The following are some of the multiple questions from the August 2018 New York State Common Core Algebra I Regents exam.

### August 2018 Algebra I, Part II

Each correct answer is worth up to 2 credits. Partial credit can be given. Work must be shown or explained.

**25.** *
Explain how to determine the zeros of f(x) = (x + 3)(x -1)(x - 8).
*

State the zeros of the function.

**Answer: **

To find the zeroes of a function in factored form, set each of the factors equal to 0 and solve for x.

The zeroes of this function are:

x + 3 = 0 OR x - 1 = 0 OR x - 8 = 0

x = -3 OR x = 1 OR x = 8

**26.** *
Four relations are shown below.
*

*
*

State which relation(s) are functions.

Explain why the other relation(s) are not functions.
**Answer: **

III and IV are functions.

I is not a function because it fails the vertical line test. When x = 6, there are two values of y, 4 and 6. This is not allowed.

II is not a function because the x values 1 and 2 are repeated with different y values.

Added information: In III, the x values in the table do not repeat. In IV, the quadratic equation given is a parabola, which is a function. There is only one possible outcome when you square a number.

**27.** *
The table below represents the height of a bird above the ground during flight, with P(t)
representing height in feet and t representing time in seconds.
*

*
Calculate the average rate of change from 3 to 9 seconds, in feet per second.
*
**Answer: **

P(3) = 6.26 and P(9) = 3.41.

The average rate of change is (3.41 - 6.26) / (9 - 3) = -2.85 / 6 = -0.475 feet per second.

**28.** *
Is the solution to the quadratic equation written below rational or irrational? Justify your answer.
*

*0 = 2x*^{2} + 3x - 10
*
*
**Answer: **

Check the discriminant: *b*^{2} - 4ac

3^{2} - 4(2)(-10) = 9 - (-80) = 89

The square root of 89 is an irrational number, so the solutions to the equation are irrational.

**29.** *
The formula for converting degrees Fahrenheit (F) to degrees Kelvin (K) is:
*

*K = (5/9) (F + 459.67)**
Solve for F, in terms of K.
*
**Answer: **

K = (5/9) (F + 459.67)

(9/5) K = F + 459.67

(9/5) K - 459.67 = F
**30.** *
Solve the following equation by completing the square:
*

*x*^{2} + 4x = 2*
*
**Answer: **

Take half of 4 (which is 2) and square it (which is 4). Add that to both sides.

x^{2} + 4x = 2

x^{2} + 4x + 4 = 2 + 4

Now factor

(x + 2)^{2} = 6

x + 2 = __+__ SQRT (6)

x = -2 __+__ SQRT (6)
Since the wanted an exact value, leave the answer in fraction form because it's a repeating decimal. DO NOT ROUND.

**31.** *
The students in Mrs. Lankford’s 4th and 6th period Algebra classes took the same test. The results
of the scores are shown in the following table:
*

*
*

Based on these data, which class has the largest spread of test scores? Explain how you arrived at
your answer.
**Answer: **

The 4th Period has the larger spread.

The sigma value (the second column), which is used to find variance and standard deviation, is greater for 4th period, so the spread will be greater.

Also, if you calculate the Interquartile Range, Q_{3} - Q_{1}, you will get the following results:

4th Period: 87.5 - 69 = 18.5

6th Period: 88 - 71.5 = 16.5

4th Period has the greater IQR so it has the greater spread of the data.

Personal note: I'm not fond of this question. A little to vague. However, they chose data that would give the same answer whichever method was chosen. I'm not sure if calculating only the range would have been good enough, but if you search online, you will see that the range is listed as one measure of the spread of data. (Certainly, the others are *better* measures, but it doesn't mean the other method isn't one.)

**32.** *
Write the first five terms of the recursive sequence defined below.
*

*a*_{1} = 0

a_{n} = 2(a_{n - 1})^{2} - 1, for n > 1
*
*
**Answer: **

a_{1} = 0

a_{2} = 2(0)^{2} - 1 = 0 - 1 = -1

a_{3} = 2(-1)^{2} - 1 = 2(1) - 1 = 1

a_{4} = 2(1)^{2} - 1 = 2(1) - 1 = 1

a_{5} = 2(1)^{2} - 1 = 2(1) - 1 = 1

0, -1, 1, 1, 1

Note: every term after the 3rd will be 1.

**End of Part II**

How did you do?

Questions, comments and corrections welcome.