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Algebra Help: Constructing Linear And Exponential Functions

Review real example questions for Constructing Linear And Exponential Functions in Algebra.

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A bacteria culture starts with 200 bacteria and increases by 10% each hour. Write a function $P(t)$ for the population after $t$ hours.

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Question 1

A bacteria culture starts with 200 bacteria and increases by 10% each hour. Write a function $P(t)$ for the population after $t$ hours.

$P(t)=200(0.90)^t$
$P(t)=210t$
$P(t)=10(200)^t$
$P(t)=200(1.10)^t$ (correct answer)

Explanation: This question tests your ability to construct exponential functions from given information like descriptions. From context descriptions, listen for the clues: 'per,' 'each,' or 'constant rate' suggest linear (use that rate as slope), while 'percent growth,' 'doubles,' or 'halves' suggest exponential (convert to growth/decay factor). 'Starting with' or 'initial' tells you the y-intercept (linear) or initial value (exponential). Starts with 200 (initial a=200), increases by 10% each hour means growth factor b=1+0.10=1.10, so P(t)=200(1.10)^t. Choice A correctly constructs the exponential by using initial 200 and factor 1.10, giving P(t)=200(1.10)^t. Perfect! A mistake could be using decay like 0.90, but since it's increase, add to 1 for the factor. Context tip: if the problem says 'starts at [value],' that's your y-intercept (linear) or initial value (exponential). If it says 'increases by [number] each time,' that's slope (linear). If it says 'increases by [percent] each time' or 'multiplies by [number],' that's exponential with that as your rate or factor. The language tells you exactly what you need! For exponential from a table: (1) Find the initial value by looking at x = 0 (if in table) or work backward using the pattern, (2) Find the common ratio by dividing consecutive y-values: y₂/y₁ (should be the same for all consecutive pairs), (3) Write y = a·b^x with your values. Example: if y goes 5, 10, 20 as x goes 0, 1, 2, then a = 5, b = 10/5 = 2, so y = 5·2^x!

Question 2

A bacteria culture starts with 500 bacteria and increases by 10% each hour. Let $t$ be the number of hours. Write the exponential function $P(t)$ that models the population.

$P(t) = 500(1.10)^t$ (correct answer)
$P(t) = 500(0.10)^t$
$P(t) = 510t$
$P(t) = 500(10)^t$

Explanation: This question tests your ability to construct exponential functions from given information like descriptions. From context descriptions, 'percent growth' suggests exponential (convert to growth factor). The bacteria starts with 500 (initial value a = 500) and increases by 10% each hour. To convert 10% growth to a growth factor: 100% + 10% = 110% = 1.10. So b = 1.10. Therefore, P(t) = 500(1.10)^t. Choice A correctly constructs P(t) = 500(1.10)^t by identifying the initial population as 500 and converting 10% growth to the factor 1.10. Perfect! Choice B incorrectly uses 0.10 as the base - remember, for growth, we need 1 + growth rate, not just the growth rate itself. Context tip: if it says 'increases by [percent] each time,' that's exponential with growth factor = 1 + (percent as decimal). For 10% growth, b = 1 + 0.10 = 1.10. The language tells you exactly what you need!

Question 3

Find the exponential function $y=a\cdot b^x$ that passes through the points $(0, 5)$ and $(2, 45)$ .

$y=45\cdot 3^x$
$y=5\cdot 3^x$ (correct answer)
$y=5\cdot 9^x$
$y=5\cdot (\sqrt{9})^x$

Explanation: This question tests your ability to construct exponential functions from given information like points. For exponential functions, we need the initial value a (the y-value when x = 0) and the growth or decay factor b (what we multiply by each time x increases by 1). If we have two points where x differs by 1, the ratio of their y-values gives us b directly: if (1, 6) and (2, 18) are on the function, then b = 18/6 = 3, and checking x = 0 or using a given point gives us a. With points (0,5) and (2,45), a=5 (at x=0); then 45=5*b^2, so b^2=9, b=3 (positive for growth), giving y=5·3^x. Choice C correctly constructs the exponential function by finding the initial a=5 and solving for b=3 from the second point, giving the function y=5·3^x. Perfect! You might forget to take the root for non-unit difference in x, leading to other choices, but dividing the exponent helps—great job practicing! For exponential from a table: (1) Find the initial value by looking at x = 0 (if in table) or work backward using the pattern, (2) Find the common ratio by dividing consecutive y-values: y₂/y₁ (should be the same for all consecutive pairs), (3) Write y = a·b^x with your values. Example: if y goes 5, 10, 20 as x goes 0, 1, 2, then a = 5, b = 10/5 = 2, so y = 5·2^x!

Question 4

A population of rabbits doubles every 4 months. If there are currently 80 rabbits (at month 0), which function models the population $P(t)$ after $t$ months?

$P(t) = 80 \cdot 2^{4t}$
$P(t) = 80 + 2^{t/4}$
$P(t) = 80 \cdot 4^{t/2}$
$P(t) = 80 \cdot 2^{t/4}$ (correct answer)

Explanation: When you encounter exponential growth problems, you need to identify three key components: the initial value, the growth factor, and the time period for that growth. This question tells you the population doubles (growth factor = 2) every 4 months, starting with 80 rabbits. The correct answer is D: $P(t) = 80 \cdot 2^{t/4}$ . Here's why this works: since the population doubles every 4 months, after 4 months you have $80 \cdot 2^{4/4} = 80 \cdot 2^1 = 160$ rabbits. After 8 months: $80 \cdot 2^{8/4} = 80 \cdot 2^2 = 320$ rabbits. The exponent $t/4$ ensures that every 4 months, you get the next whole number power of 2. Choice A uses $2^{4t}$ , which makes the population grow far too quickly—after just 1 month, you'd have $80 \cdot 2^4 = 1,280$ rabbits instead of the gradual growth that should occur. Choice B adds $2^{t/4}$ to 80 instead of multiplying, which represents linear growth plus a small exponential term, not true exponential growth where the entire population doubles. Choice C uses base 4 with exponent $t/2$ . While $4^{t/2}$ does equal $2^t$ algebraically, this doesn't match our doubling period. After 4 months, you'd get $80 \cdot 4^2 = 1,280$ rabbits, which is incorrect. Remember: in exponential growth problems, if something doubles every $n$ time units, your exponent should be $t/n$ so that when $t = n$ , the exponent equals 1 and you get exactly one doubling.

Question 5

A rental car company charges a flat fee plus a per-mile rate. Two customers provide the following data: Customer A drove 120 miles and paid $89, while Customer B drove 200 miles and paid $129. What is the per-mile rate the company charges?

$0.40 per mile
$0.50 per mile (correct answer)
$0.60 per mile
$0.74 per mile

Explanation: Let the flat fee be $b$ and the per-mile rate be $m$ . We have two equations: $120m + b = 89$ and $200m + b = 129$ . Subtracting the first from the second: $80m = 40$ , so $m = 0.50$ . Choice A results from incorrectly calculating $40 ÷ 100 = 0.40$ . Choice C comes from using $129 - 89 = 40$ divided by $120 - 200 = -80$ and taking the absolute value. Choice D comes from dividing total cost by total miles for one customer: $89 ÷ 120 ≈ 0.74$ .

Question 6

A savings account earns compound interest where the balance triples every 5 years. If the account starts with $200, what function represents the balance $ $B(t)$ $after$ $t$ $ years, and what will be the balance after 15 years?

$B(t) = 200 \cdot 3^{5t}$ ; Balance after 15 years: $5400
$B(t) = 200 \cdot 3^{t/5}$ ; Balance after 15 years: $5200
$B(t) = 200 \cdot 3^{t/5}$ ; Balance after 15 years: $5400 (correct answer)
$B(t) = 200 \cdot 3^{t/5}$ ; Balance after 15 years: $1800

Explanation: When you encounter compound interest problems where an amount grows by a specific factor over a set time period, you need to build an exponential function that captures this repeated growth pattern. Since the balance triples every 5 years, you can think of this as: after 5 years, the balance is multiplied by 3; after 10 years (two 5-year periods), it's multiplied by $3^2 = 9$ ; after 15 years (three 5-year periods), it's multiplied by $3^3 = 27$ . The pattern shows that after $t$ years, the balance is multiplied by $3^{t/5}$ because $t/5$ gives you the number of 5-year periods that have passed. This gives you the function $B(t) = 200 \cdot 3^{t/5}$ . To find the balance after 15 years: $B(15) = 200 \cdot 3^{15/5} = 200 \cdot 3^3 = 200 \cdot 27 = 5400$ . Choice A incorrectly uses $3^{5t}$ as the exponent, which would mean the balance gets multiplied by 3 every $\frac{1}{5}$ year instead of every 5 years—this creates explosive, unrealistic growth. Choice B has the correct function but calculates $3^3$ as 26 instead of 27, leading to $200 \times 26 = 5200$ . Choice D also has the right function but miscalculates $3^3$ as 9 instead of 27, giving $200 \times 9 = 1800$ . Remember: in exponential growth problems, always identify the time period for the given growth factor, then use that to determine what goes in your exponent. The growth factor's time period becomes the denominator when time is in the exponent.

Question 7

A bacteria culture starts with 50 bacteria. Every 3 hours, the population triples. After observing this pattern, a researcher needs to predict the population after 15 hours. Which function correctly models the bacteria population $P(t)$ after $t$ hours?

$P(t) = 50 \cdot 3^{t/3}$ (correct answer)
$P(t) = 50 \cdot 3^{3t}$
$P(t) = 50 + 3^{t/3}$
$P(t) = 50 \cdot 3^t$

Explanation: Since the population triples every 3 hours, we need an exponential function where the base 3 is raised to a power that equals 1 when t = 3. The function $P(t) = 50 \cdot 3^{t/3}$ satisfies this: when t = 3, we get $P(3) = 50 \cdot 3^{3/3} = 50 \cdot 3^1 = 150$ , which is triple the initial population. Choice B has the wrong exponent (would triple every 1/3 hour). Choice C uses addition instead of multiplication for exponential growth. Choice D would triple every hour, not every 3 hours.

Question 8

An arithmetic sequence has its 3rd term equal to 14 and its 7th term equal to 26. Which linear function represents the $n$ th term of this sequence?

$a_n = 3n + 2$
$a_n = 4n + 2$
$a_n = 3n + 5$ (correct answer)
$a_n = 2n + 8$

Explanation: When you encounter an arithmetic sequence problem, remember that these sequences have a constant difference between consecutive terms, making them linear functions. You need to find both the first term and the common difference to write the formula. Given information: $a_3 = 14$ and $a_7 = 26$ . First, find the common difference ( $d$ ). Since there are 4 steps from the 3rd to 7th term, you have: $d = \frac{a_7 - a_3}{7 - 3} = \frac{26 - 14}{4} = \frac{12}{4} = 3$ . Now find the first term using $a_3 = a_1 + 2d$ : $14 = a_1 + 2(3)$ , so $a_1 = 14 - 6 = 8$ . The general formula is $a_n = a_1 + (n-1)d = 8 + (n-1)(3) = 8 + 3n - 3 = 3n + 5$ . This matches answer choice C. Let's check why the other answers fail: Answer A ( $a_n = 3n + 2$ ) has the correct common difference of 3, but gives $a_3 = 11$ and $a_7 = 23$ , which don't match our given values. Answer B ( $a_n = 4n + 2$ ) has an incorrect common difference of 4. Answer D ( $a_n = 2n + 8$ ) also has the wrong common difference of 2. Study tip: Always verify your answer by substituting back into the original conditions. For arithmetic sequences, you can find the common difference using any two given terms by dividing the difference in values by the difference in positions.

Question 9

A bacteria culture starts with 200 bacteria and grows by 10% each hour. Write a function $P(t)$ for the population after $t$ hours. Write in the form $P(t)=a(1+r)^t$ .

$P(t)=200(1.10)^t$ (correct answer)
$P(t)=210(1.10)^t$
$P(t)=200(1.10t)$
$P(t)=200(0.90)^t$

Explanation: This question tests your ability to construct exponential functions from given information like descriptions. From context descriptions, listen for the clues: 'per,' 'each,' or 'constant rate' suggest linear (use that rate as slope), while 'percent growth,' 'doubles,' or 'halves' suggest exponential (convert to growth/decay factor). 'Starting with' or 'initial' tells you the y-intercept (linear) or initial value (exponential). The culture 'starts with 200' (initial a=200) and 'grows by 10% each hour' (growth factor 1+0.10=1.10), so P(t)=200(1.10)^t. Choice A correctly constructs the exponential function by using the starting amount as a=200 and converting the percent growth to the factor 1.10, giving the function P(t)=200(1.10)^t. Perfect! A common mix-up is using addition instead of multiplication for growth, but 'percent' signals exponential—keep practicing to spot it! Context tip: if the problem says 'starts at [value],' that's your y-intercept (linear) or initial value (exponential). If it says 'increases by [number] each time,' that's slope (linear). If it says 'increases by [percent] each time' or 'multiplies by [number],' that's exponential with that as your rate or factor. The language tells you exactly what you need!

Question 10

Find the exponential function $y=a\cdot b^x$ that passes through $(0,9)$ and $(2,81)$ .

$y=9\cdot 9^x$
$y=81\cdot 3^x$
$y=9\cdot 3^x$ (correct answer)
$y=9\cdot \left(\tfrac{1}{3}\right)^x$

Explanation: This question tests your ability to construct exponential functions from given information like points. For exponential functions, we need the initial value a (the y-value when x = 0) and the growth or decay factor b (what we multiply by each time x increases by 1). If we have two points where x differs by 1, the ratio of their y-values gives us b directly: if (1, 6) and (2, 18) are on the function, then b = 18/6 = 3, and checking x = 0 or using a given point gives us a. With (0,9) and (2,81), a=9; then 81=9*b^2, b^2=9, b=3 (positive for growth), so y=9·3^x. Choice B correctly constructs the function by finding initial 9 and base 3, giving y=9·3^x. Perfect! Using b=9 might happen if misdividing, but solve for b by dividing y2 by y1 and adjusting for the x-difference. For exponential from a table: (1) Find the initial value by looking at x = 0 (if in table) or work backward using the pattern, (2) Find the common ratio by dividing consecutive y-values: y₂/y₁ (should be the same for all consecutive pairs), (3) Write y = a·b^x with your values. Example: if y goes 5, 10, 20 as x goes 0, 1, 2, then a = 5, b = 10/5 = 2, so y = 5·2^x!