Algebra 2 Problems of the Day (Algebra 2 Regents, June 2012)

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, June 2012

Part I: Each correct answer will receive 2 credits.


11. Which graph represents the function log ₂ x = y? position?

Answer: 1) (see image)

The log₂ x function is the inverse of the exponential function 2^x, a reflection across the line y = x.

When x = 1 then y = 0 because when 0 is the exponent of 2 then the result is 1.

12. A circle is drawn to represent a pizza with a 12 inch diameter. The circle is cut into eight congruent pieces. What is the length of the outer edge of any one piece of this circle?

1) 3π/4
2) π
3) 3π/2
4) 3π

Answer: 3) 3π/2

The circumference of the pizza is 12π. One eighth of that is 12π/8, which simplifies to 3π/2.

13. What is the solution set for the equation √(5x + 29) = x + 3?

1) {4}
2) {-5}
3) {4, 5}
4) {-5, 4}

Answer: 1) {4}

You can work this out (see below) or you can check if 4, -5 or 5 work.

√(5x + 29) ?= x + 3
√(5(4) + 29) ?= (4) + 3
√(49) ?= 7
True, x = 4 is a solution. Eliminate Choice (2).

√(5x + 29) ?= x + 3
√(5(5) + 29) ?= (5) + 3
√(54) ?= 8
False, x = 5 is NOT a solution. Eliminate Choice (3).

√(5x + 29) ?= x + 3
√(5(-5) + 29) ?= (-5) + 3
√(4) ?= -2
False, x = 2 is NOT a solution. Eliminate Choice (4).

Choice (1) is the solution.

Alternatively, you could solve the equation:

√(5x + 29) = x + 3
5x + 29 = (x + 3)²
5x + 29 = x² + 6x + 9
x² + x - 20 = 0
(x + 5)(x - 4) = 0
x + 5 = 0 or x - 4 = 0
x = -5 or x = 4

Next check for extraneous values. As shown above x = -5 is NOT a solution. Discard it. Then x = 4 is the only solution.

14. When factored completely, x³ + 3x² - 4x - 12 equals

1) (x + 2)(x - 2)(x - 3)
2) (x + 2)(x - 2)(x + 3)
3) (x² - 4)(x + 3)
4) (x² - 4)(x - 3)

Answer: 2) (x + 2)(x - 2)(x + 3)

It says "factored completely" and (x² - 4) is not factored completely. It can be broken down further. Eliminate Choices (3) and (4).

If you factor the polynomial by grouping:

x³ + 3x² - 4x - 12
= x³ - 4x + 3x² - 12
= x(x² - 4) + 3(x - 4)
= (x² - 4)(x + 3)
= (x + 2)(x - 2)(x + 3)

15. What is the middle term in the expansion of (x/2 - 2y)⁶?

1) 20x³y³
2) -15/4 x⁴y²
3) -20x³y³
4) 15/4 x⁴y²

Answer: 3) -20x³y³

The expansion has 7 terms, starting with the x⁶y⁰ term and ending with the x⁰y⁶ term. The middle term would be the fourth one and it would be in the form x³y³, so eliminate Choices (2) and (4).

The coefficient of the fourth term would be ₆C₃ * (1/2)³ * (-2)³ = (20)(1/8)(-1/8) = -20.

More to come. Comments and questions welcome.

More Regents problems.

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