August 2023 Algebra 2, Part II

This exam was adminstered in August 2023.

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Algebra 2 August 2023

Part II: Each correct answer will receive 2 credits. Partial credit can be earned. One mistake (computational or conceptual) will lose 1 point. A second mistake will lose the other point. It is sometimes possible to get 1 point for a correct answer with no correct work shown.


25. Factor the expression 2x³ - 3x² - 18x + 27 completely.

Answer:

Factor by grouping, and then factor the quadratic you get after the first step.

There are two ways to group, and either should work in any question of this kind.

2x³ - 3x² - 18x + 27
(2x³ - 3x²) - (18x - 27)
x²(2x - 3) - 9(2x - 3)
(x² - 9)(2x - 3)
(x + 3)(x - 3)(2x - 3)

You can also switch the two middle terms around. This is just the way I learned it, so I usually do it, especially if it helps me avoid factoring out a minus sign.

2x³ - 18x - 3x² + 27
(2x³ - 18x)(3x² - 27)
2x(x² - 9) - 3(x² - 9)
(2x - 3)(x² - 9)
(2x - 3)(x + 3)(x - 3)

26. Algebraically determine the values of x that satisfy the system of equations shown below:

y = x² + 8x - 5
y = 8x - 4

Answer:

Set the two statements equal to each other. Solve the quadratic equation for x. You don't need to find y.

y = x² + 8x - 5
y = 8x - 4

x² + 8x - 5 = 8x - 4
x² - 1 = 0
(x - 1)(x + 1) = 0
x - 1 = 0 or x + 1 = 0
x = 1 or x = -1

27. Solve the equation 3x² + 5x + 8 = 0. Write your solution in a + bi form.

Answer:

When they tell you "a + bi" form, you know that it won't be something that you can factor easily and that you need to use the quadratic formula.

3x² + 5x + 8 = 0

x = (-b + √(b² - 4ac) ) / (2a)

x = (-5 + √(5² - 4(3)(8)) ) / (2(3))

x = (-5 + √(25 - 96) ) / (6))

x = (-5 + √(-71) ) / (6))

x = -5/6 + i √(71)/6

28. On the coordinate plane below, sketch at least one cycle of a cosine function with a midline at y = -2, an amplitude of 3, and a period of π/2.

Answer:

Cosine doesn't start at the midline. Cosine starts at the midline plus the amplitude, which in this case is -2 + 3 = 1. The minumim will occur at y = -2 - 3 = -5 when x = (1/2)(π/2) = π/4.

Your sketch should look like this:

Remember that a sketch doesn't have to be perfect, but it shouldn't have any obvious errors, such as crossing the axis at the wrong place.

29. Given i is the imaginary unit, simplify (5xi³ - 4i)² as a polynomial in standard form.

Answer:

Use FOIL, the Distributive Property or the Box Method/Area Model. Remember that i² = -1, i³ = -i, i⁴ = 1, i⁵ = i, and so forth.

(5xi³ - 4i)²
25x²i⁶ - 40xi⁴ + 16i²
-25x² - 40x - 16

30. Consider the parabola given by y = 1/4 x² + x + 8 vertex (-2,7) and focus (-2,8). Use this information to explain how to determine the equation of the directrix.

Answer:

The focus is 1 unit above the vertex, so the directrix is the horizontal line that contains the point 1 unit below the vertex, (-2,6).

The equation would be y = 6.

Remember that you have to give an explanation, not just the equation. The equation alone is half-credit.

31.Write (x √(x³) ) / (∛(x⁵)

Answer:

Remove the radicals by rewriting the expressions with fractional exponents. Then using the laws of exponents subtract the exponent of the numerator from the exponent of the denominator.

(x √(x³) ) / (∛(x⁵)

x (x ^(3/2)) / x ^(5/3)

x ^(5/2) / x ^(5/3)

x ^{(5/2 - 5/3)}

x ^{(15/6 - 10/6)}

x ^(5/6)

32. A fruit fly population can be modeled by the equation P = 10(1.27)^t, where P represents the number of fruit flies after t days. What is the average rate of change of the population, rounded to the nearest hundredth, over the interval [0,10.5]? Include appropriate units in your answer.

Answer:

Find P(0) and P(10.5), subtract the latter from the initial value and then divide by 10.5.

( 10(1.27)^10.5 - 10(1.27)⁰) / (10.5 - 0) = 10.7628...

10.76 fruit flies per day.

End of Part II

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