Question

In need to describe the end behavior of this function.

Answer 1

it's 1 to 1 !

Answer 2

\[f(x)=-2^{x+3}-1\]


A)
As x goes to negative infinity, y goes to negative one and as x goes to infinity, y increases towards infinity
B)
As x goes to negative infinity, y goes towards 0 and as x goes to infinity, y decreases to negative infinity
C)
As x goes to negative infinity, y goes towards zero and as x goes to infinity, y increases towards infinity
D)
As x goes to negative infinity, y goes to negative one and as x goes to infinity, y decrease towards infinity

I think it's A. I had a confusion because i thought that the minus next to "2" also exponentially grown. that would mean that whether it goes to pos. or neg. infinity would depend on whether x is odd or even. But only 2 is part of exponent. Am I right ?

Answer 3

No, not A, it's B

Answer 4

it goes to neg infinity when y increases because it's going to be greater and greater negative number

Answer 5

remember, \[-2^x = -1*2^x\]\[-2^x \ne (-2)^x\]

Answer 6

yeah that was my first confusion

Answer 7

and as x decreases the y goes to -1 because the " -2^(x+3)" part approaches 0.

Answer 8

Am I right ?

Answer 9

Ohh, I a thinking correctly, but chose the wrong choice. It's D.

Answer 10

Thanks everyone.

Answer 11

As x approaches negative infinity, the expression \(-2^{x+3}-1\) approaches \[\frac{-1}{2^\infty} -1 \approx -1\]

As x approaches positive infinity, the expression approaches \[-1*2^{\infty}-1\approx -\infty\]

D is correct only if you left out a word (negative) when you copied it.