Answer:
[tex]\begin{array}{c|c c c c c c}\sf{x} & \sf{-2} & \sf{-1} & \sf{0} & \sf{1} & \sf{2} & \sf{3}\\\cline{1-7} \sf{y} & \sf{16} & \sf{4} & \sf{1} & \sf{\frac{1}{4}} & \overline{\sf{\frac{1}{16}}} & \sf{\frac{1}{64}}\end{array}[/tex]
Step-by-step explanation:
Given function:
[tex]\sf y=\left(\dfrac{1}{4}\right)^x[/tex]
In order to find the missing parts of the table, substitute the x-values into the given function to find the y-values.
When x = -2:
[tex]\sf \\\implies y=\left(\dfrac{1}{4}\right)^{-2}\ \Bigg( \textsf{Apply the rule:} \left(\dfrac{1}{a}\right)^{-n}=a^n \Bigg)\\\\\implies y=4^2\\\\\implies y=\boxed{\sf16}[/tex]
When x = 0:
[tex]\sf \\\implies y=\left(\dfrac{1}{4}\right)^0\ \Bigg( \textsf{Apply the rule: }x^0=1 \Bigg)\\\\\implies y=\boxed{\sf1}[/tex]
When x = 3:
[tex]\sf \\\implies y=\left(\dfrac{1}{4}\right)^3\ \Bigg( \textsf{Apply the rule:} \left(\dfrac{x}{y}\right)^{n}=\dfrac{x^n}{y^n} \Bigg)\\\\\implies y=\dfrac{1^3}{4^3}\\\\\implies y=\sf \dfrac{1}{\boxed{64}}}[/tex]
