Question

problem # 3 - identify the x - intercept, y - intercept, asymptote, domain, range,
and end behavior for the given function.
$y=-2(3)^{x}+4$
asymptote:
circle one: increasing or decreasing
x - intercept:
y - intercept:
domain:
range:
end behavior: as $x\to-\infty,y\to$ as $x\to\infty,y\to$.

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 an exponential function \(y = a\cdot b^{x}+c\), the horizontal asymptote is \(y = c\). For \(y=(2)^{x}\), \(a = 1\), \(b=2\), \(c = 0\). So the asymptote is \(y = 0\).

Step2: Determine if the function is increasing or decreasing

Since \(b = 2>1\), as \(x\) increases, \(y=(2)^{x}\) increases. So the function is increasing.

Step3: Find the domain

The domain of an exponential function \(y = a\cdot b^{x}+c\) is all real numbers. So the domain is \((-\infty,\infty)\) or \(\mathbb{R}\).

Step4: Find the range

Since \(b^{x}>0\) for all \(x\in\mathbb{R}\) and \(a = 1>0\), \(y=(2)^{x}>0\). So the range is \((0,\infty)\).

Step5: Analyze the end - behavior

As \(x\to-\infty\), \(b^{x}=(2)^{x}\to0\) (because for \(0 < b<1\), \(b^{x}\to\infty\) as \(x\to-\infty\) and for \(b > 1\), \(b^{x}\to0\) as \(x\to-\infty\)). So as \(x\to-\infty\), \(y=(2)^{x}\to0\). As \(x\to\infty\), \(b^{x}=(2)^{x}\to\infty\) (because for \(b>1\), \(b^{x}\to\infty\) as \(x\to\infty\)). So as \(x\to\infty\), \(y=(2)^{x}\to\infty\).

Answer:

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