Chapter 15 - Section 15.1 - Functions of Several Variables from the Numerical, Algebraic, and Graphical Viewpoints - Exercises - Page 1094: 2

$$\eqalign{ & \left( a \right)1 \cr & \left( b \right)2 \cr & \left( c \right)2 \cr & \left( d \right){a^2} - 2a - 1 \cr & \left( e \right){y^2} - x - yx + 1 \cr & \left( f \right){\left( {x + h} \right)^2} - \left( {y + k} \right) - \left( {x + h} \right)\left( {y + k} \right) + 1 \cr} $$

$$\eqalign{ & f\left( {x,y} \right) = {x^2} - y - xy + 1 \cr & \left( a \right){\text{ Find }}f\left( {0,0} \right) \cr & f\left( {0,0} \right) = {\left( 0 \right)^2} - \left( 0 \right) - \left( 0 \right)\left( 0 \right) + 1 \cr & f\left( {0,0} \right) = 1 \cr & \cr & \left( b \right){\text{ Find }}f\left( {1,0} \right) \cr & f\left( {1,0} \right) = {\left( 1 \right)^2} - \left( 0 \right) - \left( 1 \right)\left( 0 \right) + 1 \cr & f\left( {1,0} \right) = 2 \cr & \cr & \left( c \right){\text{ Find }}f\left( {0, - 1} \right) \cr & f\left( {0, - 1} \right) = {\left( 0 \right)^2} - \left( { - 1} \right) - \left( 0 \right)\left( 0 \right) + 1 \cr & f\left( {0, - 1} \right) = 2 \cr & \cr & \left( d \right){\text{ Find }}f\left( {a,2} \right) \cr & f\left( {a,2} \right) = {\left( a \right)^2} - \left( 2 \right) - \left( a \right)\left( 2 \right) + 1 \cr & f\left( {a,2} \right) = {a^2} - 2 - 2a + 1 \cr & f\left( {a,2} \right) = {a^2} - 2a - 1 \cr & \cr & \left( e \right){\text{ Find }}f\left( {y,x} \right) \cr & f\left( {y,x} \right) = {y^2} - x - yx+ 1 \cr & \cr & \left( f \right){\text{ Find }}f\left( {x + h,y + k} \right) \cr & f\left( {x + h,y + k} \right) = {\left( {x + h} \right)^2} - \left( {y + k} \right) - \left( {x + h} \right)\left( {y + k} \right) + 1 \cr} $$