an equation to rep the number of miles the plane has flown during cruising altitude only in "X" minutes. The plane was moving 9.5 miles per minute as established in the first question. Wouldn't an answer of Y= 9.5x suffice? The way it was worded, in my opinion, could take the 192 ascending miles right out of the equation. An equation is written. Y is the number of cruising miles. A coefficient of 9.5 was established and X reps the minutes. F(x) is not specified. I think this satisfies the question. The word "total" was never established and there are no domain parameters the way the question is worded.

I got 1,028 for the third question. 32 minutes climbing for 192 and 836 cruising at 9.5 miles per minute for 88 minutes.

The question was poorly worded in that it is ambiguous what the word "only" applies to.

It is obviously limiting the domain, as the x value is the antecedent of the clause. It is less clear about applying to y. The setup makes a big deal about mileage before reaching cruising and the final portion wants distance from the ground, two hours into the flight, not two hours at cruising altitude.

This will definitely be a problem for ELLs, and I've seen them address less confusing problems in the past.

Consider a question about overtime pay when x is the hours of overtime worked only. The question should be clear if they are only looking for the additional amount earned in overtime or what the person pay would be.

"The setup makes a big deal about mileage before reaching cruising and the final portion wants distance from the ground, two hours into the flight, not two hours at cruising altitude."

Actually, the question said, assuming the plane goes at the crusing altitude speed...

And another part spoke about the total mileage from la to new york. So you had to account for the plane landing as well, simply double the ascending minutes and miles and subtract it from the total minutes and miles and dvide the remaiming distance by the remaining miles tk find cruising altitude speed.

And another part spoke about the total mileage from la to new york. So you had to account for the plane landing as well, simply double the ascending minutes and miles and subtract it from the total minutes and miles and dvide the remaiming distance by the remaining miles tk find cruising altitude speed.

Sorry to repeat myself BUT the question was worded poorly and the part about cruising altitude could have been interpreted as a limit only on the domain -- that is, the x variable for setting up the equation. The third part of the question seems to reflect this.

As for landing: No. The question did NOT say that the flight was two hours. It send TWO HOURS into the flight, assuming is stayed at cruising altitude. The plane isn't landing yet.

You can't fly from New York to Los Angeles in 2 hours.

The third part said, assuming the plane goes at cruising altitude speed, ... two hours, meaning you just plug in 120 for the equation in part b. The landing part i was speaking about was for part 1, which was to find the cruising altitude.

You see, in part 1, you had to assume the plane landed because it gave you the length of the total trip and timing.

And i believe the plane was going at 13.5 mpm at cruising altitude

For part 1:

1. 762 minus 184 (92 times 2 for ascending and descending)

2. Divided that by 28, because the 92 minus 32 minutes for ascending minus 32 minutes again for descending.

I meant for step 1 in part 1, 762 minus 384

Don't you agree?

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