Question

complete the table of values.

$x$	$4^{-x}$
$0$	$a$
$2$	$b$
$4$	$c$

$a = \square$ $b = \square$ $c = \square$

Explanation:

Step1: Find the value of \(a\) when \(x = 0\)

Substitute \(x = 0\) into the function \(y = 4^{-x}\). According to the exponent rule \(a^0 = 1\) (where \(a
eq0\)), we have \(4^{-0}=4^{0} = 1\). So \(a = 1\).

Step2: Find the value of \(b\) when \(x = 2\)

Substitute \(x = 2\) into the function \(y = 4^{-x}\). We get \(4^{-2}\). Using the exponent rule \(a^{-n}=\frac{1}{a^{n}}\), so \(4^{-2}=\frac{1}{4^{2}}=\frac{1}{16}\). Thus, \(b=\frac{1}{16}\).

Step3: Find the value of \(c\) when \(x = 4\)

Substitute \(x = 4\) into the function \(y = 4^{-x}\). We have \(4^{-4}\). By the exponent rule \(a^{-n}=\frac{1}{a^{n}}\), \(4^{-4}=\frac{1}{4^{4}}=\frac{1}{256}\). So \(c = \frac{1}{256}\).

Answer:

\(a = \boldsymbol{1}\), \(b=\boldsymbol{\frac{1}{16}}\), \(c=\boldsymbol{\frac{1}{256}}\)