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January 2025 Algebra I Regents — Answer Key & Worked Solutions

Every question from the January 2025 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 January 2025 Algebra I New York State Regents examination, published by NYSED. Worked solutions are written by RegentsLab.

Question 1 · 2 pts

When factored, the expression $x^3 - 36x$ is equivalent to

Worked solution
$x^3-36x = x(x^2-36)$ factor out the GCF x
$x^2-36 = (x+6)(x-6)$ difference of squares
$x^3-36x = x(x+6)(x-6)$
Answer: $x(x+6)(x-6)$

First factor out the common factor x, then factor the resulting difference of squares x^2-36.

Question 2 · 2 pts

Which equation represents the line that passes through the points $(-1,8)$ and $(4,-2)$?

Worked solution
$m = \frac{-2-8}{4-(-1)} = \frac{-10}{5} = -2$
$y - 8 = -2(x-(-1))$
$y = -2x - 2 + 8 = -2x+6$
Answer: $y=-2x+6$

Compute the slope between the two given points, then use point-slope form with one of the points to find the equation.

Question 3 · 2 pts

A geometric sequence is shown below. $\frac{1}{2}, 2, 8, 32, \ldots$ What is the common ratio?

Worked solution
$r = \frac{2}{1/2} = 4$
$\text{Check: } 2\times 4=8,\ 8\times4=32$
Answer: $4$

Divide any term by the term before it to find the common ratio of a geometric sequence.

Question 4 · 2 pts

What is the constant term of the polynomial $2x^3 - x + 5 + 4x^2$?

Worked solution
$2x^3-x+5+4x^2 = 2x^3+4x^2-x+5$ write in standard form
$\text{constant term} = 5$
Answer: $5$

The constant term is the term with no variable factor, which is 5.

Question 5 · 2 pts

A landscaping company charges a set fee for a spring cleanup, plus an hourly labor rate. The total cost is modeled by the function $C(x) = 55x + 80$. In this function, what does the 55 represent?

Worked solution
$C(x)=55x+80$ 55 is the coefficient multiplying x (hours), so it is the rate per hour; 80 is the fixed set fee
Answer: the hourly labor rate for a cleanup

In a linear cost function C(x)=mx+b, the coefficient m multiplying the variable (hours) represents the rate per unit, here the hourly labor rate.

Question 6 · 2 pts

Which expression is equivalent to $(5x^2 - 2x + 4) - (3x^2 + 3x - 1)$?

Worked solution
$(5x^2-2x+4)-(3x^2+3x-1)$
$=5x^2-2x+4-3x^2-3x+1$ distribute the negative sign
$=2x^2-5x+5$ combine like terms
Answer: $2x^2-5x+5$

Distribute the subtraction across the second polynomial, then combine like terms.

Question 7 · 2 pts

A system of inequalities is graphed on the set of axes below. Which point is a solution to this system?

alg1 question 7 figure
Worked solution
$\text{Test each point against both shaded regions shown in the graph}$
$\text{Only }(4,2)\text{ lies in the overlap of both shaded regions}$
Answer: $(4,2)$

The solution to a system of inequalities is the region where both shaded areas overlap; only (4,2) falls in that double-shaded region.

Question 8 · 2 pts

In an arithmetic sequence, the first term is 25 and the third term is 15. What is the tenth term in this sequence?

Worked solution
$a_3=a_1+2d \Rightarrow 15=25+2d \Rightarrow d=-5$
$a_{10}=a_1+9d=25+9(-5)=25-45=-20$
Answer: $-20$

Find the common difference using the first and third terms, then use it to find the tenth term.

Question 9 · 2 pts

When the formula $p = 2l + 2w$ is solved for $w$, the result is

Worked solution
$p=2l+2w$
$p-2l=2w$ subtract 2l from both sides
$w=\frac{p-2l}{2}$ divide both sides by 2
Answer: $w=\dfrac{p-2l}{2}$

Isolate the term with w by subtracting 2l from both sides, then divide by 2.

Question 10 · 2 pts

Market Street Pizza kept a record of pizza sales for the month of February. The results are shown in the table below. | Type | Plain | Veggie | Meat Only | The Works | |---|---|---|---|---| | Thin Crust | 300 | 80 | 120 | 100 | | Deep-dish | 200 | 25 | 105 | 70 | Of all the pizzas sold in February, what percent were plain, deep-dish pizzas?

alg1 question 10 figure
Worked solution
$\text{Total Thin Crust}=300+80+120+100=600$
$\text{Total Deep-dish}=200+25+105+70=400$
$\text{Total pizzas}=600+400=1000$
$\frac{200}{1000}=0.20=20\%$
Answer: $20\%$

Divide the number of plain deep-dish pizzas (200) by the grand total of all pizzas sold (1000) to get the percentage.

Question 11 · 2 pts

When solving $-2(3x-5) = \frac{9}{2}x - 2$ for $x$, the solution is

Worked solution
$-2(3x-5)=\frac{9}{2}x-2$
$-6x+10=\frac{9}{2}x-2$ distribute -2
$10+2=\frac{9}{2}x+6x$ add 6x and 2 to both sides
$12=\frac{21}{2}x$
$x=\frac{24}{21}=\frac{8}{7}$
Answer: $\frac{8}{7}$

Distribute the -2, collect x-terms on one side and constants on the other, then solve for x.

Question 12 · 2 pts

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

Worked solution
$x^a \cdot x^{a+b} = x^{a+(a+b)} = x^{2a+b}$ product of powers property: add exponents
Answer: $x^a \cdot x^{a+b}$

Using the product of powers rule, adding the exponents a and (a+b) gives 2a+b, matching the original expression.

Question 13 · 2 pts

The inputs and outputs of a function are shown in the table below. | x | f(x) | |---|---| | 0 | 0.0625 | | 1 | 0.125 | | 2 | 0.25 | | 3 | 0.5 | | 4 | 1 | | 5 | 2 | This function can best be described as

alg1 question 13 figure
Worked solution
$\frac{0.125}{0.0625}=2,\ \frac{0.25}{0.125}=2,\ \frac{0.5}{0.25}=2,\ \ldots$ each output is double the previous output
$\text{Constant ratio between consecutive outputs} \Rightarrow \text{exponential}$
Answer: exponential

Since each output is a constant multiple (2) of the previous output rather than a constant difference, the function is exponential.

Question 14 · 2 pts

Stephanie is solving the equation $x^2 - 12 = 7x - 8$. Her first step is shown below. Given: $x^2 - 12 = 7x - 8$ Step 1: $x^2 - 4 = 7x$ Which property justifies her first step?

Worked solution
$x^2-12=7x-8$
$x^2-12+8=7x-8+8$ add 8 to both sides of the equation
$x^2-4=7x$
Answer: addition property of equality

Adding the same quantity (8) to both sides of the equation to eliminate the -8 term is justified by the addition property of equality.

Question 15 · 2 pts

What is the sum of $8\sqrt{3}$ and $\sqrt{3}$?

Worked solution
$8\sqrt{3}+\sqrt{3}=8\sqrt{3}+1\sqrt{3}=(8+1)\sqrt{3}=9\sqrt{3}$
Answer: $9\sqrt{3}$

Like radicals combine the way like terms do, by adding their coefficients while keeping the radical the same.

Question 16 · 2 pts

The dot plots below represent test scores for 20 students on a math test. The mode for this math test is 80 and the median is 85. Which dot plot correctly represents this data?

alg1 question 16 figure
Worked solution
$\text{Mode }=80 \Rightarrow \text{the value 80 must have the most dots (tallest stack)}$
$\text{Median }=85 \Rightarrow \text{with 20 data points, the average of the 10th and 11th values is 85}$
$\text{Only dot plot I satisfies both conditions}$
Answer: I

The correct dot plot must show 80 as the tallest stack (the mode) while the middle two values of the ordered 20 data points average to 85 (the median).

Question 17 · 2 pts

A function is graphed on the set of axes below. The domain of this function is

alg1 question 17 figure
Worked solution
$\text{The graph shows an open circle at the left endpoint, at }x=-2$
$\text{The graph extends to the right indefinitely (arrow)}$
$\text{Since the endpoint is an open circle, }x=-2\text{ is not included}$
Answer: $\{x \mid x>-2\}$

An open circle on the graph's left endpoint means that x-value is excluded from the domain, so the domain is all x greater than that value.

Question 18 · 2 pts

Which ordered pair is a solution to the equation $y - 1 = 2\left(x + \frac{1}{4}\right)$?

Worked solution
$y-1=2\left(x+\frac{1}{4}\right)=2x+0.5$
$y=2x+1.5$
$\text{Test }(1.25,4):\ y=2(1.25)+1.5=2.5+1.5=4$ matches given y-value
Answer: $(1.25,4)$

Simplify the equation to slope-intercept form y=2x+1.5, then substitute each ordered pair to find the one that satisfies the equation.

Question 19 · 2 pts

Elena's fastest time for the 50-meter dash is 7 seconds. She wants to know how fast this is in inches per minute. Which expression can Elena use for a correct conversion?

Worked solution
$\text{Speed} = \frac{50\text{ meters}}{7\text{ sec}}$ start with meters per second, since we want a rate not its reciprocal
$\frac{50\text{ meters}}{7\text{ sec}}\cdot\frac{60\text{ sec}}{1\text{ min}}$ convert seconds to minutes, canceling sec
$\cdot\frac{39.37\text{ in}}{1\text{ meter}}$ convert meters to inches, canceling meters, leaving inches per minute
Answer: $\dfrac{50\text{ meters}}{7\text{ sec}} \cdot \dfrac{60\text{ sec}}{1\text{ min}} \cdot \dfrac{39.37\text{ in}}{1\text{ meter}}$

To convert meters per second to inches per minute, the units must cancel properly: seconds must cancel with seconds and meters with meters, leaving inches over minutes.

Question 20 · 2 pts

The table below shows the highest temperatures recorded in August for several years in one town. | Year | Temperature (°F) | |---|---| | 1990 | 86 | | 1991 | 78 | | 1992 | 84 | | 1993 | 95 | | 1994 | 81 | | 1995 | 77 | | 1996 | 88 | | 1997 | 93 | The interquartile range of these data is

alg1 question 20 figure
Worked solution
$\text{Sorted: }77,78,81,84,86,88,93,95$
$\text{Lower half: }77,78,81,84 \Rightarrow Q_1=\frac{78+81}{2}=79.5$
$\text{Upper half: }86,88,93,95 \Rightarrow Q_3=\frac{88+93}{2}=90.5$
$IQR=Q_3-Q_1=90.5-79.5=11$
Answer: $11$

Order the data, split into lower and upper halves to find Q1 and Q3, then subtract Q1 from Q3 to get the interquartile range.

Question 21 · 2 pts

The function $f(x) = x^2$ is multiplied by $k$, where $k < -1$. Which graph could represent $g(x) = kf(x)$?

alg1 question 21 figure
Worked solution
$g(x) = kf(x) = kx^2, \quad k < -1$
$\text{Since } k < 0,\ g(x) \text{ opens downward (reflection over the x-axis).}$
$\text{Since } |k| > 1,\ g(x) \text{ is narrower (vertically stretched) than } f(x) = x^2.$

Multiplying x^2 by a constant less than -1 reflects the parabola over the x-axis (so it opens downward) and stretches it vertically, making it narrower than the parent function, matching choice 4.

Question 22 · 2 pts

Which graph is the solution to the inequality $6.4 - 4x \ge -2.8$?

alg1 question 22 figure
Worked solution
$6.4 - 4x \ge -2.8$
$-4x \ge -2.8 - 6.4 = -9.2$
$x \le \dfrac{-9.2}{-4} = 2.3 \quad (\text{inequality flips when dividing by a negative number})$

Dividing both sides by -4 reverses the inequality symbol, giving x ≤ 2.3, which is graphed with a closed circle at 2.3 and shading to the left.

Question 23 · 2 pts

The number of fish in a pond is eight more than the number of frogs. The total number of fish and frogs in the pond is at least 20. If $x$ represents the number of frogs, which inequality can be used to represent this situation?

Worked solution
$\text{frogs} = x, \quad \text{fish} = x + 8$
$\text{fish} + \text{frogs} \ge 20 \Rightarrow (x+8) + x \ge 20$
$2x + 8 \ge 20$

Fish equal x + 8, and the sum of fish and frogs must be at least 20, giving 2x + 8 ≥ 20.

Question 24 · 2 pts

Which graph below represents a function that is always *decreasing* over the entire interval $-3 < x < 3$?

alg1 question 24 figure
Worked solution
$\text{A function is "always decreasing" on an interval if, for every pair of x-values in that interval, a larger x gives a smaller } f(x).$
$\text{Choices 1 and 3 increase then decrease (have a peak); choice 2 increases throughout.}$
$\text{Only choice 4 falls continuously across the entire interval } -3 < x < 3.$

Only the graph that falls from left to right over the whole interval, without turning upward anywhere, qualifies as always decreasing on -3 < x < 3, which is graph 4.

Question 25 · 2 pts · constructed response

The graph below models Sally's drive to the store. State an interval when Sally is traveling at a constant speed. Explain your reasoning.

alg1 question 25 figure
Model answer: Sally travels at a constant speed on the interval from t = 5 minutes to t = 9 minutes.
Worked solution
$\text{From } t=5 \text{ to } t=9 \text{ minutes, the graph is a horizontal (flat) segment at } y = 35 \text{ mph}.$
$\text{A horizontal segment on a speed-vs-time graph means the speed is not changing, i.e., constant speed.}$
Answer: Sally travels at a constant speed on the interval from t = 5 minutes to t = 9 minutes.

Constant speed corresponds to a flat (horizontal) portion of a speed-time graph; here that occurs from 5 to 9 minutes, where speed stays at 35 mph.

Question 26 · 2 pts · constructed response

Graph the function $f(x) = x^2 + 4x + 3$. State the equation of the axis of symmetry of $f(x)$.

Model answer: The axis of symmetry is $x = -2$.
Worked solution
$f(x) = x^2 + 4x + 3, \quad a = 1,\ b = 4$
$x = \dfrac{-b}{2a} = \dfrac{-4}{2(1)} = -2$
$\text{Check by factoring: } f(x) = (x+1)(x+3), \text{ roots at } x=-1, x=-3, \text{ midpoint } = \dfrac{-1+(-3)}{2} = -2$
Answer: The axis of symmetry is $x = -2$.

The axis of symmetry of a parabola in the form ax^2+bx+c is x = -b/2a, which gives x = -2 for this function; this is confirmed as the midpoint of the roots -1 and -3.

Question 27 · 2 pts · constructed response

The function $f(x)$ is shown in the table below. | x | 0 | 3 | 2 | 6 | 1 | 5 | 4 | m | |---|---|---|---|---|---|---|---|---| | f(x) | 6 | 2 | 7 | 5 | 8 | 4 | 3 | 9 | State an appropriate value for $m$ in the table, so that $f(x)$ remains a function. Explain your reasoning.

alg1 question 27 figure
Model answer: m can be any value other than 0, 1, 2, 3, 4, 5, or 6 (for example, m = 7).
Worked solution
$\text{For a relation to be a function, every input (domain value) must be paired with exactly one output.}$
$\text{The existing domain values are } \{0,1,2,3,4,5,6\}.$
$\text{If } m \text{ equals any of these repeated values, } m \text{ would have two different outputs, violating the function definition.}$
$\text{Therefore } m \text{ must be a number not already in the domain, e.g., } m = 7.$
Answer: m can be any value other than 0, 1, 2, 3, 4, 5, or 6 (for example, m = 7).

A function cannot assign two different outputs to the same input, so m must differ from every x-value already listed (0 through 6); choosing m = 7 keeps the relation a function.

Question 28 · 2 pts · constructed response

Solve $x^2 + 8x = 33$ for $x$ by completing the square.

Model answer: x = 3 or x = -11
Worked solution
$x^2 + 8x = 33$
$x^2 + 8x + 16 = 33 + 16 \quad \left(\left(\tfrac{8}{2}\right)^2 = 16\right)$
$(x+4)^2 = 49$
$x + 4 = \pm 7$
$x = 3 \ \text{ or } \ x = -11$
Answer: x = 3 or x = -11

Adding 16 (half of 8, squared) to both sides completes the square, allowing the equation to be solved by taking the square root of both sides.

Question 29 · 2 pts · constructed response

If $f(x) = \dfrac{-3x-5}{2}$, algebraically determine the value of $x$ when $f(x) = -22$.

Model answer: x = 13
Worked solution
$-22 = \dfrac{-3x-5}{2}$
$-44 = -3x - 5$
$-39 = -3x$
$x = 13$
Answer: x = 13

Setting f(x) equal to -22, multiplying both sides by 2, and solving the resulting linear equation gives x = 13.

Question 30 · 2 pts · constructed response

Rationalize the denominator of the fraction below. Express the solution in simplest form. $\dfrac{4}{\sqrt{2}}$

Model answer: $2\sqrt{2}$
Worked solution
$\dfrac{4}{\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{4\sqrt{2}}{2}$
$= 2\sqrt{2}$
Answer: $2\sqrt{2}$

Multiplying numerator and denominator by √2 removes the radical from the denominator, and simplifying gives 2√2.

Question 31 · 4 pts · constructed response

Alex had $1.70 in nickels and dimes on his desk. There were 25 coins in all. Write a system of equations that could be used to determine both the number of nickels, $n$, and the number of dimes, $d$, that Alex had. Use your system of equations to algebraically determine both the number of nickels and the number of dimes that he had.

Model answer: Alex had 16 nickels and 9 dimes.
Worked solution
$n + d = 25$
$0.05n + 0.10d = 1.70$
$n = 25 - d$
$0.05(25-d) + 0.10d = 1.70$
$1.25 - 0.05d + 0.10d = 1.70$
$0.05d = 0.45 \Rightarrow d = 9$
$n = 25 - 9 = 16$
Answer: Alex had 16 nickels and 9 dimes.

The system relates total coins (n + d = 25) and total value (0.05n + 0.10d = 1.70); substitution yields d = 9 dimes and n = 16 nickels.

Question 32 · 4 pts · constructed response

The table below shows the average heart rate, $x$, and Calories burned, $y$, for seven men on an Olympic rowing team during a one-hour workout class. | Average Heart Rate (x) | 135 | 147 | 150 | 144 | 146 | 153 | 143 | |---|---|---|---|---|---|---|---| | Calories Burned (y) | 725 | 812 | 866 | 761 | 825 | 863 | 737 | Write the linear regression equation that models these data, rounding all values to the nearest tenth. State the correlation coefficient, rounded to the nearest tenth. State what the correlation coefficient suggests about the linear fit of these data.

alg1 question 32 figure
Model answer: y = 9.1x - 527.6; r ≈ 0.9, indicating a strong positive linear relationship (a good linear fit).
Worked solution
$\text{Using linear regression on the 7 data pairs (calculator/statistics):}$
$\text{slope } b \approx 9.1, \quad \text{y-intercept } a \approx -527.6$
$y = 9.1x - 527.6$
$\text{correlation coefficient } r \approx 0.9$
$\text{Since } r \text{ is close to } 1, \text{ the data show a strong positive linear correlation, meaning the linear model fits the data well.}$
Answer: y = 9.1x - 527.6; r ≈ 0.9, indicating a strong positive linear relationship (a good linear fit).

Running linear regression on the heart rate/calories data gives the equation y = 9.1x - 527.6 with r ≈ 0.9, showing a strong, positive linear relationship, so the linear model is a good fit for the data.

Question 33 · 4 pts · constructed response

Using the quadratic formula, solve $x^2 + 4x - 3 = 0$. Express your solution in simplest radical form.

Model answer: $x = -2 + \sqrt{7}$ or $x = -2 - \sqrt{7}$
Worked solution
$a=1,\ b=4,\ c=-3$
$x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} = \dfrac{-4 \pm \sqrt{16+12}}{2}$
$= \dfrac{-4 \pm \sqrt{28}}{2} = \dfrac{-4 \pm 2\sqrt{7}}{2}$
$x = -2 \pm \sqrt{7}$
Answer: $x = -2 + \sqrt{7}$ or $x = -2 - \sqrt{7}$

Substituting a=1, b=4, c=-3 into the quadratic formula and simplifying the radical √28 = 2√7 gives x = -2 ± √7.

Question 34 · 4 pts · constructed response

Solve the following system of equations algebraically for all values of $x$ and $y$. $y = x^2 - 7x + 12$ $y = 2x - 6$

Model answer: (3, 0) and (6, 6)
Worked solution
$x^2 - 7x + 12 = 2x - 6$
$x^2 - 9x + 18 = 0$
$(x-3)(x-6) = 0$
$x = 3 \ \text{ or } \ x = 6$
$x=3: y = 2(3)-6 = 0; \quad x=6: y = 2(6)-6 = 6$
Answer: (3, 0) and (6, 6)

Setting the two expressions for y equal produces a quadratic equation that factors to give x = 3 and x = 6; substituting back into y = 2x - 6 gives the solutions (3, 0) and (6, 6).

Question 35 · 6 pts · constructed response

Anna plans to spend $30 on balloons and party hats for her daughter's birthday party. Including tax, balloons cost $2 each and party hats cost $1.50 each. The number of party hats Anna needs is twice as many as the number of balloons. If $x$ represents the number of balloons and $y$ represents the number of party hats, write a system of equations that can be used to represent this situation. Graph your system of equations on a set of axes where the x-axis shows the number of balloons from 0 to 20 and the y-axis shows the number of party hats from 0 to 20. State the coordinates of the point of intersection of your lines. Explain what each coordinate means in the context of the problem.

Model answer: System: 2x + 1.5y = 30 and y = 2x. Point of intersection: (6, 12). This means Anna buys 6 balloons and 12 party hats to spend exactly $30, satisfying the requirement of twice as many hats as balloons.
Worked solution
$2x + 1.5y = 30 \quad \text{(total cost equation)}$
$y = 2x \quad \text{(twice as many hats as balloons)}$
$2x + 1.5(2x) = 30$
$2x + 3x = 30$
$5x = 30 \Rightarrow x = 6$
$y = 2(6) = 12$
$\text{Point of intersection: } (6, 12)$
Answer: System: 2x + 1.5y = 30 and y = 2x. Point of intersection: (6, 12). This means Anna buys 6 balloons and 12 party hats to spend exactly $30, satisfying the requirement of twice as many hats as balloons.

Substituting y = 2x into the cost equation 2x + 1.5y = 30 gives x = 6 balloons and y = 12 party hats; the intersection point (6, 12) represents the unique combination of balloons and hats that costs exactly $30 while keeping hats at twice the number of balloons.

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