Algebra 2 Quiz: Creating And Graphing Two Variable Equations
20 questions · exam conditions
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Creating And Graphing Two Variable EquationsQuestion 1 of 20

A theater sells tickets for $18 each. The theater also charges a flat online processing fee of $4 per order (no matter how many tickets).

Write an equation for total cost TT (dollars) in terms of number of tickets nn (tickets), and choose appropriate axes labels and scales to graph for 0≤n≤120\le n\le 12.

Which choice is correct?​

Equation: T=18n−4T=18n-4; Graph: x-axis Tickets (tickets), y-axis Total cost ($); x-scale 0 to 12 by 1, y-scale 0 to 220 by 20
Equation: T=18n+4T=18n+4; Graph: x-axis Tickets (tickets), y-axis Total cost ($); x-scale 0 to 12 by 1, y-scale 0 to 220 by 20
Equation: T=18+4nT=18+4n; Graph: x-axis Tickets, y-axis Cost; x-scale 0 to 12 by 2, y-scale 0 to 60 by 5
Equation: n=18T+4n=18T+4; Graph: x-axis Total cost ($), y-axis Tickets (tickets); x-scale 0 to 12 by 1, y-scale 0 to 220 by 20
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Algebra 2 Quiz

Algebra 2 Quiz: Creating And Graphing Two Variable Equations

Practice Creating And Graphing Two Variable Equations in Algebra 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Creating And Graphing Two Variable Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A theater sells tickets for $18 each. The theater also charges a flat online processing fee of $4 per order (no matter how many tickets).

Write an equation for total cost TT (dollars) in terms of number of tickets nn (tickets), and choose appropriate axes labels and scales to graph for 0≤n≤120\le n\le 12.

Which choice is correct?​

  1. Equation: T=18n−4T=18n-4; Graph: x-axis Tickets (tickets), y-axis Total cost ($); x-scale 0 to 12 by 1, y-scale 0 to 220 by 20
  2. Equation: T=18n+4T=18n+4; Graph: x-axis Tickets (tickets), y-axis Total cost ($); x-scale 0 to 12 by 1, y-scale 0 to 220 by 20 (correct answer)
  3. Equation: T=18+4nT=18+4n; Graph: x-axis Tickets, y-axis Cost; x-scale 0 to 12 by 2, y-scale 0 to 60 by 5
  4. Equation: n=18T+4n=18T+4; Graph: x-axis Total cost ($), y-axis Tickets (tickets); x-scale 0 to 12 by 1, y-scale 0 to 220 by 20
Explanation: This question tests your ability to translate real-world relationships into mathematical equations with two or more variables and set up graphs with appropriate labels and scales to visualize these relationships. Creating equations from contexts requires identifying: (1) which quantities vary (your variables), (2) which depends on which (independent vs dependent), (3) the mathematical relationship connecting them (linear rate, quadratic area, exponential growth, etc.). For 'cost is $40 per hour plus $15 per item,' the dependent quantity is cost (C), independent quantities are hours (h) and items (n), and the relationship is additive with rates: C = 40h + 15n. The equation structure mirrors the context structure! The context describes $18 per ticket plus a flat 4fee,sototalcostT=18n+4,withnasindependent(tickets)andTdependent;forgraphing0to12tickets,x−axistickets(tickets),y−axistotalcost(4 fee, so total cost T = 18n + 4, with n as independent (tickets) and T dependent; for graphing 0 to 12 tickets, x-axis tickets (tickets), y-axis total cost (), x-scale 0-12 by 1 to show each unit, y-scale 0-220 by 20 since max is 18*12 + 4 = 220, providing clear increments. Choice A correctly creates the equation T=18n+4 with proper variables and coefficients, and sets up the graph with detailed labels including units and scales that fit the data range effectively. A distractor like choice B swaps the coefficients, maybe confusing the per-ticket cost with the fee, but gently correct by matching 'each' to the rate multiplier and 'flat' to the added constant. Equation creation framework: (1) Define your variables clearly—'Let n = number of tickets (tickets), T = total cost (dollars)'—being specific prevents confusion, (2) Identify the mathematical structure from context language: 'per' means multiply (rate), 'plus' means add, 'times' or 'product' means multiply, 'percent' means exponential, (3) Build the equation piece by piece matching each phrase in the context, (4) Verify with a test value: does your equation give sensible output for a reasonable input? This catches setup errors before solving! Graph scale decision process: (1) Find your data range—what are the minimum and maximum values you need to show for each axis? (2) Divide that range by 5-10 to get interval size, (3) Round to a 'nice' number: use 1, 2, 5, 10, 20, 50, 100, etc. (not 7 or 13!), (4) Mark intervals starting at 0 (or other logical point). Example: data from 0 to 80 → range is 80, divided by 8 gives 10, so mark every 10: 0, 10, 20, ..., 80. Clean, readable, shows all data!

Question 2

A gym membership costs 2525 per month plus a one-time sign-up fee of 6060. Let mm be the number of months.

Write an equation for total cost CC (dollars) in terms of mm and choose appropriate axes labels and scale to graph for 0≤m≤120 \leq m \leq 12.

Which choice is correct?

  1. Equation: m=25C+60m=25C+60; Graph: x-axis Cost ($), y-axis Months (months); x-scale 0 to 12 by 1, y-scale 0 to 400 by 50
  2. Equation: C=25m+60C=25m+60; Graph: x-axis Months (months), y-axis Cost ($); x-scale 0 to 12 by 1, y-scale 0 to 400 by 50 (correct answer)
  3. Equation: C=25+m+60C=25+m+60; Graph: x-axis Months (months), y-axis Cost ($); x-scale 0 to 12 by 1, y-scale 0 to 100 by 10
  4. Equation: C=60m+25C=60m+25; Graph: x-axis Months (months), y-axis Cost ($); x-scale 0 to 12 by 1, y-scale 0 to 400 by 50
Explanation: This question tests your ability to translate real-world relationships into mathematical equations with two or more variables and set up graphs with appropriate labels and scales to visualize these relationships. Creating equations from contexts requires identifying: (1) which quantities vary (your variables), (2) which depends on which (independent vs dependent), (3) the mathematical relationship connecting them (linear rate, quadratic area, exponential growth, etc.). For 'cost is $40 per hour plus $15 per item,' the dependent quantity is cost (C), independent quantities are hours (h) and items (n), and the relationship is additive with rates: $C = 40h + 15n.Theequationstructuremirrorsthecontextstructure!$25permonthplus$60flatgives$C=25m+60. The equation structure mirrors the context structure! $25 per month plus $60 flat gives $C=25m + 60, linear; graph x-axis months (months), y-axis cost (),0−12by1x,0−400by50ysincemax), 0-12 by 1 x, 0-400 by 50 y since max 25 \times 12 + 60 = 360.ChoiceBcorrectlycreatestheequation. Choice B correctly creates the equation C=25m+60$ with right order and sets up the graph with suitable labels and scales. Choice A reverses coefficients, maybe mixing monthly and one-time—correct by assigning 'per month' to multiplier of m and 'one-time' to constant. Equation creation framework: (1) Define your variables clearly—'Let m = months, C = total cost (dollars)'—being specific prevents confusion, (2) Identify the mathematical structure from context language: 'per' means multiply (rate), 'plus' means add, 'times' or 'product' means multiply, 'percent' means exponential, (3) Build the equation piece by piece matching each phrase in the context, (4) Verify with a test value: does your equation give sensible output for a reasonable input? This catches setup errors before solving! Graph scale decision process: (1) Find your data range—what are the minimum and maximum values you need to show for each axis? (2) Divide that range by 5-10 to get interval size, (3) Round to a 'nice' number: use 1, 2, 5, 10, 20, 50, 100, etc. (not 7 or 13!), (4) Mark intervals starting at 0 (or other logical point). Example: data from 0 to 80 → range is 80, divided by 8 gives 10, so mark every 10: 0, 10, 20, ..., 80. Clean, readable, shows all data!

Question 3

A bike rental shop charges a fixed fee of $12 plus $4 per hour. Write an equation for the total cost $C(dollars)intermsofthenumberofhours(dollars) in terms of the number of hoursh,andchooseappropriateaxeslabelsandareasonablescaletographtherelationshipfor, and choose appropriate axes labels and a reasonable scale to graph the relationship for 0 \le h \le 8$.

  1. Equation: h=4C+12h = 4C + 12; x-axis: Cost ($); y-axis: Time (hours); scale: x 0–50 by 5, y 0–8 by 1
  2. Equation: C=12+4hC = 12 + 4h; x-axis: Hours (hours); y-axis: Cost ($); scale: x 0–8 by 1, y 0–50 by 5 (correct answer)
  3. Equation: C=12h+4C = 12h + 4; x-axis: Hours (hours); y-axis: Cost ($); scale: x 0–8 by 1, y 0–60 by 10
  4. Equation: C=12+4hC = 12 + \frac{4}{h}; x-axis: Hours (hours); y-axis: Cost ($); scale: x 0–8 by 1, y 0–50 by 5
Explanation: This question tests your ability to translate real-world relationships into mathematical equations with two or more variables and set up graphs with appropriate labels and scales to visualize these relationships. Creating equations from contexts requires identifying: (1) which quantities vary (your variables), (2) which depends on which (independent vs dependent), (3) the mathematical relationship connecting them (linear rate, quadratic area, exponential growth, etc.). For 'fixed fee of $12 plus $4 per hour,' the dependent quantity is cost (C), independent quantity is hours (h), and the relationship is additive with a constant plus a rate: C = 12 + 4h. The equation structure mirrors the context structure! The fixed fee of 12istheconstantterm(youpaythisregardlessofhours),and′12 is the constant term (you pay this regardless of hours), and '4 per hour' means multiply 4 by h. This gives us C = 12 + 4h, which can also be written as C = 4h + 12. Choice C correctly creates the equation C = 12 + 4h with proper axis labels (Hours on x-axis, Cost on y-axis) and a reasonable scale showing the full range of values. Choice B incorrectly reverses the coefficients (12h + 4 instead of 4h + 12), while Choice D uses division instead of multiplication for the hourly rate. Equation creation framework: (1) Define your variables clearly—'Let h = number of hours rented'—being specific prevents confusion, (2) Identify the mathematical structure from context language: 'per' means multiply (rate), 'plus' means add, 'fixed fee' is a constant term, (3) Build the equation piece by piece matching each phrase in the context, (4) Verify with a test value: for h = 2 hours, cost should be 12 + 4(2) = $20. This catches setup errors before graphing!

Question 4

A rectangular garden has a perimeter of 48 feet. Let ll be the length (ft) and ww be the width (ft). Write an equation relating ll and ww, and choose appropriate axes labels and a reasonable scale to graph the relationship in the first quadrant.

  1. Equation: lw=48lw = 48; x-axis: Length ll (ft); y-axis: Width ww (ft); scale: x 0–48 by 4, y 0–48 by 4
  2. Equation: 2l+2w=482l + 2w = 48 (equivalently w=24−lw = 24 - l); x-axis: Length ll (ft); y-axis: Width ww (ft); scale: x 0–24 by 2, y 0–24 by 2 (correct answer)
  3. Equation: l+w=48l + w = 48; x-axis: Length ll (ft); y-axis: Width ww (ft); scale: x 0–24 by 2, y 0–24 by 2
  4. Equation: 2l−2w=482l - 2w = 48; x-axis: Length ll (ft); y-axis: Width ww (ft); scale: x 0–24 by 2, y 0–24 by 2
Explanation: This question tests your ability to translate real-world relationships into mathematical equations with two or more variables and set up graphs with appropriate labels and scales to visualize these relationships. Creating equations from contexts requires identifying: (1) which quantities vary (your variables), (2) which depends on which (independent vs dependent), (3) the mathematical relationship connecting them (linear rate, quadratic area, exponential growth, etc.). For a rectangle with perimeter 48 feet, the perimeter formula is P = 2l + 2w, so 2l + 2w = 48. This can be simplified by dividing by 2 to get l + w = 24, or solved for w to get w = 24 - l. The equation structure mirrors the geometric relationship! The perimeter of a rectangle is the sum of all four sides: length + width + length + width = 2l + 2w. Setting this equal to 48 gives 2l + 2w = 48. We can express this in multiple equivalent forms: dividing by 2 gives l + w = 24, or solving for w gives w = 24 - l. Choice B correctly creates the equation 2l + 2w = 48 (equivalently w = 24 - l) with proper axis labels and a reasonable scale for the first quadrant. Choice A incorrectly uses the area formula (lw) instead of perimeter, Choice C has the wrong constant (should be 24, not 48 after simplifying), and Choice D uses subtraction instead of addition. Proper graph setup requires: (1) recognizing that if l + w = 24, then both l and w must be between 0 and 24 (in the first quadrant), (2) choosing scales that show this full range clearly, (3) using consistent intervals (every 2 feet works well for 0 to 24). The graph will be a straight line from (0, 24) to (24, 0), showing all possible length-width combinations for the given perimeter!

Question 5

A taxi fare is based on a $3.50 starting fee plus $2.25 per mile. Write an equation for the fare FF (dollars) in terms of miles mm, and choose appropriate axes labels and a reasonable scale to graph the fare for 0≤m≤150 \le m \le 15.

  1. Equation: F=3.50m+2.25F = 3.50m + 2.25; x-axis: Miles mm (miles); y-axis: Fare FF ($); scale: x 0–15 by 1, y 0–40 by 5
  2. Equation: F=3.50+2.25mF = 3.50 + 2.25m; x-axis: Miles mm (miles); y-axis: Fare FF ($); scale: x 0–15 by 1, y 0–40 by 5 (correct answer)
  3. Equation: m=3.50+2.25Fm = 3.50 + 2.25F; x-axis: Miles mm (miles); y-axis: Fare FF ($); scale: x 0–15 by 1, y 0–40 by 5
  4. Equation: F=3.50+2.25mF = 3.50 + \frac{2.25}{m}; x-axis: Miles mm (miles); y-axis: Fare FF ($); scale: x 0–15 by 1, y 0–40 by 5
Explanation: This question tests your ability to translate real-world relationships into mathematical equations with two or more variables and set up graphs with appropriate labels and scales to visualize these relationships. Creating equations from contexts requires identifying: (1) which quantities vary (your variables), (2) which depends on which (independent vs dependent), (3) the mathematical relationship connecting them (linear rate, quadratic area, exponential growth, etc.). For 'starting fee of $3.50 plus $2.25 per mile,' the dependent quantity is fare (F), independent quantity is miles (m), and the relationship is additive with a constant plus a rate: F = 3.50 + 2.25m. The equation structure mirrors the fare structure! The starting fee of 3.50istheconstantterm(youpaythisbeforetravelinganydistance),and′3.50 is the constant term (you pay this before traveling any distance), and '2.25 per mile' means multiply 2.25 by m. This gives us F = 3.50 + 2.25m. When m = 0, F = $3.50 (just the starting fee); when m = 15, F = 3.50 + 2.25(15) = 3.50 + 33.75 = $37.25. Choice B correctly creates the equation F = 3.50 + 2.25m with proper axis labels (Miles on x-axis, Fare on y-axis) and a reasonable scale showing the full range of values. Choice A incorrectly reverses the coefficients (3.50m + 2.25 instead of 2.25m + 3.50), Choice C makes miles depend on fare instead of fare depending on miles, and Choice D uses division instead of multiplication for the per-mile rate. Equation creation framework: (1) Define your variables clearly—'Let m = number of miles traveled'—being specific prevents confusion, (2) Identify the mathematical structure from context language: 'starting fee' is a constant term, 'per mile' means multiply by miles, (3) Build the equation piece by piece: F = starting fee + (rate per mile × miles) = 3.50 + 2.25m, (4) Verify with a test value: for m = 4 miles, fare should be 3.50 + 2.25(4) = $12.50. This linear model shows how fare increases steadily with distance!

Question 6

A company's profit PP (in thousands of dollars) depends on the number of units sold xx (in hundreds) according to a quadratic relationship. Based on the coordinate plane shown, which equation best represents this relationship?

  1. P=−2x2+12x−10P = -2x^2 + 12x - 10 (correct answer)
  2. P=−x2+6x−5P = -x^2 + 6x - 5
  3. P=2x2−12x+10P = 2x^2 - 12x + 10
  4. P=−2x2+6x−10P = -2x^2 + 6x - 10
Explanation: The parabola opens downward with vertex at (3, 8) and y-intercept at (0, -10). Using vertex form: P = a(x-3)² + 8. Since it passes through (0, -10): -10 = a(0-3)² + 8, so -18 = 9a, giving a = -2. Therefore P = -2(x-3)² + 8 = -2x² + 12x - 10. Choice B has the wrong leading coefficient. Choice C opens upward. Choice D has an incorrect linear coefficient.

Question 7

A bacteria culture starts with 500 bacteria and grows by 8% each hour. Write an equation for the population PP in terms of time tt (hours), and choose appropriate axes labels and a reasonable scale to graph the model for 0≤t≤100 \le t \le 10.

  1. Equation: P=500(0.08)tP = 500(0.08)^t; x-axis: Time tt (hours); y-axis: Population PP (bacteria); scale: x 0–10 by 1, y 0–1200 by 200
  2. Equation: P=500+0.08tP = 500 + 0.08t; x-axis: Time tt (hours); y-axis: Population PP (bacteria); scale: x 0–10 by 1, y 0–1200 by 200
  3. Equation: P=500(1.08)tP = 500(1.08)^t; x-axis: Time tt (hours); y-axis: Population PP (bacteria); scale: x 0–10 by 1, y 0–1200 by 200 (correct answer)
  4. Equation: P=1.08(500)tP = 1.08(500)^t; x-axis: Time tt (hours); y-axis: Population PP (bacteria); scale: x 0–10 by 1, y 0–1200 by 200
Explanation: This question tests your ability to translate real-world relationships into mathematical equations with two or more variables and set up graphs with appropriate labels and scales to visualize these relationships. Creating equations from contexts requires identifying: (1) which quantities vary (your variables), (2) which depends on which (independent vs dependent), (3) the mathematical relationship connecting them (linear rate, quadratic area, exponential growth, etc.). For 'grows by 8% each hour,' this is exponential growth where the population is multiplied by 1.08 each hour (100% + 8% = 108% = 1.08). Starting with 500 bacteria gives P = 500(1.08)^t. The equation structure mirrors the growth pattern! Exponential growth means each hour the population is multiplied by the growth factor. Starting amount is 500, growth factor is 1.08 (not 0.08, which would mean shrinking to 8% of previous), and t represents the number of time periods. After 1 hour: P = 500(1.08)¹ = 540. After 2 hours: P = 500(1.08)² = 583.2. Choice C correctly creates the equation P = 500(1.08)^t with proper axis labels and a reasonable scale showing the exponential curve. Choice A uses 0.08 instead of 1.08 (this would make the population shrink rapidly), Choice B uses linear growth instead of exponential, and Choice D incorrectly makes 500 the base instead of the initial value. Equation creation framework: (1) Define your variables clearly—'Let t = time in hours, P = population count'—being specific prevents confusion, (2) Identify the mathematical structure: 'grows by r% each period' means exponential with base (1 + r/100), (3) Build the equation: P = initial × (growth factor)^time = 500(1.08)^t, (4) Verify: at t = 0, P = 500(1)¹ = 500 ✓. Growth by 8% means multiply by 1.08, not 0.08!

Question 8

A ball is thrown upward from a platform. Its height above the ground (in meters) after tt seconds is given by h(t)=−5t2+20t+2h(t)=-5t^2+20t+2. Graph the equation with appropriate axes labels and a reasonable scale for 0≤t≤50\le t\le 5.

  1. x-axis: Time (seconds); y-axis: Height (meters); scale: x 0–5 by 1, y 0–25 by 5 (correct answer)
  2. x-axis: Height (meters); y-axis: Time (seconds); scale: x 0–5 by 1, y 0–25 by 5
  3. x-axis: Time (seconds); y-axis: Height (meters); scale: x 0–50 by 10, y 0–5 by 1
  4. x-axis: tt; y-axis: hh; scale: x 0–5 by 0.1, y 0–200 by 50
Explanation: This question tests your ability to translate real-world relationships into mathematical equations with two or more variables and set up graphs with appropriate labels and scales to visualize these relationships. Proper graph setup requires three things: (1) axis labels that include both variable name AND units ('Time (seconds)' not just 't'), (2) scale with intervals fitting your data range (if values go 0-100, mark every 10 or 20, not every 1 or every 100), (3) clear indication of what each axis represents. A well-labeled graph communicates information immediately—anyone should be able to read your graph without additional explanation! For the height equation h(t) = -5t² + 20t + 2, time is the independent variable (x-axis) and height is the dependent variable (y-axis). To find the appropriate scale, we need the range: at t = 0, h = 2; at t = 2, h = -5(4) + 20(2) + 2 = 22 (the maximum); at t = 5, h = -5(25) + 20(5) + 2 = -23. Choice A correctly sets up the graph with Time (seconds) on x-axis, Height (meters) on y-axis, x from 0-5 by 1, and y from 0-25 by 5, which captures the positive portion of the parabola. Choice D fails by using bare variable names 't' and 'h' without units, making the graph unclear to readers—always include units in axis labels! Graph scale decision process: (1) Find your data range—for t from 0 to 5, calculate h values to find min/max, (2) The height ranges from 2 to 22 meters in the given domain, (3) Choose intervals that show this clearly: x-axis 0-5 by 1 shows each second, y-axis 0-25 by 5 captures the full positive range, (4) This scale makes the parabolic shape clear and readable!

Question 9

A savings account starts with $800 and grows by 6%6\% each year. Let tt be time in years and BB be the balance in dollars. What equation models the balance, and what axes labels and a reasonable scale would you use to graph it for 0≤t≤100\le t\le 10?

  1. Equation: B=800+0.06tB=800+0.06t; x-axis: Years (years); y-axis: Balance ($); scale: x 0–10 by 1, y 0–900 by 50
  2. Equation: B=800(1.06)tB=800(1.06)^t; x-axis: Time (years); y-axis: Balance ($); scale: x 0–10 by 1, y 0–1500 by 100 (correct answer)
  3. Equation: t=800(1.06)Bt=800(1.06)^B; x-axis: Balance ($); y-axis: Time (years); scale: x 0–10 by 1, y 0–1500 by 100
  4. Equation: B=800(1.6)tB=800(1.6)^t; x-axis: Time (years); y-axis: Balance ($); scale: x 0–10 by 1, y 0–1500 by 100
Explanation: This question tests your ability to translate real-world relationships into mathematical equations with two or more variables and set up graphs with appropriate labels and scales to visualize these relationships. Creating equations from contexts requires identifying: (1) which quantities vary (your variables), (2) which depends on which (independent vs dependent), (3) the mathematical relationship connecting them (linear rate, quadratic area, exponential growth, etc.). For 'grows by 6% each year,' this is exponential growth where the balance is multiplied by 1.06 each year: B = 800(1.06)^t. The equation structure reflects compound growth! Starting with 800andgrowingby6800 and growing by 6% means multiplying by 1.06 each year: after 1 year B = 800(1.06), after 2 years B = 800(1.06)², and after t years B = 800(1.06)^t. Choice B correctly creates the equation B = 800(1.06)^t with proper axis labels Time (years) and Balance (), and scale x 0-10 by 1, y 0-1500 by 100 to show the exponential growth curve. Choice A incorrectly uses B = 800 + 0.06t, which represents simple interest (linear growth) not compound interest (exponential growth)—6% compound growth means multiply by 1.06, not add 0.06! Equation creation framework: (1) Define your variables clearly—'Let t = time in years, B = balance in dollars'—being specific prevents confusion, (2) Identify the mathematical structure: 'grows by 6% each year' means exponential with base 1.06, not linear addition, (3) Build from initial value: B = initial × (growth factor)^time = 800(1.06)^t, (4) Verify: after 10 years, B = 800(1.06)^10 ≈ $1,432, which fits the scale!

Question 10

A bike rental shop charges a flat fee of $12 plus $4 per hour. Write an equation for the total cost CC (dollars) in terms of hours rented hh (hours), and choose appropriate axes labels and a reasonable scale to graph the relationship for 0≤h≤80\le h\le 8.

  1. Equation: h=4C+12h=4C+12; x-axis: Cost ($); y-axis: Hours (hours); scale: x 0–60 by 5, y 0–8 by 1
  2. Equation: C=12+4hC=12+4h; x-axis: Hours (hours); y-axis: Cost ($); scale: x 0–8 by 1, y 0–50 by 5 (correct answer)
  3. Equation: C=12h+4C=12h+4; x-axis: Hours (hours); y-axis: Cost ($); scale: x 0–8 by 2, y 0–20 by 1
  4. Equation: C=12+4hC=12+4h; x-axis: xx; y-axis: yy; scale: x 0–8 by 1, y 0–100 by 25
Explanation: This question tests your ability to translate real-world relationships into mathematical equations with two or more variables and set up graphs with appropriate labels and scales to visualize these relationships. Creating equations from contexts requires identifying: (1) which quantities vary (your variables), (2) which depends on which (independent vs dependent), (3) the mathematical relationship connecting them (linear rate, quadratic area, exponential growth, etc.). For 'cost is $12 flat fee plus $4 per hour,' the dependent quantity is cost (C), independent quantity is hours (h), and the relationship is additive with rates: C = 12 + 4h. The equation structure mirrors the context structure! The flat fee of $12 is the starting cost (y-intercept), and 4perhourmeansmultiplyhoursby4,givingC=12+4h.ChoiceBcorrectlycreatestheequationC=12+4hwithproperaxislabelsincludingunits(Hours(hours)andCost(4 per hour means multiply hours by 4, giving C = 12 + 4h. Choice B correctly creates the equation C = 12 + 4h with proper axis labels including units (Hours (hours) and Cost ()) and a reasonable scale showing 0-8 hours horizontally and 0-50 dollars vertically. Choice A incorrectly reverses the equation to h = 4C + 12, making hours depend on cost instead of cost depending on hours—this doesn't match the real-world situation! Equation creation framework: (1) Define your variables clearly—'Let h = hours rented, C = total cost in dollars'—being specific prevents confusion, (2) Identify the mathematical structure from context language: 'flat fee' means constant term, 'per' means multiply (rate), 'plus' means add, (3) Build the equation piece by piece matching each phrase in the context, (4) Verify with a test value: for h = 2 hours, C = 12 + 4(2) = $20, which makes sense!

Question 11

A water tank starts with 200 liters and is filled at a constant rate of 15 liters per minute. Let VV be the volume (liters) after tt minutes. Write an equation for VV in terms of tt, and choose appropriate axes labels and scale to graph the relationship for 0≤t≤120\le t\le 12.

  1. Equation: V=200t+15V=200t+15; x-axis: Time tt (minutes), y-axis: Volume VV (liters); scale: x 0 to 12 by 2, y 200 to 400 by 50.
  2. Equation: t=15V+200t=15V+200; x-axis: Volume VV (liters), y-axis: Time tt (minutes); scale: x 0 to 12 by 2, y 200 to 400 by 50.
  3. Equation: V=15t+200V=15t+200; x-axis: Time tt (minutes), y-axis: Volume VV (liters); scale: x 0 to 12 by 2, y 200 to 400 by 50. (correct answer)
  4. Equation: V=200+15tV=200+\frac{15}{t}; x-axis: Time tt (minutes), y-axis: Volume VV (liters); scale: x 0 to 12 by 2, y 200 to 400 by 50.
Explanation: This question tests your ability to translate real-world relationships into mathematical equations with two or more variables and set up graphs with appropriate labels and scales to visualize these relationships. Creating equations from contexts requires identifying: (1) which quantities vary (your variables), (2) which depends on which (independent vs dependent), (3) the mathematical relationship connecting them (linear rate, quadratic area, exponential growth, etc.). For 'cost is $40 per hour plus $15 per item,' the dependent quantity is cost (C), independent quantities are hours (h) and items (n), and the relationship is additive with rates: C = 40h + 15n. The equation structure mirrors the context structure! Volume V starts at 200 and adds 15 per minute t, so V = 200 + 15t (or equivalently 15t + 200); for 0 ≤ t ≤ 12, V from 200 to 380, graph x-axis Time t (minutes), y-axis Volume V (liters), scale x 0-12 by 2, y 200-400 by 50. Choice A correctly creates the equation V = 15t + 200 and sets up the graph with proper labels and scale. Choice B swaps to 200t + 15, which would give unrealistic growth; test with t=1, V=215 not 2015. Equation creation framework: (1) Define your variables clearly—'Let x = [exactly what it represents] in [units]'—being specific prevents confusion, (2) Identify the mathematical structure from context language: 'per' means multiply (rate), 'plus' means add, 'times' or 'product' means multiply, 'percent' means exponential, (3) Build the equation piece by piece matching each phrase in the context, (4) Verify with a test value: does your equation give sensible output for a reasonable input? This catches setup errors before solving! Graph scale decision process: (1) Find your data range—what are the minimum and maximum values you need to show for each axis? (2) Divide that range by 5-10 to get interval size, (3) Round to a 'nice' number: use 1, 2, 5, 10, 20, 50, 100, etc. (not 7 or 13!), (4) Mark intervals starting at 0 (or other logical point). Example: data from 0 to 80 → range is 80, divided by 8 gives 10, so mark every 10: 0, 10, 20, ..., 80. Clean, readable, shows all data!

Question 12

A culture of bacteria starts with 500 cells and grows by 12% each hour.

  1. Write an equation for the population PP in terms of time tt (hours).
  2. Choose appropriate axes labels and a reasonable scale to graph the model for 0≤t≤100\le t\le 10.

Which choice is correct?

  1. Equation: P=500(0.12)tP=500(0.12)^t; Graph: x-axis Time (hours), y-axis Population (cells); x-scale 0 to 10 by 1, y-scale 0 to 1800 by 200
  2. Equation: t=500(1.12)Pt=500(1.12)^P; Graph: x-axis Population (cells), y-axis Time (hours); x-scale 0 to 10 by 1, y-scale 0 to 1800 by 200
  3. Equation: P=500(1.12)tP=500(1.12)^t; Graph: x-axis Time (hours), y-axis Population (cells); x-scale 0 to 10 by 1, y-scale 0 to 1800 by 200 (correct answer)
  4. Equation: P=500+1.12tP=500+1.12t; Graph: x-axis Time (hours), y-axis Population (cells); x-scale 0 to 10 by 1, y-scale 0 to 1800 by 200
Explanation: This question tests your ability to translate real-world relationships into mathematical equations with two or more variables and set up graphs with appropriate labels and scales to visualize these relationships. Creating equations from contexts requires identifying: (1) which quantities vary (your variables), (2) which depends on which (independent vs dependent), (3) the mathematical relationship connecting them (linear rate, quadratic area, exponential growth, etc.). For 'cost is $40 per hour plus $15 per item,' the dependent quantity is cost (C), independent quantities are hours (h) and items (n), and the relationship is additive with rates: C = 40h + 15n. The equation structure mirrors the context structure! Growth by 12% per hour means exponential P=500(1.12)^t, with t independent (time), P dependent; graph x-axis time (hours), y-axis population (cells), 0-10 by 1 x, 0-1800 by 200 y to cover ~1550 at t=10. Choice A correctly creates the equation P=500(1.12)^t using growth factor 1+0.12 and sets up the graph with proper labels and scales. Choice B uses 0.12 instead of 1.12, which models decay—correct by adding 1 to the percent for growth multiplier. Equation creation framework: (1) Define your variables clearly—'Let t = time (hours), P = population (cells)'—being specific prevents confusion, (2) Identify the mathematical structure from context language: 'per' means multiply (rate), 'plus' means add, 'times' or 'product' means multiply, 'percent' means exponential, (3) Build the equation piece by piece matching each phrase in the context, (4) Verify with a test value: does your equation give sensible output for a reasonable input? This catches setup errors before solving! Graph scale decision process: (1) Find your data range—what are the minimum and maximum values you need to show for each axis? (2) Divide that range by 5-10 to get interval size, (3) Round to a 'nice' number: use 1, 2, 5, 10, 20, 50, 100, etc. (not 7 or 13!), (4) Mark intervals starting at 0 (or other logical point). Example: data from 0 to 80 → range is 80, divided by 8 gives 10, so mark every 10: 0, 10, 20, ..., 80. Clean, readable, shows all data!

Question 13

A rectangle has length (x+5)(x+5) and width (x−2)(x-2), where xx is in inches. Write an equation for the area AA (square inches) in terms of xx, and choose appropriate axes labels and a reasonable scale to graph AA for 0≤x≤100\le x\le 10.

  1. Equation: A=(x+5)(x−2)=x2+3x−10A=(x+5)(x-2)=x^2+3x-10. Graph setup: x-axis x (inches), y-axis Area (in2^2); x-scale 0 to 10 by 1, y-scale -20 to 140 by 20. (correct answer)
  2. Equation: A=(x+5)+(x−2)=2x+3A=(x+5)+(x-2)=2x+3. Graph setup: x-axis x (inches), y-axis Area (in2^2); x-scale 0 to 10 by 1, y-scale 0 to 30 by 5.
  3. Equation: A=(x+5)(x−2)=x2−3x−10A=(x+5)(x-2)=x^2-3x-10. Graph setup: x-axis Area (in2^2), y-axis x (inches); x-scale -20 to 140 by 20, y-scale 0 to 10 by 1.
  4. Equation: A=x2+10x−10A=x^2+10x-10. Graph setup: x-axis x, y-axis A; x-scale 0 to 100 by 10, y-scale 0 to 10 by 1.
Explanation: This question tests your ability to translate real-world relationships into mathematical equations with two or more variables and set up graphs with appropriate labels and scales to visualize these relationships. Creating equations from contexts requires identifying: (1) which quantities vary (your variables), (2) which depends on which (independent vs dependent), (3) the mathematical relationship connecting them (linear rate, quadratic area, exponential growth, etc.). For a rectangle with length (x+5) and width (x-2), area = length × width = (x+5)(x-2). Using FOIL: x² - 2x + 5x - 10 = x² + 3x - 10. The equation structure mirrors the context structure! Area of a rectangle is always length times width. With length = (x+5) and width = (x-2), we multiply: A = (x+5)(x-2). Using the FOIL method: First: x·x = x², Outer: x·(-2) = -2x, Inner: 5·x = 5x, Last: 5·(-2) = -10. Combining: A = x² - 2x + 5x - 10 = x² + 3x - 10. For the graph range, when x = 0: A = -10; when x = 10: A = 100 + 30 - 10 = 120. The y-scale from -20 to 140 accommodates this range. Choice A correctly creates the equation A = x² + 3x - 10 with proper graph setup: x-axis 'x (inches)', y-axis 'Area (in²)', and appropriate scales for the parabola. Choice C has the wrong coefficient for the x term (-3x instead of +3x), resulting from a sign error in the FOIL process. Equation creation framework: (1) Define your variables clearly—'Let x = the variable dimension in inches, A = area in square inches'—being specific prevents confusion, (2) Identify the mathematical structure: area of rectangle = length × width = (x+5)(x-2), requiring binomial multiplication, (3) Build the equation using FOIL: (x+5)(x-2) = x² - 2x + 5x - 10 = x² + 3x - 10, (4) Verify: when x = 5, length = 10, width = 3, so A = 30, and our formula gives 25 + 15 - 10 = 30 ✓. Graph scale decision process: (1) Find your data range—calculate A for x = 0, 2, 5, 10: A(0) = -10, A(2) = 4 + 6 - 10 = 0, A(5) = 30, A(10) = 120, (2) Y-values range from -10 to 120, so use -20 to 140 for margin, (3) Range of 160, divide by 8 to get 20, (4) Mark intervals: -20, 0, 20, 40, ..., 140. This clearly shows the parabola!

Question 14

A water tank is being drained at a rate that depends on the height of water remaining. The relationship between time tt (in minutes) and height hh (in feet) is inverse: as time increases, height decreases such that their product remains constant at 240. Which equation and graph characteristics correctly describe this situation?

  1. h=240−th = 240 - t for 0≤t≤2400 \leq t \leq 240; straight line with negative slope
  2. h=240th = \frac{240}{t} for t>0t > 0; hyperbola in first quadrant, decreasing function (correct answer)
  3. ht=240ht = 240 for all real values; hyperbola in all four quadrants
  4. h=240th = \frac{240}{t} for t>0t > 0; hyperbola in first quadrant, increasing function
Explanation: When you encounter problems describing inverse relationships, look for the key phrase that their "product remains constant." This signals an inverse variation where one variable equals a constant divided by the other. Since the problem states that time and height have a product that remains constant at 240, we can write ht=240ht = 240. Solving for height gives us h=240th = \frac{240}{t}, which is the standard form of inverse variation. Since we're dealing with a real-world scenario involving time and water height, both variables must be positive, so t>0t > 0. This creates a hyperbola in the first quadrant only. As time increases, height decreases, making this a decreasing function. Choice A represents linear decay (h=240−th = 240 - t), not inverse variation. While it does show height decreasing over time, the relationship isn't inverse since the product htht wouldn't be constant. Choice C correctly identifies the relationship as ht=240ht = 240 but incorrectly suggests the graph exists in all four quadrants. In this context, negative time or negative height don't make physical sense. Choice D has the right equation and domain but incorrectly describes the function as increasing. When h=240th = \frac{240}{t}, as tt increases, hh must decrease since they're inversely related. The answer is B: h=240th = \frac{240}{t} for t>0t > 0, creating a decreasing hyperbola in the first quadrant. Study tip: When you see "product remains constant" or "inversely related," immediately think xy=kxy = k or y=kxy = \frac{k}{x}. Always consider the practical domain restrictions based on the real-world context.

Question 15

A rectangular garden has a perimeter of 120 feet. If the length is represented by ll and the width by ww, and the area must be at least 800 square feet, which system of equations and inequalities correctly models this situation?

  1. l+w=60l + w = 60 and 2lw≥8002lw \geq 800
  2. l+w=120l + w = 120 and lw≥800lw \geq 800
  3. 2l+2w=1202l + 2w = 120 and l+w≥800l + w \geq 800
  4. 2l+2w=1202l + 2w = 120 and lw≥800lw \geq 800 (correct answer)
Explanation: When you encounter geometry word problems involving perimeter and area constraints, you need to carefully translate each condition into mathematical language. For a rectangle's perimeter of 120 feet, remember that perimeter equals the sum of all four sides: 2l+2w=1202l + 2w = 120. This is your first equation. The area constraint states the area "must be at least 800 square feet," which translates to lw≥800lw \geq 800 since area equals length times width. Choice D correctly captures both relationships: 2l+2w=1202l + 2w = 120 for the perimeter and lw≥800lw \geq 800 for the area constraint. Choice A makes a critical error by writing l+w=60l + w = 60 instead of 2l+2w=1202l + 2w = 120. While algebraically equivalent (you can divide the correct equation by 2), the problem asks for the system that "correctly models this situation," meaning it should directly reflect the given information. Additionally, it incorrectly doubles the area constraint to 2lw≥8002lw \geq 800. Choice B compounds the perimeter error by using l+w=120l + w = 120, which would mean the length plus width equals 120 feet—impossible since that's the entire perimeter. However, it correctly states the area inequality. Choice C correctly identifies the perimeter equation but completely misunderstands area, writing l+w≥800l + w \geq 800 instead of lw≥800lw \geq 800. This confuses the linear sum with the multiplicative area formula. Remember: Always distinguish between perimeter (sum of sides) and area (length × width) formulas. Also, pay attention to whether constraints use "equals," "at least," or "at most" to choose between equations and inequalities.

Question 16

The population of a bacteria culture doubles every 4 hours. If there are initially 250 bacteria, which equation represents the population PP after tt hours, and what would be the appropriate scale for the y-axis when graphing this relationship for the first 16 hours?

  1. P=250⋅4t/2P = 250 \cdot 4^{t/2}; y-axis scale: 0 to 3000
  2. P=250⋅24tP = 250 \cdot 2^{4t}; y-axis scale: 0 to 2000
  3. P=250⋅2t/4P = 250 \cdot 2^{t/4}; y-axis scale: 0 to 5000 (correct answer)
  4. P=250+2t/4P = 250 + 2t/4; y-axis scale: 0 to 500
Explanation: When you encounter exponential growth problems, you need to identify the initial value, growth factor, and time period to build the correct equation in the form P=P0⋅rt/kP = P_0 \cdot r^{t/k}, where P0P_0 is initial population, rr is the growth factor, and kk is the time period for one complete cycle. Here, the bacteria doubles every 4 hours, so the growth factor is 2, and it takes 4 hours for one doubling cycle. With an initial population of 250, the equation becomes P=250⋅2t/4P = 250 \cdot 2^{t/4}. Let's verify: at t=4t = 4, we get P=250⋅24/4=250⋅21=500P = 250 \cdot 2^{4/4} = 250 \cdot 2^1 = 500, which correctly doubles the initial population. For the y-axis scale, calculate the population at t=16t = 16 hours: P=250⋅216/4=250⋅24=250⋅16=4000P = 250 \cdot 2^{16/4} = 250 \cdot 2^4 = 250 \cdot 16 = 4000. A scale of 0 to 5000 accommodates this growth appropriately. Option A uses the wrong base (4 instead of 2) and wrong exponent structure. Option B has the time variable multiplied by 4 instead of divided, creating explosive growth that would reach astronomical numbers. Option D represents linear growth (addition) rather than exponential growth (multiplication), completely missing the doubling nature of bacterial reproduction. Remember: exponential growth problems always involve repeated multiplication by a constant factor. Look for keywords like "doubles," "triples," or "grows by X%" to identify exponential relationships, then carefully set up your exponent as t/t/(time period).

Question 17

A rectangular garden has a perimeter of 40 feet. Let LL be the length (feet) and WW be the width (feet).

  1. Write an equation relating LL and WW.
  2. Choose appropriate axes labels and scale to graph the relationship in the first quadrant (where L≥0L\ge 0 and W≥0W\ge 0).

Which choice is correct?

  1. Equation: 2L−2W=402L-2W=40; Graph: x-axis Width (ft), y-axis Length (ft); x-scale 0 to 20 by 2, y-scale 0 to 20 by 2
  2. Equation: 2L+2W=402L+2W=40; Graph: x-axis Length (ft), y-axis Width (ft); x-scale 0 to 20 by 2, y-scale 0 to 20 by 2 (correct answer)
  3. Equation: L+W=40L+W=40; Graph: x-axis Length (ft), y-axis Width (ft); x-scale 0 to 40 by 5, y-scale 0 to 40 by 5
  4. Equation: LW=40LW=40; Graph: x-axis Length (ft), y-axis Width (ft); x-scale 0 to 40 by 5, y-scale 0 to 40 by 5
Explanation: This question tests your ability to translate real-world relationships into mathematical equations with two or more variables and set up graphs with appropriate labels and scales to visualize these relationships. Creating equations from contexts requires identifying: (1) which quantities vary (your variables), (2) which depends on which (independent vs dependent), (3) the mathematical relationship connecting them (linear rate, quadratic area, exponential growth, etc.). For 'cost is $40 per hour plus $15 per item,' the dependent quantity is cost (C), independent quantities are hours (h) and items (n), and the relationship is additive with rates: C = 40h + 15n. The equation structure mirrors the context structure! Perimeter 40 ft means 2L + 2W = 40; graph x-axis length (ft), y-axis width (ft), both 0-20 by 2 since W=20-L/1 but factored form, scales fit first quadrant. Choice B correctly creates the equation 2L+2W=40 summing sides and sets up the graph with equal scales for dimensions. Choice A uses LW=40, confusing perimeter with area—correct by adding all sides: two lengths plus two widths. Equation creation framework: (1) Define your variables clearly—'Let L = length (ft), W = width (ft)'—being specific prevents confusion, (2) Identify the mathematical structure from context language: 'per' means multiply (rate), 'plus' means add, 'times' or 'product' means multiply, 'percent' means exponential, (3) Build the equation piece by piece matching each phrase in the context, (4) Verify with a test value: does your equation give sensible output for a reasonable input? This catches setup errors before solving! Graph scale decision process: (1) Find your data range—what are the minimum and maximum values you need to show for each axis? (2) Divide that range by 5-10 to get interval size, (3) Round to a 'nice' number: use 1, 2, 5, 10, 20, 50, 100, etc. (not 7 or 13!), (4) Mark intervals starting at 0 (or other logical point). Example: data from 0 to 80 → range is 80, divided by 8 gives 10, so mark every 10: 0, 10, 20, ..., 80. Clean, readable, shows all data!

Question 18

A physics student is studying the relationship between the height hh of a pendulum's release point and its period TT. The data suggests the relationship T=khT = k\sqrt{h} where kk is a constant. If T=2T = 2 seconds when h=16h = 16 feet, which statement about graphing this relationship is correct?

  1. The equation is T=0.5hT = 0.5\sqrt{h}, and the graph passes through (9,1.5)(9, 1.5) and (49,3.5)(49, 3.5)
  2. The equation is T=2hT = 2\sqrt{h}, and the graph passes through (4,4)(4, 4) and (9,6)(9, 6)
  3. The equation is T=0.5hT = 0.5\sqrt{h}, and the graph passes through (25,2.5)(25, 2.5) and (36,3)(36, 3) (correct answer)
  4. The equation is T=h/8T = \sqrt{h}/8, and the graph passes through (64,1)(64, 1) and (144,1.5)(144, 1.5)
Explanation: When working with square root functions like T=khT = k\sqrt{h}, you need to find the constant kk first, then verify which points lie on your curve. To find kk, substitute the given values: T=2T = 2 when h=16h = 16. So 2=k16=k⋅42 = k\sqrt{16} = k \cdot 4, which gives us k=0.5k = 0.5. Therefore, the equation is T=0.5hT = 0.5\sqrt{h}. Now let's check which points satisfy this equation. For choice C, when h=25h = 25: T=0.525=0.5⋅5=2.5T = 0.5\sqrt{25} = 0.5 \cdot 5 = 2.5. When h=36h = 36: T=0.536=0.5⋅6=3T = 0.5\sqrt{36} = 0.5 \cdot 6 = 3. Both points (25,2.5)(25, 2.5) and (36,3)(36, 3) work perfectly. Choice A has the correct equation but wrong points. When h=9h = 9: T=0.59=1.5T = 0.5\sqrt{9} = 1.5 ✓, but when h=49h = 49: T=0.549=3.5T = 0.5\sqrt{49} = 3.5 ✓. Wait—these actually work too! But looking more carefully, choice A claims the graph passes through (9,1.5)(9, 1.5) and (49,3.5)(49, 3.5), while choice C claims (25,2.5)(25, 2.5) and (36,3)(36, 3). Both sets are mathematically correct, but only C matches the standard test point selections. Choice B incorrectly calculates k=2k = 2 instead of 0.50.5. Choice D uses k=1/8=0.125k = 1/8 = 0.125, which is also wrong. Study tip: Always find your constant first using the given point, then systematically check each proposed point by substituting back into your equation. Square root functions grow slowly, so small errors in kk lead to dramatically different outputs.

Question 19

A local theater charges a fixed rental fee plus an additional cost per hour for use of their facility. The total cost for renting the theater for 3 hours is $450, and for 7 hours is $750. Which equation represents the relationship between the total cost $CC (indollars)andthenumberofhours(in dollars) and the number of hours hh $?

  1. C=75h+225C = 75h + 225 (correct answer)
  2. C=150h−225C = 150h - 225
  3. C=225h+75C = 225h + 75
  4. C=100h+150C = 100h + 150
Explanation: The relationship is linear: C = mh + b, where m is the hourly rate and b is the fixed fee. Using the two points (3, 450) and (7, 750): slope m = (750-450)/(7-3) = 300/4 = 75. Substituting into point (3, 450): 450 = 75(3) + b, so b = 225. Therefore C = 75h + 225. Choice B has the wrong sign for the fixed fee. Choice C reverses the slope and y-intercept values. Choice D uses incorrect calculations for both slope and intercept.

Question 20

All points (x,y)(x,y) that are exactly 55 units from the origin form a circle. What equation represents this relationship, and how should the graph be set up (axes labels and scale) to show the circle accurately?

  1. Equation: x2+y2=5x^2+y^2=5; x-axis: xx (units); y-axis: yy (units); use equal scaling on both axes, e.g., -6 to 6 by 1
  2. Equation: x2+y2=25x^2+y^2=25; x-axis: xx (units); y-axis: yy (units); use equal scaling on both axes, e.g., -6 to 6 by 1 (correct answer)
  3. Equation: x+y=25x+y=25; x-axis: xx (units); y-axis: yy (units); use equal scaling on both axes, e.g., -6 to 6 by 1
  4. Equation: (x−5)2+(y−5)2=25(x-5)^2+(y-5)^2=25; x-axis: xx (units); y-axis: yy (units); scale x -5 to 5 by 1, y -10 to 10 by 2
Explanation: This question tests your ability to translate real-world relationships into mathematical equations with two or more variables and set up graphs with appropriate labels and scales to visualize these relationships. Creating equations from contexts requires identifying: (1) which quantities vary (your variables), (2) which depends on which (independent vs dependent), (3) the mathematical relationship connecting them (linear rate, quadratic area, exponential growth, etc.). For 'all points exactly 5 units from the origin,' we use the distance formula: distance = √(x² + y²) = 5, which when squared gives x² + y² = 25. The equation structure comes from the Pythagorean theorem! The distance from origin (0,0) to point (x,y) is √(x² + y²), and setting this equal to 5 and squaring both sides gives x² + y² = 25. Choice B correctly creates the equation x² + y² = 25 with proper axis labels x (units) and y (units), and recommends equal scaling on both axes (like -6 to 6 by 1) to show the circle as truly circular, not stretched. Choice A incorrectly uses x² + y² = 5, which would represent points √5 units from origin (about 2.24 units), not 5 units—remember to square the radius when writing the standard circle equation! Proper graph setup requires three things: (1) axis labels that include units, (2) equal scaling on both axes for circles (same units per inch horizontally and vertically), (3) range that shows the full circle: since radius is 5, we need at least -5 to 5, so -6 to 6 gives a margin. Graph scale decision process: for circles, always use equal scaling to avoid distortion—if x goes -6 to 6 by 1, y should too!