Use the graph of the function f to create a table of values for the given points.Then create a second table that can be used to find f^-1.
$Use the graph of the function f to create a table of values for the given points.Then create a second table that can be used to find f<sup>-1</sup>. A) \begin{array}{|l|l|l|l|l|} \hline x & 1 & 4 & 7 & 8 \\ \hline y & 1 & 4 & 7 & 9 \\ \hline \end{array} \begin{array}{|l|l|l|l|l|} \hline x & 1 & 4 & 7 & 9 \\ \hline f^{-1}(x) & 1 & 4 & 7 & 8 \\ \hline \end{array} B) \begin{array}{|l|l|l|l|l|} \hline x & 1 & 4 & 7 & 9 \\ \hline y & 1 & 4 & 7 & 8 \\ \hline \end{array} \begin{array}{|l|l|l|l|l|} \hline x & 1 & 4 & 7 & 8 \\ \hline f^{-1}(x) & 1 & 4 & 7 & 9 \\ \hline \end{array} C) \begin{array}{l} \begin{array} { | l | l | l | l | l | } \hline x & 1 & 4 & 7 & 8 \\ \hline y & 1 & 4 & 7 & 9 \\ \hline \end{array}\\\\ \begin{array} { | l | l | l | l | l | } \hline x & 1 & 4 & 7 & 9 \\ \hline f ^ { - 1 } ( x ) & - 1 & - 4 & - 7 & - 8 \\ \hline \end{array} \end{array} D) \begin{array}{l} \begin{array} { | l | l | l | l | l | } \hline x & - 1 & - 4 & - 7 & - 9 \\ \hline y & 1 & 4 & 7 & 8 \\ \hline \end{array}\\\\ \begin{array} { | l | l | l | l | l | } \hline x & 1 & 4 & 7 & 8 \\ \hline f ^ { - 1 } ( x ) & 1 & 4 & 7 & 9 \\ \hline \end{array} \end{array} E) \begin{array}{l} \begin{array} { | l | l | l | l | l | } \hline x & - 1 & - 4 & - 7 & - 9 \\ \hline y & 1 & 4 & 7 & 8 \\ \hline \end{array}\\\\ \begin{array} { | l | l | l | l | l | } \hline x & 1 & 4 & 7 & 9 \\ \hline f ^ { - 1 } ( x ) & - 1 & - 4 & - 7 & - 8 \\ \hline \end{array} \end{array}$
A)

\begin{array}{|l|l|l|l|l|}\hline x & 1 & 4 & 7 & 8 \\\hline y & 1 & 4 & 7 & 9 \\\hline\end{array}

\begin{array}{|l|l|l|l|l|}\hline x & 1 & 4 & 7 & 9 \\\hline f^{-1}(x) & 1 & 4 & 7 & 8 \\\hline\end{array}

\begin{array}{|l|l|l|l|l|}\hline x & 1 & 4 & 7 & 9 \\\hline y & 1 & 4 & 7 & 8 \\\hline\end{array}

\begin{array}{|l|l|l|l|l|}\hline x & 1 & 4 & 7 & 8 \\\hline f^{-1}(x) & 1 & 4 & 7 & 9 \\\hline\end{array}

\begin{array}{l}\begin{array} { | l | l | l | l | l | } \hline x & 1 & 4 & 7 & 8 \\\hline y & 1 & 4 & 7 & 9 \\\hline\end{array}\\\\\begin{array} { | l | l | l | l | l | } \hline x & 1 & 4 & 7 & 9 \\\hline f ^ { - 1 } ( x ) & - 1 & - 4 & - 7 & - 8 \\\hline\end{array}\end{array}

\begin{array}{l}\begin{array} { | l | l | l | l | l | } \hline x & - 1 & - 4 & - 7 & - 9 \\\hline y & 1 & 4 & 7 & 8 \\\hline\end{array}\\\\\begin{array} { | l | l | l | l | l | } \hline x & 1 & 4 & 7 & 8 \\\hline f ^ { - 1 } ( x ) & 1 & 4 & 7 & 9 \\\hline\end{array}\end{array}

\begin{array}{l}\begin{array} { | l | l | l | l | l | } \hline x & - 1 & - 4 & - 7 & - 9 \\\hline y & 1 & 4 & 7 & 8 \\\hline\end{array}\\\\\begin{array} { | l | l | l | l | l | } \hline x & 1 & 4 & 7 & 9 \\\hline f ^ { - 1 } ( x ) & - 1 & - 4 & - 7 & - 8 \\\hline\end{array}\end{array}

Write the Standard Form of the Equation of the Circle (x+3)2+(y−2)2=50( x + 3 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 50(x+3)2+(y−2)2=50

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Write the Standard Form of the Equation of the Circle $( x + 3 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 50$

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