find the horizontal asymptote of y=-2(1/3)^x + 8. type your answer as an equation. no spaces, no caps. ex) y=8

Explanation:

Step1: Recall exponential - function property

For an exponential function of the form $y = a\cdot b^{x}+k$ ($b>0,b\neq1$), when $|b|\lt1$, $\lim_{x\rightarrow+\infty}b^{x}=0$. Here $a=-2$, $b = \frac{1}{3}$, and $k = 8$.

Step2: Find the limit as $x\rightarrow+\infty$

We find $\lim_{x\rightarrow+\infty}y=\lim_{x\rightarrow+\infty}\left(-2\left(\frac{1}{3}\right)^{x}+8\right)$. Since $\lim_{x\rightarrow+\infty}\left(\frac{1}{3}\right)^{x}=0$, then $\lim_{x\rightarrow+\infty}\left(-2\left(\frac{1}{3}\right)^{x}+8\right)=-2\times0 + 8=8$.

Step3: Determine the horizontal asymptote

The horizontal asymptote of the function $y=-2\left(\frac{1}{3}\right)^{x}+8$ is $y = 8$.

Answer:

y=8