Question

independent practice:
problem # 1 - identify the asymptote,
domain, range, and end behavior.
$y=(2)^{x}$
asymptote:
circle one: increasing or decreasing
domain:
range:
end behavior: as $x\to-\infty,y\to$. as $x\to\infty,y\to$.

Explanation:

Step1: Find the asymptote

For the exponential function \(y = a^{x}\) (\(a>0,a
eq1\)), the horizontal asymptote is \(y = 0\).

Step2: Determine if the function is increasing or decreasing

For \(y=(2)^{x}\), since \(a = 2>1\), the function is increasing.

Step3: Find the domain

The domain of an exponential function \(y = a^{x}\) is all real numbers. In interval notation, \(x\in(-\infty,\infty)\).

Step4: Find the range

Since \(2^{x}>0\) for all real \(x\), the range is \(y\in(0,\infty)\).

Step5: Analyze the end - behavior

As \(x\to-\infty\), \(y = 2^{x}=\frac{1}{2^{-x}}\to0\). As \(x\to\infty\), \(y = 2^{x}\to\infty\).

Answer:

asymptote: \(y = 0\)
Circle one: increasing
domain: \((-\infty,\infty)\)
range: \((0,\infty)\)
end behavior: As \(x\to-\infty,y\to0\). As \(x\to\infty,y\to\infty\)