Match the function with its graph (labeled I-VI). Givereasons a) f(x,y)=|x|+|y| b) f(x,y)=|xy| c) f(x,y)=1÷1+x2+y2 d) f(x,y)=(x2−y2)2 e) f(x,y)=(x−y)2 f) f(x,y)=sin⁡(|x|+|y|)

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Match the function with its graph (labeled I-VI). Givereasons    a) f(x,y)=|x|+|y|    b) f(x,y)=|xy|    c) f(x,y)=1÷1+x2+y2    d) f(x,y)=(x2−y2)2    e) f(x,y)=(x−y)2    f) f(x,y)=sin⁡(|x|+|y|)

i1ziZ · Accepted Answer

Match the function with its graph (labeled I-VI). Givereasonsa) f(x,y)=|x|+|y|Think of f(x,y) as Z: soyou have Z = |x| when y=0 and Z=|y| when x=0 so the only graph thisfits to is VI.b) f(x,y)=|xy|Here when x=0 then Z=0. andwhen y=0, z=0 so it must be graph V.c) f(x,y)=1÷1+x2+y2Here when x=0 then Z=0. andwhen y=0, z=0 so it must be graph V.11+x2+y2goes to when x = 011+y2and when y = 011+x2This causes z to be close to zero when x and yare large which leads to graph I.d) f(x,y)=(x2−y2)2when x=0,z=y4When y=0,z=x4But when z=0,y=±x so that fits graph IV.e) f(x,y)=(x−y)2when x=0,z=y2When y=0,z=x2But when z=0, y=x so that fits graph IIf) f(x,y)=sin⁡(|x|+|y|)The oscillating nature ofthe trig should give this one away even if it wasn&#039;t the last oneon the list. So the answer is III, due to the fact when X=0 thenyou have z=sin(|y|) and when y=0 z=sin(|x|).Hope this helps.

Match the function with its graph (labeled I-VI). Givereasons a) f(x,y) = |x|+|y| b) f(x,y) = |xy| c) f(x,y) = 1\div1+x^{2}+y^{2} d) f(x,y) = (x^{2} -y^{2})^{2} e) f(x,y) = (x-y)^{2} f) f(x,y) = \sin(|x|+|y|)

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