Give a rule for the piecewise-defined function. Then give the domain and range.
- $Give a rule for the piecewise-defined function. Then give the domain and range. - A) \text { } f(x)=\left\{\begin{array}{ll} -6 & \text { if } x \leq-2 \\ -1 & \text { if } x>-2 \end{array} ; \text { Domain: }(\infty, \infty) \text {, Range: }\{-6,-1\}\right. B) f ( x ) = \left\{ \begin{array} { l l } - 6 & \text { if } x < - 2 \\ - 1 & \text { if } x \geq - 2 \end{array} ; \right. Domain: \{ - 6 , - 1 \} , Range: ( x , \infty ) C) f ( x ) = \left\{ \begin{array} { l l } - 6 & \text { if } x \leq - 2 \\ - 1 & \text { if } x > - 2 \end{array} ; \right. Domain: \{ - 6 , - 1 \} , Range: ( \infty , \infty ) D) f ( x ) = \left\{ \begin{array} { l l } - 6 & \text { if } x < - 2 \\ - 1 & \text { if } x \geq - 2 \end{array} ; \right. Domain: ( \infty , \infty ) , Range: \{ - 6 , - 1 \}$
A)

\text { } f(x)=\left\{\begin{array}{ll}-6 & \text { if } x \leq-2 \\-1 & \text { if } x>-2\end{array} ; \text { Domain: }(\infty, \infty) \text {, Range: }\{-6,-1\}\right.

f ( x ) = \left\{ \begin{array} { l l } - 6 & \text { if } x < - 2 \\ - 1 & \text { if } x \geq - 2 \end{array} ; \right.

Domain:

\{ - 6 , - 1 \}

, Range:

( x , \infty )

f ( x ) = \left\{ \begin{array} { l l } - 6 & \text { if } x \leq - 2 \\ - 1 & \text { if } x > - 2 \end{array} ; \right.

Domain:

\{ - 6 , - 1 \}

, Range:

( \infty , \infty )

f ( x ) = \left\{ \begin{array} { l l } - 6 & \text { if } x < - 2 \\ - 1 & \text { if } x \geq - 2 \end{array} ; \right.

Domain:

( \infty , \infty )

, Range:

\{ - 6 , - 1 \}

Decide Whether or Not the Equation Has a Circle as Its

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