The following are some of the multiple questions from the August 2019 New York State Common Core Algebra I Regents exam.

### August 2019 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.** *
If g(x) = -4x ^{2} - 3x + 2, determine g(2).
*

**Answer: **

Substituion:

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

= -4(4) - (-6) + 2

= -16 + 6 + 2 = -8

One point off for one computational error, such as (-2)^{2} = -4, or similar.

**26.** *
A student is in the process of solving an equation. The original equation and the first step are
shown below.
*

Original: 3a + 6 = 2 - 5a + 7

Step one: 3a + 6 = 2 + 7 - 5a

Original: 3a + 6 = 2 - 5a + 7

Step one: 3a + 6 = 2 + 7 - 5a

*Which property did the student use for the first step? Explain why this property is correct.*

**Answer: **

The student used the Commutative Property to switch the order of the last to terms, which allows the student to Combine Like Terms in the next step.

**27.** *
On the set of axes below, graph the line whose equation is 2y = -3x - 2.
This linear equation contains the point (2,k). State the value of k
*

**Answer:**

Divide the equation by 2 to get it in slope-intercept form. You will see that the y-intercept is -1 and the slope is -3/2. You could use the graphing calculator at this point, or you can plot the point (0, 11) and then make you next point down 3 spaces and 2 to the right, and then repeat. When you get to the edge of the graph, go back the other way: 3 spaces up and 2 to the left, and repeat. And the arrows.

You can plainly see that the point (2, k) is (2, -4), so k = -4.

If you graphed this incorrectly, but your answer for k matches your graph, you will get credit for that portion.

If you used the equation 2y = -3(2) - 2 to find the y-coordinate, you will probably get the credit.

**28.** *
The formula a = (v _{f} - v_{i}) / t is used to calculate acceleration as the change in velocity over the period of time.
*

Solve the formula for the final velocity, v_{f}, in terms of initial velocity, v_{i}, acceleration, a, and time, t.

**Answer: **

This is a real-world problem: calculating a final velocity when given an initial velocity, acceleration and a period of time.

Isolate the variable by moving everything else to the other side of the equation.

a = (v_{f} - v_{i}) / t

at =v_{f} - v_{i}

at + v_{i} = v_{f}

**29.** *
Solve 3/5 x + 1/3 < 4/5 x - 1/3 for x.
*

**Answer: **

You can eliminate the fractions, if you want, by multiplying the entire equation by 15. Or you can simplify it a little first.

3/5 x + 1/3 < 4/5 x - 1/3

2/3 < 1/5 x

(15)(2/3) < (1/5 x)(15)

10 < 3x

10/3 < x

**30.** *
Is the product of two irrational numbers always irrational? Justify your answer.
*

**Answer: **

No, sometimes the answer is rational. SQRT(2) * SQRT(2) = 2, which is rational.

**31.** *
Solve 6x ^{2} - 42 = 0 for the exact values of x.
*

**Answer: **

Exact values of x means that you do NOT estimate irrational numbers. Do not round anything.

6x^{2} - 42 = 0

6x^{2} = 42

x^{2} = 7

x = __+__SQRT(7)

If you go any further than that, you will lose a point, so STOP.

If you used the Quadratic Formula, you would have gotten something like __+__ SQRT(1008) / 12. This is OKAY because that is an Exact value, which can be simplified into 12 * SQRT(7) / 12, but you wouldn't be required to do this. However, if you did attempt to simplify it, you would have to get it right.

**32.** *
Graph the function:
*

*h(x) = { 2x - 3, x < 0*

x

x

^{2}- 4x - 5, 0__<__x__<__5

**Answer: **

ONLY Graph up to x = 5 and no farther. No arrow. The left side, however, does get an arrow. The left portion ends with an OPEN circle. The parabola starts with a CLOSED circle. You could end it with a closed circle as well, if you want, or just a point.

**End of Part II**

How did you do?

Questions, comments and corrections welcome.

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