Question

1. $y = \left(\frac{1}{4}\

ight)^x$
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: Determine the range

For an exponential function \(y = a^{x}\) (\(a>0,a
eq1\)), the range is \((0,\infty)\). Here \(a=\frac{1}{4}\), so the range of \(y = (\frac{1}{4})^{x}\) is \((0,\infty)\).

Step2: Analyze the end - behavior

As \(x
ightarrow\infty\), for \(y = (\frac{1}{4})^{x}\), we know that if \(0 < a<1\) and \(y=a^{x}\), then \(\lim_{x
ightarrow\infty}a^{x}=0\). So as \(x
ightarrow\infty\), \(f(x)
ightarrow0\).
As \(x
ightarrow-\infty\), if \(0 < a<1\) and \(y = a^{x}\), then \(y=a^{x}=\frac{1}{a^{-x}}\), and \(\lim_{x
ightarrow-\infty}a^{x}=\infty\) (since \(a^{-x}\) with \(0 < a<1\) and \(x
ightarrow-\infty\) is equivalent to \(b^{t}\) where \(b=\frac{1}{a}>1\) and \(t=-x
ightarrow\infty\)). So as \(x
ightarrow-\infty\), \(f(x)
ightarrow\infty\).

Step3: Find the \(y -\)intercept

The \(y -\)intercept occurs when \(x = 0\). Substitute \(x = 0\) into \(y=(\frac{1}{4})^{x}\), we get \(y=(\frac{1}{4})^{0}=1\) (using the rule \(a^{0}=1,a
eq0\)).

Step4: Find the asymptote

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

Answer:

Domain: \((-\infty,\infty)\)
Range: \((0,\infty)\)
End Behavior: As \(x
ightarrow\infty\), \(f(x)
ightarrow0\); As \(x
ightarrow-\infty\), \(f(x)
ightarrow\infty\)
\(y -\)intercept: \((0,1)\)
Asymptote: \(y = 0\)