Answer
a) $f(0,0)=4.$
b) $f(0,1)=0.$
c) $f(2,3)=-36.$
d) $f(1,y) =3-4y^2.$
e) $f(x,0)= 4-x^2.$
f) $f(t,1) = -t^2.$
Work Step by Step
We will simply substitute $x$ from the first coordinate and $y$ from the second coordinate in the given point into the given function $f(x,y) = 4-x^2-4y^2$.
a) $f(0,0)=4-0^2-4\times0^2 =4.$
b) $f(0,1)=4-0^2-4\times 1^2 = 4-4=0.$
c) $f(2,3)=4-2^2-4\times 3^2 = 4-4-36=-36.$
d) $f(1,y) = 4-1^2-4y^2=3-4y^2.$
e) $f(x,0)=4-x^2-4\times0^2 = 4-x^2.$
f) $f(t,1) = 4-t^2-4\times 1^2 = -t^2.$
