Constructing Linear and Exponential Functions - Algebra 2

A bacteria culture starts with 200 bacteria and grows by 10% each hour. Let $t$ be the number of hours. Which exponential function models the population $P(t)$ ?
Answer: $P(t)=200(1.10)^t$ This question tests your ability to construct linear or exponential functions from given information like points, tables, graphs, or descriptions of relationships. To construct a linear function from two points, find the slope m = (y₂ - y₁)/(x₂ - x₁), then find the y-intercept b by substituting one point into y = mx + b and solving for b. Once you have m and b, you've got your function! For exponential functions, find the initial value a (the y-value when x = 0, or work backward if needed) and the growth/decay factor b (divide consecutive y-values: b = y₂/y₁ when x increases by 1). Then write f(x) = a· $b^x$ . For growth of 10% per hour starting at 200, it's exponential with a=200 and b=1.10, so P(t)=200 $(1.10)^t$ . Choice B correctly constructs the exponential function with initial value 200 and growth factor 1.10 from the description. A distractor like choice A uses decay $0.10^t$ instead of growth—remember growth is 1 + rate, so 1.10 for 10%. Exponential construction strategy: (1) Find initial value: if you have x = 0 in data, that y is your a; otherwise calculate backward using the pattern, (2) Find base: divide consecutive y-values (with x differing by 1): b = $y_{x+1}$ / $y_x$ —should be constant for exponential, (3) Write f(x) = a· $b^x$ , (4) Verify with all data points. Example: points (0, 100) and (1, 110) and (2, 121) → a = 100, b = 110/100 = 1.1 (check: 121/110 = 1.1 ✓), so f(x) = 100· $(1.1)^x$ . The ratio test both identifies the type and gives you the base!
A population of rabbits grows according to an exponential model. The population triples every 4 months. If there are currently 20 rabbits, which function gives the population after $m$ months?
Answer: $P(m) = 20 \cdot 3^{m/4}$ Since the population triples every 4 months, we need $P(4) = 3P(0)$ . The general form is $P(m) = P_0 \cdot b^{m/k}$ where $k$ is the time period for the growth factor $b$ . Here, $P_0 = 20$ , $b = 3$ , and $k = 4$ , giving $P(m) = 20 \cdot 3^{m/4}$ . This ensures $P(4) = 20 \cdot 3^1 = 60$ and $P(8) = 20 \cdot 3^2 = 180$ . Choice A triples every month, too fast. Choice C is linear growth. Choice D uses incorrect parameters.
A geometric sequence is given: $3, 6, 12, 24, \dots$ Let $g(n)$ give the $n$ th term with $g(1)=3$ . Which exponential function matches this sequence?
Answer: $g(n)=3\cdot 2^{n-1}$ This question tests your ability to construct linear or exponential functions from given information like points, tables, graphs, or descriptions of relationships. From a table, first determine which type: calculate differences between consecutive y-values (if constant → linear with slope = that difference), and calculate ratios (if constant → exponential with base = that ratio). This identification step is crucial—you can't construct the right function if you don't know which type it is! Once identified, extract the parameters (slope and intercept for linear, initial value and base for exponential) and write the formula. For the geometric sequence with constant ratio 2 and g(1)=3, adjust to g(n)=3· $2^{n-1}$ to fit the exponent starting at n=1. Choice A correctly constructs the exponential function with a=3 and b=2 adjusted for n-1 from the sequence. A distractor like choice D uses b=3, but 3· $3^{1-1}$ =3, then 3· $3^1$ =9≠6—check the common ratio matches b. Exponential construction strategy: (1) Find initial value: if you have x = 0 in data, that y is your a; otherwise calculate backward using the pattern, (2) Find base: divide consecutive y-values (with x differing by 1): b = $y_{x+1}$ / $y_x$ —should be constant for exponential, (3) Write f(x) = a· $b^x$ , (4) Verify with all data points. Example: points (0, 100) and (1, 110) and (2, 121) → a = 100, b = 110/100 = 1.1 (check: 121/110 = 1.1 ✓), so f(x) = 100· $(1.1)^x$ . The ratio test both identifies the type and gives you the base!
A car rental company charges a flat fee plus a rate per mile. After driving 120 miles, the total cost is $68. After driving 200 miles, the total cost is $92. Which function represents the total cost $C(m)$ for driving $m$ miles?
Answer: $C(m) = 0.30m + 32$ Using the two points (120, 68) and (200, 92), the slope is $\frac{92-68}{200-120} = \frac{24}{80} = 0.30$ per mile. Using point-slope form with (120, 68): $C(m) = 0.30m + b$ . Substituting: $68 = 0.30(120) + b$ , so $68 = 36 + b$ , giving $b = 32$ . Therefore $C(m) = 0.30m + 32$ . Choice B uses an incorrect slope calculation. Choice C reverses the coefficient and constant. Choice D uses an incorrect flat fee calculation.
The geometric sequence is $5, 15, 45, 135, \dots$ where the term number is $x$ (so $x=1$ gives 5). Write an exponential function $g(x)$ that gives the $x$ th term.
Answer: $g(x)=5\cdot 3^{x-1}$ This question tests your ability to construct linear or exponential functions from given information like points, tables, graphs, or descriptions of relationships. For exponential functions from geometric sequences, the common ratio is the base b, and you need to account for the fact that the sequence starts at x = 1, not x = 0. The sequence 5, 15, 45, 135, ... has a common ratio of 3 (15/5=3, 45/15=3, etc.), so b = 3. Since g(1) = 5, we have 5 = a·3¹ = 3a, which gives us a = 5/3. Therefore, g(x) = (5/3)· $3^x$ = 5· $3^{x-1}$ . Choice C correctly constructs g(x) = 5· $3^{x-1}$ , which gives the first term 5 when x = 1 and has common ratio 3. Choice A incorrectly writes g(x) = 5· $3^x$ , which would give g(1) = 15, not 5—this shifts the sequence by one position. Geometric sequence to exponential function strategy: (1) Find common ratio r between consecutive terms—that's your base b, (2) For sequences starting at x = 1, use g(x) = (first term)· $b^{x-1}$ , or find a such that g(1) = first term, (3) Write the function, (4) Verify all terms. Here: g(x) = 5· $3^{x-1}$ . Verify: g(1) = 5·3⁰ = 5 ✓, g(2) = 5·3¹ = 15 ✓, g(3) = 5·3² = 45 ✓, g(4) = 5·3³ = 135 ✓!
An exponential function satisfies $f(0)=7$ and $f(1)=14$ . Write the function in the form $f(x)=a\cdot b^x$ .
Answer: $f(x)=7\cdot 2^x$ This question tests your ability to construct linear or exponential functions from given information like points, tables, graphs, or descriptions of relationships. To construct an exponential function from two points, find the initial value a (the y-value when x = 0, or work backward if needed) and the growth/decay factor b (divide consecutive y-values: b = y₂/y₁ when x increases by 1). Then write f(x) = a· $b^x$ . We're given f(0) = 7 and f(1) = 14. Since f(0) = 7, we have a = 7 (because a·b⁰ = a·1 = a). To find b, we use f(1) = 14: 14 = 7·b¹, so b = 14/7 = 2. Therefore, f(x) = 7· $2^x$ . Choice B correctly constructs f(x) = 7· $2^x$ with initial value 7 and base 2 from the given conditions. Choice A incorrectly has initial value 14 instead of 7, which would give f(0) = 14, not 7. Exponential construction strategy: (1) Find initial value: if you have x = 0 in data, that y is your a; otherwise calculate backward using the pattern, (2) Find base: divide consecutive y-values (with x differing by 1): b = $y_{x+1}$ / $y_x$ —should be constant for exponential, (3) Write f(x) = a· $b^x$ , (4) Verify with all data points. Let's verify: f(0) = 7·2⁰ = 7 ✓ and f(1) = 7·2¹ = 14 ✓!
A geometric sequence is given: $3,\ 6,\ 12,\ 24,\dots$ Let $g(n)$ be the $n$ th term with $g(1)=3$ . Which exponential function gives $g(n)$ ?
Answer: $g(n)=3\cdot 2^{n-1}$ This question tests your ability to construct linear or exponential functions from given information like points, tables, graphs, or descriptions of relationships. From a table, first determine which type: calculate differences between consecutive y-values (if constant → linear with slope = that difference), and calculate ratios (if constant → exponential with base = that ratio). This identification step is crucial—you can't construct the right function if you don't know which type it is! Once identified, extract the parameters (slope and intercept for linear, initial value and base for exponential) and write the formula. For the geometric sequence 3, 6, 12, 24,..., the common ratio is 6/3=2 (check: 12/6=2, 24/12=2). Since g(1)=3 and we multiply by 2 each time, we have g(n) = 3· $2^{n-1}$ . Let's verify: g(1) = 3· $2^{1-1}$ = 3· $2^0$ = 3·1 = 3 ✓, g(2) = 3· $2^{2-1}$ = 3· $2^1$ = 6 ✓, g(3) = 3· $2^{3-1}$ = 3· $2^2$ = 12 ✓. Choice B correctly constructs g(n) = 3· $2^{n-1}$ with first term 3 and common ratio 2 from the geometric sequence. Choice A incorrectly models this as linear, choice C uses the wrong exponent (it should be n-1, not n), and choice D uses the wrong common ratio (3 instead of 2). Exponential construction strategy: (1) Find initial value: if you have x = 0 in data, that y is your a; otherwise calculate backward using the pattern, (2) Find base: divide consecutive y-values (with x differing by 1): b = $y_{x+1}$ / $y_x$ —should be constant for exponential, (3) Write f(x) = a· $b^x$ , (4) Verify with all data points. Example: points (0, 100) and (1, 110) and (2, 121) → a = 100, b = 110/100 = 1.1 (check: 121/110 = 1.1 ✓), so f(x) = 100· $(1.1)^x$ . The ratio test both identifies the type and gives you the base!
A linear function has $y$ -intercept 6 and passes through the point $(4,14)$ . Which function is correct?
Answer: $f(x)=2x+6$ This question tests your ability to construct linear or exponential functions from given information like points, tables, graphs, or descriptions of relationships. To construct a linear function from two points, find the slope m = (y₂ - y₁)/(x₂ - x₁), then find the y-intercept b by substituting one point into y = mx + b and solving for b. Once you have m and b, you've got your function! For exponential functions, find the initial value a (the y-value when x = 0, or work backward if needed) and the growth/decay factor b (divide consecutive y-values: b = y₂/y₁ when x increases by 1). Then write f(x) = a· $b^x$ . Given y-intercept 6 (so b = 6) and point (4,14), we need to find the slope. Using y = mx + b with the known values: 14 = m(4) + 6, so 14 = 4m + 6, which gives 8 = 4m, therefore m = 2. The function is f(x) = 2x + 6. Choice A correctly constructs f(x) = 2x + 6 with slope 2 and y-intercept 6 from the given information. Choice B has slope and intercept reversed (6x + 2), choice C has the wrong slope (1/2 instead of 2), and choice D has the wrong y-intercept (-6 instead of 6). Linear construction from two points recipe: (1) Find slope: m = (y₂ - y₁)/(x₂ - x₁), (2) Find y-intercept: substitute either point and m into y = mx + b, solve for b, (3) Write function: f(x) = [m]x + [b], (4) Verify: check that both original points work in your function. Example: (1, 4) and (3, 10) → m = (10-4)/(3-1) = 3, then 4 = 3(1) + b gives b = 1, so f(x) = 3x + 1. Check: f(1) = 4 ✓, f(3) = 10 ✓!
Write the exponential function $k(x)=a\cdot b^x$ that passes through $(1,9)$ and $(3,81)$ . Which formula is correct?
Answer: $k(x)=3^{x+1}$ This question tests your ability to construct linear or exponential functions from given information like points, tables, graphs, or descriptions of relationships. To construct a linear function from two points, find the slope m = (y₂ - y₁)/(x₂ - x₁), then find the y-intercept b by substituting one point into y = mx + b and solving for b. Once you have m and b, you've got your function! For exponential functions, find the initial value a (the y-value when x = 0, or work backward if needed) and the growth/decay factor b (divide consecutive y-values: b = y₂/y₁ when x increases by 1). Then write f(x) = a· $b^x$ . For exponential k(x) = a· $b^x$ through (1,9) and (3,81), let's set up equations: 9 = a· $b^1$ and 81 = a· $b^3$ . Dividing the second by the first: 81/9 = ( $a·b^3$ )/( $a·b^1$ ) = $b^2$ , so 9 = $b^2$ , giving b = 3. Substituting back: 9 = a· $3^1$ = 3a, so a = 3. Therefore k(x) = 3· $3^x$ = $3^{x+1}$ . Choice C correctly constructs k(x) = $3^{x+1}$ which equals 3· $3^x$ from the given points. Choice A gives 9· $3^x$ which doesn't pass through (1,9), choice B reverses the values incorrectly, and choice D gives $3^{2x}$ which also doesn't fit the points. Exponential construction strategy: (1) Find initial value: if you have x = 0 in data, that y is your a; otherwise calculate backward using the pattern, (2) Find base: divide consecutive y-values (with x differing by 1): b = $y_{x+1}$ / $y_x$ —should be constant for exponential, (3) Write f(x) = a· $b^x$ , (4) Verify with all data points. Example: points (0, 100) and (1, 110) and (2, 121) → a = 100, b = 110/100 = 1.1 (check: 121/110 = 1.1 ✓), so f(x) = 100· $(1.1)^x$ . The ratio test both identifies the type and gives you the base!
An exponential function has the form $g(x)=a\cdot b^x$ and passes through $(0,5)$ and $(2,45)$ . Which function is correct?
Answer: $g(x)=5\cdot 3^x$ This question tests your ability to construct linear or exponential functions from given information like points, tables, graphs, or descriptions of relationships. To construct a linear function from two points, find the slope m = (y₂ - y₁)/(x₂ - x₁), then find the y-intercept b by substituting one point into y = mx + b and solving for b. Once you have m and b, you've got your function! For exponential functions, find the initial value a (the y-value when x = 0, or work backward if needed) and the growth/decay factor b (divide consecutive y-values: b = y₂/y₁ when x increases by 1). Then write f(x) = a· $b^x$ . For points (0,5) and (2,45), a = 5 (from x=0), then 45 = 5 * $b^2$ gives $b^2$ = 9 so b=3, yielding g(x) = 5· $3^x$ . Choice B correctly constructs the exponential function with a=5 and b=3 from the points. A common distractor like choice A uses b=9, but that would give g(2)=5*81=405 ≠45—remember to solve for b using the second point after finding a. Exponential construction strategy: (1) Find initial value: if you have x = 0 in data, that y is your a; otherwise calculate backward using the pattern, (2) Find base: divide consecutive y-values (with x differing by 1): b = $y_{x+1}$ / $y_x$ —should be constant for exponential, (3) Write f(x) = a· $b^x$ , (4) Verify with all data points. Example: points (0, 100) and (1, 110) and (2, 121) → a = 100, b = 110/100 = 1.1 (check: 121/110 = 1.1 ✓), so f(x) = 100· $(1.1)^x$ . The ratio test both identifies the type and gives you the base!
A radioactive substance has a half-life of 5 years. If there are initially 800 grams of the substance, which function models the amount remaining after $t$ years?
Answer: $A(t) = 800(0.5)^{t/5}$ For exponential decay with half-life, $A(t) = A_0(0.5)^{t/h}$ where $h$ is the half-life. With $A_0 = 800$ and $h = 5$ , we get $A(t) = 800(0.5)^{t/5}$ . This gives $A(5) = 800(0.5)^1 = 400$ , which is correct for one half-life. Choice A would halve the substance every year, not every 5 years. Choice C uses the wrong base. Choice D represents linear decay, not exponential.
A water tank contains 200 liters at time $t=0$ hours and the amount increases by 10% each hour. Write an exponential function $A(t)$ that models the amount of water (in liters) after $t$ hours.
Answer: $A(t)=200(1.1)^t$ This question tests your ability to construct linear or exponential functions from given information like points, tables, graphs, or descriptions of relationships. For exponential functions, find the initial value a (the y-value when x = 0, or work backward if needed) and the growth/decay factor b (divide consecutive y-values: b = y₂/y₁ when x increases by 1). Then write f(x) = a· $b^x$ . Here, the initial amount a=200 at t=0, and it increases by 10% each hour, so growth factor b=1.1, giving A(t)=200* $(1.1)^t$ . Choice B correctly constructs the exponential with a=200 and b=1.1 from the description. Choice A uses decay b=0.9, which would decrease the amount—remember that increase means b>1. Exponential construction strategy: (1) Find initial value: if you have x = 0 in data, that y is your a; otherwise calculate backward using the pattern, (2) Find base: divide consecutive y-values (with x differing by 1): b = $y_{x+1}$ / $y_x$ —should be constant for exponential, (3) Write f(x) = a· $b^x$ , (4) Verify with all data points. Example: points (0, 100) and (1, 110) and (2, 121) → a = 100, b = 110/100 = 1.1 (check: 121/110 = 1.1 ✓), so f(x) = 100· $(1.1)^x$ . The ratio test both identifies the type and gives you the base!
Find the linear function $f(x)=mx+b$ with slope $m=-3$ that passes through the point $(2,5)$ . Which function is correct?
Answer: $f(x)=-3x+11$ This question tests your ability to construct linear or exponential functions from given information like points, tables, graphs, or descriptions of relationships. To construct a linear function from a slope and a point, use the point-slope form or substitute directly into y = mx + b to find b. We're given slope m = -3 and point (2,5). Substituting into f(x) = -3x + b: 5 = -3(2) + b, which gives us 5 = -6 + b, so b = 11. Therefore, f(x) = -3x + 11. Choice A correctly constructs f(x) = -3x + 11 with the given slope -3 and passing through point (2,5). Choice C incorrectly has b = -11 instead of b = 11, likely from a sign error when solving for b. Linear construction from slope and point recipe: (1) Start with f(x) = mx + b where m is given, (2) Substitute the given point (x₀, y₀) to get y₀ = mx₀ + b, (3) Solve for b: b = y₀ - mx₀, (4) Write the complete function. For m = -3 and (2,5): b = 5 - (-3)(2) = 5 - (-6) = 5 + 6 = 11, so f(x) = -3x + 11. Verify: f(2) = -3(2) + 11 = -6 + 11 = 5 ✓!
A geometric sequence has its first term $a_1 = 6$ and its fourth term $a_4 = 162$ . Which function represents the $n$ th term of this sequence?
Answer: $a_n = 6 \cdot 3^{n-1}$ When you encounter a geometric sequence problem, remember that every term follows the pattern $a_n = a_1 \cdot r^{n-1}$ , where $a_1$ is the first term and $r$ is the common ratio. Your job is to find the common ratio using the given information. Since $a_1 = 6$ and $a_4 = 162$ , you can substitute into the formula: $a_4 = a_1 \cdot r^{4-1} = 6 \cdot r^3 = 162$ . Solving for $r$ : $r^3 = \frac{162}{6} = 27$ , so $r = 3$ . Therefore, the formula is $a_n = 6 \cdot 3^{n-1}$ , which is answer choice D. Let's check why the other options fail. Choice A gives $a_n = 6 \cdot 27^{n-1}$ . This would make $a_4 = 6 \cdot 27^3$ , which is astronomically large—far beyond 162. Choice B gives $a_n = 6 \cdot 3^{n+1}$ . Here, $a_1 = 6 \cdot 3^2 = 54$ , not 6, so this doesn't match our first term. Choice C gives $a_n = 2 \cdot 3^n$ . This makes $a_1 = 2 \cdot 3^1 = 6$ (correct first term), but $a_4 = 2 \cdot 3^4 = 162$ (seemingly correct). However, this formula gives $a_2 = 18$ and $a_3 = 54$ , making the ratios inconsistent with a geometric sequence starting at 6. Always verify your geometric sequence formula by checking that it produces the correct first term and maintains a constant ratio between consecutive terms. The standard form $a_n = a_1 \cdot r^{n-1}$ is your most reliable starting point.
Determine whether the relationship is linear or exponential, then write the function. $x$ : 1, 2, 3, 4 $y$ : 2, 6, 18, 54
Answer: $y=2\cdot 3^{x-1}$ This question tests your ability to construct linear or exponential functions from given information like points, tables, graphs, or descriptions of relationships. From a table, first determine which type: calculate differences between consecutive y-values (if constant → linear with slope = that difference), and calculate ratios (if constant → exponential with base = that ratio). This identification step is crucial—you can't construct the right function if you don't know which type it is! Differences vary (4,12,36), but ratios are constant 3 (6/2=3, etc.), so exponential with base 3; adjusted form y=2* $3^{x-1}$ fits starting at x=1 with y=2. Choice A correctly constructs the exponential with adjusted parameters from the table. Choice B uses y=2* $3^x$ without shift, giving 6 at x=1≠2; incorporate the exponent shift for non-zero starting x. Exponential construction strategy: (1) Find initial value: if you have x = 0 in data, that y is your a; otherwise calculate backward using the pattern, (2) Find base: divide consecutive y-values (with x differing by 1): b = $y_{x+1}$ / $y_x$ —should be constant for exponential, (3) Write f(x) = a· $b^x$ , (4) Verify with all data points. Example: points (0, 100) and (1, 110) and (2, 121) → a = 100, b = 110/100 = 1.1 (check: 121/110 = 1.1 ✓), so f(x) = 100· $(1.1)^x$ . The ratio test both identifies the type and gives you the base!
A line has slope $m=3$ and passes through the point $(1,5)$ . Which linear function is correct?
Answer: $f(x)=3x+2$ This question tests your ability to construct linear or exponential functions from given information like points, tables, graphs, or descriptions of relationships. To construct a linear function from two points, find the slope m = (y₂ - y₁)/(x₂ - x₁), then find the y-intercept b by substituting one point into y = mx + b and solving for b. Once you have m and b, you've got your function! For exponential functions, find the initial value a (the y-value when x = 0, or work backward if needed) and the growth/decay factor b (divide consecutive y-values: b = y₂/y₁ when x increases by 1). Then write f(x) = a· $b^x$ . Given slope m=3 and point (1,5), substitute: 5=3*1 + b gives b=2, so f(x)=3x+2. Choice A correctly constructs the linear function with m=3 and b=2 from the slope and point. A distractor like choice B uses b=5, but that would be f(1)=3+5=8≠5—always solve for b after plugging in the point. Linear construction from two points recipe: (1) Find slope: m = (y₂ - y₁)/(x₂ - x₁), (2) Find y-intercept: substitute either point and m into y = mx + b, solve for b, (3) Write function: f(x) = [m]x + [b], (4) Verify: check that both original points work in your function. Example: (1, 4) and (3, 10) → m = (10-4)/(3-1) = 3, then 4 = 3(1) + b gives b = 1, so f(x) = 3x + 1. Check: f(1) = 4 ✓, f(3) = 10 ✓!
Write the exponential function of the form $f(x)=a\cdot b^x$ that passes through $(1,8)$ and $(3,32)$ .
Answer: $f(x)=4\cdot 2^x$ This question tests your ability to construct linear or exponential functions from given information like points, tables, graphs, or descriptions of relationships. For exponential functions, find the initial value a (the y-value when x = 0, or work backward if needed) and the growth/decay factor b (divide consecutive y-values: b = y₂/y₁ when x increases by 1). Then write f(x) = a· $b^x$ . To construct a linear function from two points, find the slope m = (y₂ - y₁)/(x₂ - x₁), then find the y-intercept b by substituting one point into y = mx + b and solving for b. Once you have m and b, you've got your function! Using (1,8) and (3,32), divide 32/8 = $b^2$ gives b=2, then 8=a* $2^1$ solves a=4, so f(x)=4* $2^x$ . Choice C correctly constructs the exponential function with initial value 4 and base 2 from the points. A distractor like choice B might miscalculate a by using a different point—solve the system step by step! Exponential construction strategy: (1) Find initial value: if you have x = 0 in data, that y is your a; otherwise calculate backward using the pattern, (2) Find base: divide consecutive y-values (with x differing by 1): b = $y_{x+1}$ / $y_x$ —should be constant for exponential, (3) Write f(x) = a· $b^x$ , (4) Verify with all data points. Example: points (0, 100) and (1, 110) and (2, 121) → a = 100, b = 110/100 = 1.1 (check: 121/110 = 1.1 ✓), so f(x) = 100· $(1.1)^x$ . The ratio test both identifies the type and gives you the base!
A bacteria culture starts with 200 bacteria and grows by 10% each hour. Let $B(t)$ be the number of bacteria after $t$ hours. Which exponential function models this situation?
Answer: $B(t)=200(1.10)^t$ This question tests your ability to construct linear or exponential functions from given information like points, tables, graphs, or descriptions of relationships. To construct a linear function from two points, find the slope m = (y₂ - y₁)/(x₂ - x₁), then find the y-intercept b by substituting one point into y = mx + b and solving for b. Once you have m and b, you've got your function! For exponential functions, find the initial value a (the y-value when x = 0, or work backward if needed) and the growth/decay factor b (divide consecutive y-values: b = y₂/y₁ when x increases by 1). Then write f(x) = a· $b^x$ . For bacteria growing by 10% each hour, this is exponential growth. Starting with 200 bacteria means a = 200. Growing by 10% means each hour we multiply by 1.10 (100% + 10% = 110% = 1.10). Therefore, B(t) = 200 $(1.10)^t$ . Choice B correctly constructs B(t) = 200 $(1.10)^t$ with initial value 200 and growth factor 1.10 from the problem description. Choice A incorrectly models this as linear growth, choice C uses 0.90 which would be 10% decay not growth, and choice D omits the initial value entirely. Exponential construction strategy: (1) Find initial value: if you have x = 0 in data, that y is your a; otherwise calculate backward using the pattern, (2) Find base: divide consecutive y-values (with x differing by 1): b = $y_{x+1}$ / $y_x$ —should be constant for exponential, (3) Write f(x) = a· $b^x$ , (4) Verify with all data points. Example: points (0, 100) and (1, 110) and (2, 121) → a = 100, b = 110/100 = 1.1 (check: 121/110 = 1.1 ✓), so f(x) = 100· $(1.1)^x$ . The ratio test both identifies the type and gives you the base!
Find the linear function $f(x)=mx+b$ with slope $m=-3$ that passes through the point $(2,5)$ . What is $f(x)$ ?
Answer: $f(x)=-3x+11$ This question tests your ability to construct linear or exponential functions from given information like points, tables, graphs, or descriptions of relationships. To construct a linear function from two points, find the slope m = (y₂ - y₁)/(x₂ - x₁), then find the y-intercept b by substituting one point into y = mx + b and solving for b. Once you have m and b, you've got your function! Given slope m=-3 and point (2,5), substitute: 5 = -3*2 + b gives b=11, so f(x)=-3x+11. Choice A correctly constructs the linear function with m=-3 and b=11 using the given slope and point. Choice B uses positive m=3, which doesn't match the given slope—remember to use the provided m directly. Linear construction from two points recipe: (1) Find slope: m = (y₂ - y₁)/(x₂ - x₁), (2) Find y-intercept: substitute either point and m into y = mx + b, solve for b, (3) Write function: f(x) = [m]x + [b], (4) Verify: check that both original points work in your function. Example: (1, 4) and (3, 10) → m = (10-4)/(3-1) = 3, then 4 = 3(1) + b gives b = 1, so f(x) = 3x + 1. Check: f(1) = 4 ✓, f(3) = 10 ✓!
The arithmetic sequence $4,\ 9,\ 14,\ 19,\dots$ defines values of a function at integer inputs with $f(1)=4$ , $f(2)=9$ , $f(3)=14$ , etc. Write a linear function $f(x)$ that extends this pattern.
Answer: $f(x)=5x-1$ This question tests your ability to construct linear or exponential functions from given information like points, tables, graphs, or descriptions of relationships. To construct a linear function from two points, find the slope m = (y₂ - y₁)/(x₂ - x₁), then find the y-intercept b by substituting one point into y = mx + b and solving for b. Once you have m and b, you've got your function! For this arithmetic sequence with common difference 5, the linear function has slope m=5, and using f(1)=4: 4=5*1 + b gives b=-1, so f(x)=5x-1. Choice A correctly constructs the linear function with m=5 and b=-1 to extend the sequence. Choice B uses m=4, which doesn't match the difference—f(2)=4*2+1=9 ok but f(3)=12+1=13≠14; verify with multiple terms. Linear construction from two points recipe: (1) Find slope: m = (y₂ - y₁)/(x₂ - x₁), (2) Find y-intercept: substitute either point and m into y = mx + b, solve for b, (3) Write function: f(x) = [m]x + [b], (4) Verify: check that both original points work in your function. Example: (1, 4) and (3, 10) → m = (10-4)/(3-1) = 3, then 4 = 3(1) + b gives b = 1, so f(x) = 3x + 1. Check: f(1) = 4 ✓, f(3) = 10 ✓!

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