Question

need help pls!

Answer 1

ok

Answer 2

nooonooooo

Answer 3

So what I came up with:
\[\large y=7^{(x-1)}\]

You can verify by pluggin in values: x=0
\[\large y= 7^{-1} = \frac{1}{7} \]
x= 1
\[\large y=7^0 =1\]
x= 2
\[\large y=7^1=7 \]

Answer 4

If you want it with a base of 1/7 you could factor out a negative:

\[\large y=7^{-(-x+1)} = \frac{ 1}{7^{(1-x)}} \]

Answer 5

Honestly, you can start with any base, it doesn't necessarily have to be the value at x=0.

Answer 6

As long as it works, of course. Starting with the value at x=0 is just a way to have an idea of what it could be.

Answer 7

@luigi0210 so.. which of the answers should i try

Answer 8

theres only one correct answer btw, i shouldve mentioned that earlier

Answer 9

They are both the same function basically. But if you're looking for a function involving the base at x=0, then the fraction.

Answer 10

it told me some parts of my expression were correct, and some were wrong

Answer 11

@luigi0210

Answer 12

lol try just the fraction.

Answer 13

it told me it was incorrect, is there any way you would like me to try it?

Answer 14

Try just the 7^ (x-1)

If it's not that... then either I need to reevaluate my math skills or the answer key is wrong lol

Answer 15

thank god theres a "try another one" option

Answer 16

@luigi0210 HELP THICK BOI

Answer 17

That is literally the same function too, if you graph them side-by-side.. you should complain loll

Answer 18

question mark

Answer 19

But okay, using the "format" they want I came up with this:

\[\large 8*(2)^x \]

Answer 20

At -1, it becomes 8*(1/2)
at 0 it is 8*1
at 1 it is 8*2
at 2 it is 8*4

Answer 21

LOL ITS WRONG

Answer 22

i think its me.... idk

Answer 23

Ya know what, let's ask someone smarter lol
@jhonyy9 @narad

Answer 24

hes probably playing red dead, check his discord status