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Question

Three Algebra 1 students are comparing how fast their social media posts have spread. Their results are shown in the following data:

Amber shared her photo with 3 people. They continued to share it, so the number of shares increases every day as shown by the function. f(x) = 3(4)^x

Ben shared his post with 2 friends. Each of those friends shares with 3 more every day, so the number of shares triples every day. Day: 0 - # of shares: 2 / Day: 2 - # of shares: 6 / Day: 3 - # of shares: 18

Carter shared his post with 10 friends, who each share with only 2 people every day.

4.Using the functions for each student, predict how many shares each student's post will have recieved on Day 3 and Day 10. Justify your answers.

Answer 1

Amber already has her situation modeled with a function:

f(x) = 3(4)^x

Ben's situation is also an exponential equation fitting y = a•b^x with the initial value of 2, which means a = 2. And the multiplier is 3 (triples daily) but we need the value on day 2 to be 6, therefore the exponent needs to be "x - 1". We'll let Ben's function be g(x). So g(x) = 2(3)^x-1 The verbiage says the first day is day 0 and the second day is day 2 which isn't consistent. We need to call the first day "day 1", then the second day is day 2 and the third day is day 3. Checking our function: g(1) = 2(3)^1-1 = 2(3)⁰ = 2(1) = 2 and g(2) = 2(3)^2-1 = 2(3)¹ = 2(3) = 6 and g(3) = 2(3)^3-1 = 2(3)² = 2(9) = 18 so this seems to check out.

Carter's situation again is an exponential function but with a multiplier of 2 and an initial value of 10 so (using what we did with Ben's situation), letting Carter be h(x) we get h(x) = 10(2)^x-1

Amber: f(x) = 3(4)^x

so f(3) = 3(4)³ = 3(64) = 192

and f(10) = 3(4)¹⁰ = 3(1048576) = 3145728

Ben: g(x) = 2(3)^x-1

so g(3) = 2(3)^3-1 = 2(3)² = 2(9) = 18

and g(10) = 2(3)^10-1 = 2(3)⁹ = 2(19683) = 39366

Carter: h(x) = 10(2)^x-1

so h(3) = 10(2)^3-1 = 10(2)² = 10(4) = 40

and h(10) = 10(2)^10-1 = 10(2)⁹ = 10(512) = 5120