Question

1. $y = \left(\frac{1}{2}\

ight)^x + 1$
domain: $(-\infty, \infty)$
range:
end behavior:
as $x \to \infty$, $f(x) \to $
as $x \to -\infty$, $f(x) \to $
y-intercept:
asymptotes:

Explanation:

Step1: Find the range

For the function \(y = (\frac{1}{2})^{x}-1\), since \((\frac{1}{2})^{x}>0\) for all \(x\in R\), then \(y = (\frac{1}{2})^{x}-1>- 1\). So the range is \((-1,\infty)\).

Step2: Analyze the end - behavior

As \(x
ightarrow\infty\), we know that \(a^{x}\) with \(0 < a<1\) (here \(a=\frac{1}{2}\)) approaches \(0\). So \(y = (\frac{1}{2})^{x}-1
ightarrow0 - 1=-1\).
As \(x
ightarrow-\infty\), \(a^{x}\) with \(0 < a<1\) approaches \(\infty\). So \(y = (\frac{1}{2})^{x}-1
ightarrow\infty-1=\infty\).

Step3: Find the \(y\) - intercept

Set \(x = 0\), then \(y=(\frac{1}{2})^{0}-1=1 - 1=0\).

Step4: Find the asymptote

Since \(y = (\frac{1}{2})^{x}-1\) and \(\lim_{x
ightarrow\infty}(\frac{1}{2})^{x}-1=-1\), the horizontal asymptote is \(y=-1\).

Answer:

Domain: \((-\infty,\infty)\)
Range: \((-1,\infty)\)
End Behavior:
As \(x
ightarrow\infty\), \(f(x)
ightarrow - 1\)
As \(x
ightarrow-\infty\), \(f(x)
ightarrow\infty\)
\(y\) - intercept: \(0\)
Asymptote: \(y=-1\)