August 2024 Algebra I Regents — Answer Key & Worked Solutions

Every question from the August 2024 Algebra I Regents exam, with the correct answer and a full step-by-step solution. Check your work, learn the method, then convert your raw score.

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These are real questions from the August 2024 Algebra I New York State Regents examination, published by NYSED. Worked solutions are written by RegentsLab.

Question 1 · 2 pts

What is the correct factorization of $x^2 + 4x - 12$?

(A) $(x + 3)(x - 4)$
(B) $(x - 3)(x + 4)$
(C) $(x + 2)(x - 6)$
(D) $(x - 2)(x + 6)$ ✓

Worked solution

$\text{Find two numbers that multiply to } -12 \text{ and add to } 4$ those numbers are 6 and -2

$(x - 2)(x + 6) = x^2 + 6x - 2x - 12 = x^2 + 4x - 12$

Answer: $(x - 2)(x + 6)$

Since -2 and 6 multiply to -12 and sum to 4, the factorization is (x-2)(x+6); the other choices expand to different middle or constant terms.

Question 2 · 2 pts

Which situation can be modeled by a linear function?

(A) A printer can print one page every three seconds. ✓
(B) A bank account earns 0.5% interest each year, compounded annually.
(C) The number of cells in an organism doubles every four days.
(D) The attendance at a professional sports team's games decreases by 1.5% each year.

Worked solution

$\text{Linear growth: constant rate of change (not a percent/multiplicative change)}$

$\text{Printing at a constant rate of 1 page per 3 seconds is a constant additive rate} \Rightarrow \text{linear}$

$\text{The other three describe percent (multiplicative) change} \Rightarrow \text{exponential}$

Answer: A printer can print one page every three seconds.

A constant rate of pages per second is additive growth, which is linear, while percent-based growth or doubling describes exponential functions.

Question 3 · 2 pts

Which expression is equivalent to $3(x^2 - 2x + 3) - (4x^2 + 3x - 1)$?

(A) $-x^2 + x + 2$
(B) $-x^2 - 8x + 7$
(C) $-x^2 - 3x + 8$
(D) $-x^2 - 9x + 10$ ✓

Worked solution

$3(x^2-2x+3) = 3x^2 - 6x + 9$

$3x^2 - 6x + 9 - (4x^2 + 3x - 1) = 3x^2 - 6x + 9 - 4x^2 - 3x + 1$

$= -x^2 - 9x + 10$

Answer: $-x^2 - 9x + 10$

Distributing the 3 and then subtracting the second polynomial (distributing the negative sign) gives -x^2 - 9x + 10.

Question 4 · 2 pts

At Adelynn's first birthday party, each guest brought $1 in coins for her piggy bank. Guests brought nickels, dimes, and quarters for a total of $28. There were twice as many dimes as nickels and 12 more quarters than nickels. Which equation could be used to determine the number of nickels, $x$, that her guests brought to her party?

(A) $.05x + .10x + .25x = 28$
(B) $.05x + .10(2x) + .25(x + 12) = 28$ ✓
(C) $.05(2x) + .10x + .25(x + 12) = 28$
(D) $.05(x + 12) + .10(2x) + .25x = 28$

Worked solution

$\text{nickels} = x,\ \text{dimes} = 2x,\ \text{quarters} = x+12$

$.05x + .10(2x) + .25(x+12) = 28$

Answer: $.05x + .10(2x) + .25(x + 12) = 28$

Nickels are worth $0.05 each (x of them), dimes $0.10 each (2x of them), and quarters $0.25 each (x+12 of them), summing to the total $28.

Question 5 · 2 pts

A student creates a fourth-degree trinomial with a leading coefficient of 2 and a constant value of 5. The trinomial could be

(A) $2x^4 + 3x^2 + 5$ ✓
(B) $2x^4 + 5x + 3$
(C) $4x^2 - 3x + 5$
(D) $4x^3 - 5x^2 + 3$

Worked solution

$\text{Need degree 4, leading coefficient 2, constant term 5, exactly 3 terms}$

$2x^4 + 3x^2 + 5 \text{ satisfies all four conditions}$

Answer: $2x^4 + 3x^2 + 5$

Only 2x^4+3x^2+5 has degree 4, leading coefficient 2, a constant of 5, and exactly three terms (a trinomial).

Question 6 · 2 pts

When solving the equation $4x^2 - 16 = 0$, Laura wrote $4x^2 = 16$ as her first step. Which property justifies Laura's first step?

(A) distributive property of multiplication over addition
(B) multiplication property of equality
(C) commutative property of addition
(D) addition property of equality ✓

Worked solution

$4x^2 - 16 = 0$

$4x^2 - 16 + 16 = 0 + 16 \Rightarrow 4x^2 = 16$ adding 16 to both sides

Answer: addition property of equality

Adding 16 to both sides of the equation to isolate the variable term is justified by the addition property of equality.

Question 7 · 2 pts

Which expression results in an irrational number?

(A) $\sqrt{3} \cdot \sqrt{3}$
(B) $-\frac{2}{3} + \frac{1}{4}$
(C) $5 \cdot \sqrt{81}$
(D) $\frac{1}{3} + \sqrt{3}$ ✓

Worked solution

$\sqrt{3}\cdot\sqrt{3} = 3 \text{ (rational)}$

$-\frac{2}{3}+\frac{1}{4} \text{ (rational, sum of rationals)}$

$5\cdot\sqrt{81} = 5\cdot 9 = 45 \text{ (rational)}$

$\frac{1}{3} + \sqrt{3} \text{: rational + irrational = irrational}$

Answer: $\frac{1}{3} + \sqrt{3}$

A rational number plus an irrational number (like sqrt(3)) always produces an irrational number, while the other three expressions simplify to rational numbers.

Question 8 · 2 pts

Which equation has the same solutions as $x^2 + 6x - 18 = 0$?

(A) $(x + 3)^2 = 24$
(B) $(x + 3)^2 = 27$ ✓
(C) $(x + 6)^2 = 24$
(D) $(x + 6)^2 = 27$

Worked solution

$x^2 + 6x - 18 = 0 \Rightarrow x^2 + 6x = 18$

$\text{Add } \left(\frac{6}{2}\right)^2 = 9 \text{ to both sides: } x^2+6x+9 = 18+9$

$(x+3)^2 = 27$

Answer: $(x + 3)^2 = 27$

Completing the square on x^2+6x-18=0 requires adding 9 to both sides, giving (x+3)^2 = 27.

Question 9 · 2 pts

The heights, in inches, of eight football players are given below. 76, 70, 72, 70, 69, 71, 78, 74 Which box plot represents these data?

(A) Box plot (1): number line 65 to 85; left whisker at about 69, box from about 72 to 78 with the median at about 75, right whisker to about 84.
(B) Box plot (2): number line 65 to 85; left whisker at about 66, a narrow box from about 71 to 74 with the median at about 71.5, right whisker to about 84.
(C) Box plot (3): number line 65 to 85; left whisker at 69, box from 70 to 75 with the median at 71.5, right whisker to 78. ✓
(D) Box plot (4): number line 65 to 85; left whisker at about 70, a narrow box from about 71 to 74 with the median at about 72.5, right whisker to about 78.

Worked solution

$\text{Sort data: } 69, 70, 70, 71, 72, 74, 76, 78$

$\text{Min} = 69,\ \text{Max} = 78$

$\text{Median} = \frac{71+72}{2} = 71.5$

$Q_1 = \frac{70+70}{2} = 70,\quad Q_3 = \frac{74+76}{2} = 75$

Answer: the box plot with whiskers from 69 to 78 and a box from 70 to 75 (median 71.5)

Sorting the eight heights gives 69, 70, 70, 71, 72, 74, 76, 78, so the median is (71 + 72)/2 = 71.5, Q1 is the median of 69, 70, 70, 71, which is 70, and Q3 is the median of 72, 74, 76, 78, which is 75. This five-number summary of 69, 70, 71.5, 75, 78 matches the narrow box plot whose whiskers run only from 69 to 78 and whose box spans 70 to 75, with the median line sitting closer to the left edge of the box since 71.5 is nearer to 70 than to 75.

Question 10 · 2 pts

A bookstore owner recorded the number of books sold and the profit made selling the books. Books Sold: 100, Profit: $50.00 Books Sold: 250, Profit: $275.00 Books Sold: 300, Profit: $350.00 Books Sold: 350, Profit: $425.00 What is the average rate of change, in dollars per book, between 100 and 350 books sold?

(A) 0.50
(B) 0.67
(C) 1.50 ✓
(D) 2.00

Worked solution

$\text{Average rate of change} = \frac{425 - 50}{350 - 100} = \frac{375}{250}$

$= 1.50$

Answer: 1.50

The average rate of change is the change in profit divided by the change in books sold between the two given points, which equals 1.50 dollars per book.

Question 11 · 2 pts

If $f(x) = x^2$, then which function represents a shift of the graph of $f(x)$ 4 units to the right and 3 units down?

(A) $g(x) = (x + 4)^2 + 3$
(B) $j(x) = (x + 4)^2 - 3$
(C) $h(x) = (x - 4)^2 - 3$ ✓
(D) $k(x) = (x - 4)^2 + 3$

Worked solution

$\text{Shift right 4: replace } x \text{ with } (x-4)$

$\text{Shift down 3: subtract 3 from the function}$

$h(x) = (x-4)^2 - 3$

Answer: $h(x) = (x - 4)^2 - 3$

A horizontal shift right by 4 replaces x with (x-4), and a vertical shift down by 3 subtracts 3 from the whole expression.

Question 12 · 2 pts

The amount of money a plumber charges is represented by the function $p(h) = 45 + 90h$. The best interpretation of the $y$-intercept of this function is that the plumber charges

(A) $45 to come to the house ✓
(B) $45 per hour that he works
(C) $90 to come to the house
(D) $90 per hour that he works

Worked solution

$p(h) = 45 + 90h$

$\text{At } h=0,\ p(0)=45,\ \text{the flat/base fee charged regardless of hours}$

Answer: $45 to come to the house

The y-intercept (45) is the constant charge incurred at h=0 hours, representing the flat fee for the plumber to come to the house.

Question 13 · 2 pts

What is the solution to the inequality $2m - 4 \le 3(2m + 4)$?

(A) $m \le -2$
(B) $m \ge -2$
(C) $m \le -4$
(D) $m \ge -4$ ✓

Worked solution

$2m - 4 \le 3(2m+4) = 6m + 12$

$2m - 4 \le 6m + 12$

$-16 \le 4m$

$m \ge -4$

Answer: $m \ge -4$

Distributing and isolating m gives -16 <= 4m, and dividing by positive 4 (no sign flip) gives m >= -4.

Question 14 · 2 pts

A survey of students at West High School was taken to determine a theme for the prom. The results of the survey are summarized in the table below. Beach Party: Girls 86, Boys 123 Hollywood: Girls 112, Boys 77 Broadway: Girls 68, Boys 79 Approximately what percentage of the students who chose the Broadway theme were girls?

(A) 26
(B) 27
(C) 46 ✓
(D) 68

Worked solution

$\text{Total Broadway votes} = 68 + 79 = 147$

$\frac{68}{147} \approx 0.4626 = 46.26\%$

Answer: 46

Dividing the number of girls who chose Broadway (68) by the total Broadway votes (147) gives approximately 46 percent.

Question 15 · 2 pts

The sum of $2\sqrt{54}$ and $2\sqrt{6}$ is

(A) $4\sqrt{60}$
(B) $8\sqrt{15}$
(C) $7\sqrt{6}$
(D) $8\sqrt{6}$ ✓

Worked solution

$\sqrt{54} = \sqrt{9\cdot 6} = 3\sqrt{6}$

$2\sqrt{54} = 2(3\sqrt{6}) = 6\sqrt{6}$

$6\sqrt{6} + 2\sqrt{6} = 8\sqrt{6}$

Answer: $8\sqrt{6}$

Simplifying 2 times the square root of 54 to 6 times the square root of 6 allows combining like radical terms with 2 times the square root of 6 to get 8 times the square root of 6.

Question 16 · 2 pts

The functions $f(x) = x^2 - 5x - 14$ and $g(x) = x + 2$ are graphed on the same set of axes. What are the solutions to the equation $f(x) = g(x)$?

(A) $-14$ and $0$
(B) $0$ and $2$
(C) $-2$ and $8$ ✓
(D) $-2$ and $7$

Worked solution

$x^2-5x-14 = x+2$

$x^2 - 6x - 16 = 0$

$(x-8)(x+2) = 0$

$x = 8 \text{ or } x = -2$

Answer: $-2$ and $8$

Setting f(x) equal to g(x), moving all terms to one side, and factoring the resulting quadratic gives solutions x = 8 and x = -2.

Question 17 · 2 pts

If $x = 4a^2 - a + 3$ and $y = a - 5$, then which polynomial is equivalent to the product of $x$ and $y$?

(A) $-17a^2 - 2a - 15$
(B) $-17a^2 + 8a - 15$
(C) $4a^3 - 21a^2 - 2a - 15$
(D) $4a^3 - 21a^2 + 8a - 15$ ✓

Worked solution

$xy = (4a^2 - a + 3)(a-5)$

$= 4a^3 - 20a^2 - a^2 + 5a + 3a - 15$

$= 4a^3 - 21a^2 + 8a - 15$

Answer: $4a^3 - 21a^2 + 8a - 15$

Distributing each term of the first polynomial across (a-5) and combining like terms produces 4a^3 - 21a^2 + 8a - 15.

Question 18 · 2 pts

What is an equation of the line that passes through $(3,7)$ and has a slope of 2?

(A) $y - 7 = 2(x - 3)$ ✓
(B) $y - 3 = 2(x - 7)$
(C) $y + 7 = 2(x + 3)$
(D) $y + 3 = 2(x + 7)$

Worked solution

$\text{Point-slope form: } y - y_1 = m(x - x_1)$

$y - 7 = 2(x - 3)$

Answer: $y - 7 = 2(x - 3)$

Substituting the point (3,7) and slope 2 into point-slope form gives y - 7 = 2(x - 3).

Question 19 · 2 pts

A geometric sequence with a common ratio of $-3$ is

(A) $-10,-7,-4,-1, ...$
(B) $14,11,8,5, ...$
(C) $-2,-6,-18,-54, ...$
(D) $4,-12,36,-108, ...$ ✓

Worked solution

$\text{Check ratio: } \frac{-12}{4} = -3,\ \frac{36}{-12}=-3,\ \frac{-108}{36}=-3$

Answer: $4,-12,36,-108, ...$

Each term in choice 4 divided by the previous term equals -3, confirming a common ratio of -3; the sequence -2,-6,-18,-54 has ratio 3, not -3.

Question 20 · 2 pts

When the equation $6 - ax = ax - 2$ is solved for $x$ in terms of $a$, and $a \ne 0$, the result is

(A) $4a$
(B) $\frac{4}{a}$ ✓
(C) $2a$
(D) $\frac{2}{a}$

Worked solution

$6 - ax = ax - 2$

$6 + 2 = ax + ax$

$8 = 2ax$

$x = \frac{8}{2a} = \frac{4}{a}$

Answer: $\frac{4}{a}$

Collecting the ax terms on one side and constants on the other gives 8 = 2ax, and dividing by 2a yields x = 4/a.

Question 21 · 2 pts

Which function has the zeros $-1$, 3, and $-4$?

(A) $f(x) = (x + 1)(x - 3)(x - 4)$
(B) $g(x) = (x - 1)(x + 3)(x - 4)$
(C) $h(x) = (x + 1)(x - 3)(x + 4)$ ✓
(D) $k(x) = (x - 1)(x + 3)(x + 4)$

Worked solution

$\text{A zero at } x=-1 \text{ needs factor } (x+1)$

$\text{A zero at } x=3 \text{ needs factor } (x-3)$

$\text{A zero at } x=-4 \text{ needs factor } (x+4)$

$h(x)=(x+1)(x-3)(x+4)$

Answer: $h(x) = (x + 1)(x - 3)(x + 4)$

Each zero r of a polynomial corresponds to a factor (x - r), so zeros -1, 3, -4 require the factors (x+1), (x-3), and (x+4), matching choice 3.

Question 22 · 2 pts

The expression $5^{a + 2b}$ is equivalent to

(A) $5^{a} \cdot 5^{2} \cdot 5^{b}$
(B) $5^{a} \cdot 25^{b}$ ✓
(C) $25^{2ab}$
(D) $25^{a + 2b}$

Worked solution

$5^{a+2b} = 5^{a} \cdot 5^{2b}$

$5^{2b} = (5^{2})^{b} = 25^{b}$

$5^{a+2b} = 5^{a} \cdot 25^{b}$

Answer: $5^{a} \cdot 25^{b}$

Using the product rule for exponents, 5^(a+2b) splits into 5^a times 5^(2b), and 5^(2b) rewrites as 25^b since 5^2 = 25.

Question 23 · 2 pts

In an arithmetic sequence, the first term is 4 and the third term is $-2$. What is the common difference?

(A) $-1$
(B) $-2$
(C) $-3$ ✓
(D) $-6$

Worked solution

$a_3 = a_1 + 2d$

$-2 = 4 + 2d$

$-6 = 2d \Rightarrow d = -3$

Answer: $-3$

The third term equals the first term plus two common differences, so solving -2 = 4 + 2d gives d = -3.

Question 24 · 2 pts

Joe is ordering water for his swimming pool. He determines the volume of his pool to be about 3240 cubic feet. There are approximately 7.5 gallons of water in 1 cubic foot. A truck load holds 6000 gallons of water. Which expression would allow Joe to correctly calculate the number of truck loads of water he needs to fill his pool?

(A) $\dfrac{3240\ \text{ft}^3}{1\ \text{pool}} \cdot \dfrac{1\ \text{ft}^3}{7.5\ \text{gal}} \cdot \dfrac{6000\ \text{gal}}{1\ \text{truck load}}$
(B) $\dfrac{3240\ \text{ft}^3}{1\ \text{pool}} \cdot \dfrac{1\ \text{ft}^3}{7.5\ \text{gal}} \cdot \dfrac{1\ \text{truck load}}{6000\ \text{gal}}$
(C) $\dfrac{3240\ \text{ft}^3}{1\ \text{pool}} \cdot \dfrac{7.5\ \text{gal}}{1\ \text{ft}^3} \cdot \dfrac{6000\ \text{gal}}{1\ \text{truck load}}$
(D) $\dfrac{3240\ \text{ft}^3}{1\ \text{pool}} \cdot \dfrac{7.5\ \text{gal}}{1\ \text{ft}^3} \cdot \dfrac{1\ \text{truck load}}{6000\ \text{gal}}$ ✓

Worked solution

$\text{ft}^3 \text{ must cancel, so gal must be on top of the first conversion factor}$

$\text{gal must then cancel, so gal is on bottom of the second conversion factor}$

$\frac{3240\text{ ft}^3}{1}\cdot\frac{7.5\text{ gal}}{1\text{ ft}^3}\cdot\frac{1\text{ truck load}}{6000\text{ gal}} = \text{truck loads}$

Answer: $\dfrac{3240\ \text{ft}^3}{1\ \text{pool}} \cdot \dfrac{7.5\ \text{gal}}{1\ \text{ft}^3} \cdot \dfrac{1\ \text{truck load}}{6000\ \text{gal}}$

Correct dimensional analysis requires each unit in a denominator to cancel with the same unit in the next numerator, which only happens when gallons per cubic foot is multiplied (not divided) and truck loads per gallon finishes the chain, as in choice 4.

Question 25 · 2 pts · constructed response

On the set of axes below, graph $f(x) = x^2 + 4x + 1$. State the coordinates of the minimum.

Model answer: Minimum at $(-2, -3)$

Worked solution

$x = -\dfrac{b}{2a} = -\dfrac{4}{2(1)} = -2$

$f(-2) = (-2)^2 + 4(-2) + 1 = 4 - 8 + 1 = -3$

$\text{Minimum: } (-2,-3)$

$\text{Plot the parabola through points such as } (-4,1),(-3,-2),(-2,-3),(-1,-2),(0,1)$

Answer: Minimum at $(-2, -3)$

The axis of symmetry x = -b/(2a) = -2 gives the x-coordinate of the vertex, and substituting back into f(x) gives the minimum y-value of -3.

Question 26 · 2 pts · constructed response

If $f(x) = \dfrac{30x^2}{x+2}$, determine the value of $f\left(\dfrac{1}{2}\right)$.

Model answer: $f\left(\frac{1}{2}\right) = 3$

Worked solution

$f\left(\tfrac12\right) = \dfrac{30\left(\tfrac12\right)^2}{\tfrac12+2}$

$= \dfrac{30\left(\tfrac14\right)}{\tfrac52} = \dfrac{7.5}{2.5}$

$= 3$

Answer: $f\left(\frac{1}{2}\right) = 3$

Substitute x = 1/2 into the function and simplify the resulting fraction to get f(1/2) = 3.

Question 27 · 2 pts · constructed response

Explain why the relation shown in the table below is a function. | x | -1 | 0 | 1 | 2 | |---|----|---|---|---| | y | 2 | 4 | 4 | 5 | Complete the table below with values for both x and y so that this new relation is *not* a function. | x | -1 | 0 | 1 | 2 | | |---|----|---|---|---|--| | y | 2 | 4 | 4 | 5 | |

Model answer: The relation is a function because every x-value is paired with exactly one y-value. To make it not a function, add a fifth column with an x-value that repeats one already used (for example x = -1) but pairs it with a different y-value (for example y = 10).

Worked solution

$\text{A relation is a function if every input (x-value) maps to exactly one output (y-value).}$

$\text{In the table, } x=-1,0,1,2 \text{ each appear only once, each with a single } y.\text{ so it is a function.}$

$\text{To break this, repeat an x-value with a new y-value: add } x=-1,\ y=10 \text{ in the new column.}$

Answer: The relation is a function because every x-value is paired with exactly one y-value. To make it not a function, add a fifth column with an x-value that repeats one already used (for example x = -1) but pairs it with a different y-value (for example y = 10).

A function requires each x-value to correspond to only one y-value; since no x-value repeats with a different y in the original table it is a function, and repeating an existing x with a new, different y destroys that property.

Question 28 · 2 pts · constructed response

Solve algebraically for $x$: $0.05(x-3) = 0.35x - 7.5$

Model answer: $x = 24.5$

Worked solution

$0.05(x-3) = 0.35x - 7.5$

$0.05x - 0.15 = 0.35x - 7.5$

$-0.15 + 7.5 = 0.35x - 0.05x$

$7.35 = 0.3x$

$x = \dfrac{7.35}{0.3} = 24.5$

Answer: $x = 24.5$

Distribute, then collect the x-terms on one side and constants on the other, and divide to isolate x, giving x = 24.5.

Question 29 · 2 pts · constructed response

Use the quadratic formula to determine the exact roots of the equation $x^2 + 3x - 6 = 0$.

Model answer: $x = \dfrac{-3 \pm \sqrt{33}}{2}$

Worked solution

$a=1,\ b=3,\ c=-6$

$x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} = \dfrac{-3 \pm \sqrt{9-4(1)(-6)}}{2(1)}$

$= \dfrac{-3 \pm \sqrt{9+24}}{2} = \dfrac{-3 \pm \sqrt{33}}{2}$

Answer: $x = \dfrac{-3 \pm \sqrt{33}}{2}$

Substituting a=1, b=3, c=-6 into the quadratic formula and simplifying under the radical (9+24=33, which does not simplify further) gives the exact irrational roots.

Question 30 · 2 pts · constructed response

Factor $5x^3 - 80x$ completely.

Model answer: $5x(x-4)(x+4)$

Worked solution

$5x^3-80x = 5x(x^2-16)$

$x^2-16 = (x-4)(x+4) \text{ (difference of squares)}$

$5x^3-80x = 5x(x-4)(x+4)$

Answer: $5x(x-4)(x+4)$

First pull out the greatest common factor of 5x, then factor the resulting difference of squares x^2 - 16 into (x-4)(x+4).

Question 31 · 4 pts · constructed response

The owner of an ice cream stand kept track of the number of ice cream cones that were sold each day of the first week in June. She compared the ice cream sales to the average daily temperature. The data are shown in the table below. | Average Daily Temp. (x) | 72 | 75 | 81 | 78 | 77 | 76 | 80 | |---|---|---|---|---|---|---|---| | Daily Ice Cream Cone Sales (y) | 126 | 183 | 263 | 229 | 200 | 185 | 249 | State the linear regression equation for these data, rounding all values to the nearest hundredth. State the correlation coefficient, to the nearest hundredth, for the line of best fit for these data. State what this correlation coefficient indicates about the linear fit of the data.

Model answer: $y = 15.13x - 959.63$; $r \approx 0.99$, which indicates a strong, positive linear correlation (the linear model fits the data very well).

Worked solution

$n=7,\ \sum x = 539,\ \sum y = 1435,\ \sum xy = 111342,\ \sum x^2 = 41559,\ \sum y^2 = 307201$

$S_{xx} = \sum x^2 - \dfrac{(\sum x)^2}{n} = 41559 - \dfrac{539^2}{7} = 56$

$S_{xy} = \sum xy - \dfrac{(\sum x)(\sum y)}{n} = 111342 - \dfrac{539 \cdot 1435}{7} = 847$

$b = \dfrac{S_{xy}}{S_{xx}} = \dfrac{847}{56} = 15.125 \approx 15.13$

$a = \bar y - b\bar x = 205 - 15.125(77) = -959.625 \approx -959.63$

$S_{yy} = \sum y^2 - \dfrac{(\sum y)^2}{n} = 307201-\dfrac{1435^2}{7}=13026$

$r = \dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}} = \dfrac{847}{\sqrt{56\cdot13026}} \approx 0.99$

Answer: $y = 15.13x - 959.63$; $r \approx 0.99$, which indicates a strong, positive linear correlation (the linear model fits the data very well).

Using a linear regression calculation on the seven (x,y) data pairs gives a best-fit line with slope about 15.13 and y-intercept about -959.63, and the correlation coefficient of about 0.99 shows a very strong positive linear relationship between temperature and cone sales.

Question 32 · 4 pts · constructed response

Graph the system of inequalities on the set of axes below: $y > 3x - 4$ and $x + 2y \le 6$. Label the solution set $S$. Is the point $(2,2)$ a solution to the system? Justify your answer.

Model answer: No, (2,2) is not a solution because it fails the strict inequality $y > 3x-4$.

Worked solution

$\text{Graph } y=3x-4 \text{ as a dashed line (since the inequality is strict), shade above it for } y>3x-4$

$\text{Graph } x+2y=6, \text{ i.e., } y=-\tfrac12x+3, \text{ as a solid line (since } \le \text{), shade below/on it}$

$\text{Label the overlapping shaded region } S$

$\text{Check } (2,2): \ 2 > 3(2)-4 = 2 \Rightarrow 2>2 \text{ is false}$

$\text{Since the first inequality fails, } (2,2) \text{ is not a solution to the system}$

Answer: No, (2,2) is not a solution because it fails the strict inequality $y > 3x-4$.

The point (2,2) lies exactly on the boundary line y = 3x - 4, and because that inequality is strict (>), a point on the boundary does not satisfy it, so (2,2) is not in the solution set even though it does satisfy the second inequality.

Question 33 · 4 pts · constructed response

An object is launched upward at 64 feet per second from a platform 80 feet above the ground. The function $s(t)$ models the height of the object $t$ seconds after launch. If $s(t) = -16t^2 + 64t + 80$, state the vertex of $s(t)$, and explain in detail what each coordinate means in the context of the problem. After the object is launched, how many seconds does it take for the object to hit the ground? Justify your answer.

Model answer: Vertex: $(2, 144)$, meaning the object reaches its maximum height of 144 feet at 2 seconds after launch. The object hits the ground at $t = 5$ seconds.

Worked solution

$t = -\dfrac{b}{2a} = -\dfrac{64}{2(-16)} = 2$

$s(2) = -16(2)^2+64(2)+80 = -64+128+80 = 144$

$\text{Vertex } (2,144): \text{ at } t=2\text{ s the object reaches its maximum height of }144\text{ ft}$

$\text{Set } s(t)=0: -16t^2+64t+80=0$

$\text{Divide by } -16: t^2-4t-5=0 \Rightarrow (t-5)(t+1)=0$

$t=5 \text{ or } t=-1;\ \text{since } t\ge 0,\ t=5\text{ seconds}$

Answer: Vertex: $(2, 144)$, meaning the object reaches its maximum height of 144 feet at 2 seconds after launch. The object hits the ground at $t = 5$ seconds.

The vertex of the downward-opening parabola gives the time and maximum height the object reaches, while setting the height function equal to zero and solving (discarding the negative time) gives the time the object lands.

Question 34 · 4 pts · constructed response

Solve the system of equations algebraically for all values of $x$ and $y$. $y = x^2 + 4x - 1$ and $y = 2x + 7$

Model answer: $(x,y) = (-4,-1)$ and $(2,11)$

Worked solution

$x^2+4x-1 = 2x+7$

$x^2+2x-8=0$

$(x+4)(x-2)=0 \Rightarrow x=-4 \text{ or } x=2$

$x=-4: y=2(-4)+7=-1$

$x=2: y=2(2)+7=11$

Answer: $(x,y) = (-4,-1)$ and $(2,11)$

Setting the two expressions for y equal produces a quadratic equation that factors into two solutions for x, each of which is substituted back into the linear equation to find the corresponding y-values.

Question 35 · 6 pts · constructed response

Jen joined the Fan Favorite Movie Club at the local movie theater. At this theater, the cost of admission in May and June remains the same. In May, she saw 2 matinees and 3 regular-priced shows and spent $38.50. In June, she went to 6 matinees and one regular-priced show and spent $47.50. Write a system of equations to represent the cost, $m$, of a matinee ticket and the cost, $r$, of a regular-priced ticket. Jen said she spent \$5.75 on each matinee and \$9 on each regular show. Is Jen correct? Justify your answer. Use your system of equations to algebraically determine both the actual cost of each matinee ticket and the actual cost of each regular ticket.

Model answer: System: $2m+3r=38.50$ and $6m+r=47.50$. Jen is not correct. The actual costs are $m = \$6.50$ per matinee and $r = \$8.50$ per regular ticket.

Worked solution

$2m+3r=38.50 \quad (\text{May})$

$6m+r=47.50 \quad (\text{June})$

$\text{Check Jen's claim } m=5.75, r=9: 2(5.75)+3(9)=11.5+27=38.5 \checkmark$

$6(5.75)+9 = 34.5+9 = 43.5 \ne 47.50 \Rightarrow \text{Jen's values fail the June equation, so Jen is not correct}$

$\text{From June eq: } r = 47.50-6m$

$\text{Substitute into May eq: } 2m+3(47.50-6m)=38.50$

$2m+142.50-18m=38.50 \Rightarrow -16m=-104 \Rightarrow m=6.50$

$r=47.50-6(6.50)=47.50-39=8.50$

Answer: System: $2m+3r=38.50$ and $6m+r=47.50$. Jen is not correct. The actual costs are $m = \$6.50$ per matinee and $r = \$8.50$ per regular ticket.

Translating the two visits into a linear system, Jen's proposed prices satisfy the May equation but fail the June equation, showing she is incorrect; solving the system by substitution gives the true prices of $6.50 for a matinee and $8.50 for a regular-priced ticket.

August 2024 Algebra I Regents — Answer Key & Worked Solutions

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