Question

Which sequence is modeled by the graph below?

Answer 1

an = one third(27)n − 1

an = 27(one third)n − 1

an = one third(3)n − 1

an = 3(one half)n − 1

Answer 2

This seems like a trial-and-error problem, so I don't think I can offer a thorough explanation of this.

However, I will walk you through my thought process.

Immediately I noticed that these points are quite steep. If it were to represent a curve, it would have to be an exponential function, something like e^x or 2^x.

However, ALL exponential functions go through the point (0,1). This one doesn't. Instead it goes through (2,1). So, that means whatever kind of exponential function it is, it must be shifted two units to the right.

Then I looked at (3,3). Since it's an exponential function, I was thinking of 3^1. So I chose 3^x as my "test" function.

I looked at (4,9). But 3^4 = 81 which is not 9. However, remember, we said that this exponential is shifted two units to the right!

So maybe y = 3^(x-2) could work?

At (2,1) ==> 3^(2-2) = 3^0 = 1.
At (3,3) ==> 3^(3-2) = 3^1 = 3.
At (4,9) ==> 3^(4-2) = 3^2 = 9.

Success! All the y-values match.

So, these points represent y=3^(2-x).

Answer 3

Oops! Messed up the last line. It's y = 3^(x-2), NOT y=3^(2-x).

Answer 4

Cool! do I have to put this into the format of the answer options? How would I do that?

Answer 5

What are the answer options?

Answer 6

an = one third(27)n − 1

an = 27(one third)n − 1

an = one third(3)n − 1

an = 3(one half)n − 1

Answer 7

Oh my bad, didn't see those above!

Is that n raising the thing in parentheses?
Like this?
\[ a_n = 27 \left(\frac{1}{3}\right)^{n} - 1\]

Answer 8

lol, sorry it wouldn't paste the fractions!

Yes, thats exactly how the one looks.

Answer 9

Oh okay so the choices would be
\[ a_n = \frac{1}{3} 27^n - 1\]
\[ a_n = \left(\frac{1}{3}\right)^n - 1\]
\[ a_n = \frac{1}{3} 3^n - 1 \]
\[ a_n = 3\left(\frac{1}{2}\right)^n - 1\]
Yes?

Answer 10

In the first one, the 27 is in parenthesis, and in the third, the 3 is in parenthesis, so other than that, its right

Answer 11

Unfortunately none of them satisfy the points given, which is strange.

The 'n' in a_n denotes which term in the series you are it.

So for example, using the first choice: a_1 = 1/3 27^1 - 1 = 9 - 1 = 8.

Second choice: a_1 = 27 (1/3)^1 = 9 - 1 = 8.

Third choice: a_1 = 1/3 3^1 = 1 - 1 =0.

Fourth choice: a_1 = 3(1/2)^1 - 1 = 3/2 - 1 = 1/2.

Answer 12

If I had to choose one, I would go with the third choice since it actually uses 3 as the base of the exponential.

Simplifying it results in 3^(n-1) - 1.

Answer 13

Series can be treated as functions, so we could also plug in these values into these equations.

However, it won't give us corresponding points.

It's a strange problem, sorry.

Answer 14

Yea I think I agree, since the points went in intervals of 3. I'll go with that. Tricky problem! Thx for your help:D

Answer 15

My pleasure!

Answer 16

Yes the third was correct!! Ty sm!