Algebra 2 Problems of the Day (Algebra 2/Trigonometry Regents, January 2011)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

Algebra 2/Trigonometry Regents, January 2011

Part I: Each correct answer will receive 2 credits.


1. Which graph does not represent a function?

Answer: 4) [SEE IMAGE]

The last choice fails the vertical-line test, so it is not a function.

No x value can have multiple y values in a function.

2. The roots of the equation x² − 10 x + 25 = 0 are

1) imaginary
2) real and irrational
3) real, rational, and equal
4) real, rational, and unequal

Answer: 3) real, rational, and equal

If you graph the equation in your graphing calculator, you will see that it touches the x-axis at its vertex. That eliminates Choices (1) and (4).

If you check the Table of Values, you will see that the vertex is x = 5. You could also have used x = -b/(2a) to find that.

So the roots are rational. And they are equal.

If you factored, you would have had y = (x - 5)(x - 5), which tells you x = 5 is the only root.

3. Which values of x are solutions of the equation x³ + x² − 2x = 0?

1) 0, 1, 2
2) 0, 1, -2
3) 0, -1, 2
4) 0, -1, -2

Answer: 2) 0, 1, -2

Obviously x = 0 is a solution, and it's in all four choices. You can factor or you can use substitution to find the other two. Since the final sign is negative, you know it has to be (2) or (3), so if you are substituting, start with those.

x³ + x² − 2x = 0

x(x² + x − 2 = 0

x(x + 2)(x - 1) = 0

x = 0 or x = -2 or x = 1

4. In the diagram below of a unit circle, the ordered pair (−√2/2, −√2/2) represents the point where the terminal side of θ intersects the unit circle.

What is m∠θ?

1) 45
2) 135
3) 225
4) 240

Answer: 3) 225

First, the point is in Quadrant III, so the measure must be 1800 < θ < 270. This eliminates Choices (1) and (2).

Second, you should know that (√2/2, √2/2) represents 45 degrees. Add 180 to that and you have 225, which is Choice (3).

5. What is the fifteenth term of the sequence 5, −10, 20, −40, 80, . . . ?

1) −163,840
2) −81,920
3) 81,920
4) 327,680

Answer: 3) 81,920

Notice that the odd terms are positive, so the fifteenth term (15th, not 50th) must be positive. Eliminate Choices (1) and (2).

The common ratio is the sequence is -10/5 = -2, 20/-10 = -2, etc. (You will notice that it does NOT have a common difference, so it is not arithmetic.)

The fifteenth term is (5)(-2)¹⁴ = 81,920.

More to come. Comments and questions welcome.

More Regents problems.

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