Ask your own question, for FREE!
Mathematics 17 Online
OpenStudy (eriyalovesu):

PLEASE HELP! Which of the following exponential functions goes through the points (1, 12) and (2, 36)? f(x) = 4(3)−x f(x) = 3(4)−x f(x) = 3(4)x f(x) = 4(3)x

OpenStudy (agent0smith):

points (1, 12) and (2, 36) Plug in x=1, and then x=2, into each of the given functions.

OpenStudy (eriyalovesu):

KK

OpenStudy (eriyalovesu):

I just got (11,10) for the first two and than (12,24) for that last two

OpenStudy (agent0smith):

That doesn't make sense... First function: what is the value you get when you plug in x=1? And x=2?

OpenStudy (eriyalovesu):

4(3)-1 =11 4(3)-2 =10

OpenStudy (agent0smith):

Isn't the x an exponent...? What you did is not using exponents.

OpenStudy (eriyalovesu):

Right! you right! 12 for x=1 and 36 for x=2

OpenStudy (agent0smith):

Remember how to deal with negative exponents? Or you can just use a calculator, but you aren't following order of operations \[\large 4(3)^{-1}= 4\left( \frac{ 1 }{3 } \right)=\]

OpenStudy (agent0smith):

\[\large 4(3)^{-2 } = 4\left( \frac{ 1 }{ 3^2 } \right)=\]

OpenStudy (eriyalovesu):

1.3333 and 0.4444 is what my caulator gave me

OpenStudy (agent0smith):

Good, so does that look like a correct choice? Remember you're looking for the one that matches "points (1, 12) and (2, 36)"

OpenStudy (eriyalovesu):

well no that doesn't. wouldn't 3(4) be the same so A and B out out right?

OpenStudy (agent0smith):

B would not be the same. For B: what is the value you get when you plug in x=1? And x=2?

OpenStudy (eriyalovesu):

0.75 and 0.1875

OpenStudy (agent0smith):

Now do C and D.

OpenStudy (eriyalovesu):

12 and 48 for C 12 and 36 for D

OpenStudy (eriyalovesu):

therefore D is correct right?

OpenStudy (agent0smith):

Yes.

OpenStudy (eriyalovesu):

Thank you!

OpenStudy (agent0smith):

Welcome!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!