Question

Decide if the function is an exponential growth function or exponential decay function, and describe its end behavior using limits. Y=7^-x

Answer 1

Answer:

Exponential Decay

Its end behavior on the left is as follows as x approaches negative infinity y approaches positive infinity. Its end behavior on the right is as follows as x approaches positive infinity y approaches negative infinity.

Step-by-step explanation:

We can graph the function by graphing two points when x=0 and x=1.

x=0 has $y=7^{-x} =7^{0} =1$

x=1 has y=$7^{-x} =7^{-1} =\frac{1}{7}$

This function starts with higher output values and decreases over time. This is Exponential Decay. Its end behavior on the left is as follows as x approaches negative infinity y approaches positive infinity. Its end behavior on the right is as follows as x approaches positive infinity y approaches negative infinity.

Answer 2

Using limits, it is found that since $\lim_{x \rightarrow \infty} f(x) < \lim_{x \rightarrow -\infty} f(x)$, it is an exponential decay function, as it starts at infinity and ends at 0.

How we use limits to classify an exponential function as growth or decay?

An exponential function is modeled by:

$f(x) = ab^x$.

Then:

• If $\lim_{x \rightarrow \infty} f(x) < \lim_{x \rightarrow -\infty} f(x)$, it is exponential decay.

• If $\lim_{x \rightarrow \infty} f(x) < \lim_{x \rightarrow -\infty} f(x)$, it is exponential growth.

In this problem, the function is:

$y = 7^{-x}$

Hence:

$\lim_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow \infty} 7^{-x} = 7^{-\infty} = \frac{1}{7^{\infty}} = 0$

$\lim_{x \rightarrow -\infty} f(x) = \lim_{x \rightarrow -\infty} 7^{-x} = 7^{\infty} = \infty$

Hence, it is exponential decay, as it starts at infinity and ends at 0.

More can be learned about exponential functions at https://brainly.com/question/25537936