Asymptote: y = -7(4^2x-6) - 3

*I Get -10 but thats not one of my choices

a) y = -7
b) y = -6
c) y = -3
d) y = 4

Question

Asymptote: y = -7(4^2x-6) - 3

*I Get -10 but thats not one of my choices

a) y = -7
b) y = -6
c) y = -3
d) y = 4

anonymous · Answer

Hey,is this $$-7*4^{2x-6} - 3?$$

anonymous · Answer

Yea

anonymous · Answer

(what level of math are you at?)

anonymous · Answer

Well Im Doing Algebra 2 ATM But Some Of It Carries Onto more advanced maths

anonymous · Answer

Cool! one sec

anonymous · Answer

Here's how I think of it.  Let me know if this sounds too complicated.  Do you know anything about limits?

anonymous · Answer

Cant Remember TBH

anonymous · Answer

Even without the "limit" terminology...
since the answer choices you're given are "y=blahblahblah"

anonymous · Answer

you are looking at horizontal asymptotes
So you have to think of two scenarios...when X gets REALLLLY big...or x gets REALLLLY small

anonymous · Answer

Ok

anonymous · Answer

if x gets really big in this case, then the whole thing goes to infinity

anonymous · Answer

when x gets REALLY REALLY small, −7∗4^(2x−6) also gets really, really small...in fact, it gets closer and closer to zero

anonymous · Answer

So you're left with y = "0" - 3 .... an asymptote of y = -3

anonymous · Answer

That is, as x gets smaller and smaller, y gets closer and closer to -3 but never really hits -3

anonymous · Answer

Hmm Ok

anonymous · Answer

anonymous · Answer

how were you trying to do it?

anonymous · Answer

Just Calculating the asymptote