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: Analyze the function \(y = 2^{x}\)

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

Step2: Find the asymptote

Since \(c = 0\), the asymptote is \(y = 0\).

Step3: Determine if the function is increasing or decreasing

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

Step4: Find the domain

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

Step5: Find the range

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

Step6: Analyze the end - behavior

As \(x\to-\infty\), \(2^{x}\to0\) (because \(2^{x}=\frac{1}{2^{-x}}\) and as \(x\to-\infty\), \(-x\to\infty\)). As \(x\to\infty\), \(2^{x}\to\infty\) (because the base \(b = 2>1\)).

Answer:

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