Answer
$$\eqalign{
& \left( {\text{a}} \right) - 2 \cr
& \left( {\text{b}} \right) - 14 \cr
& \left( {\text{c}} \right)2 \cr
& \left( {\text{d}} \right)18 - 2{y^2} \cr} $$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 6 - 4x - 2{y^2} \cr
& \cr
& \left( {\text{a}} \right)\left( {0,2} \right) \cr
& {\text{Evaluate, substitute 0 for }}x{\text{ and 2 for }}y \cr
& f\left( {0,2} \right) = 6 - 4\left( 0 \right) - 2{\left( 2 \right)^2} = - 2 \cr
& \cr
& \left( {\text{b}} \right)\left( {5,0} \right) \cr
& {\text{Evaluate, substitute 5 for }}x{\text{ and 0 for }}y \cr
& f\left( {5,0} \right) = 6 - 4\left( 5 \right) - 2{\left( 0 \right)^2} = - 14 \cr
& \cr
& \left( {\text{c}} \right)\left( { - 1, - 2} \right) \cr
& {\text{Evaluate, substitute }} - 1{\text{ for }}x{\text{ and }} - 2{\text{ for }}y \cr
& f\left( {0,2} \right) = 6 - 4\left( { - 1} \right) - 2{\left( 2 \right)^2} = 2 \cr
& \cr
& \left( {\text{d}} \right)\left( { - 3,y} \right) \cr
& {\text{Evaluate, substitute }} - 3{\text{ for }}x \cr
& f\left( {0,2} \right) = 6 - 4\left( { - 3} \right) - 2{\left( y \right)^2} = 18 - 2{y^2} \cr} $$
