y=1/(2√(x+3)+√(x+3))=1/(3√(x+3)).
This can be rationalised be multiplying top and bottom by √(x+3):
y=√(x+3)/(3(x+3))=√(x+3)/(3x+9).
y(1)=1/(3√4)=⅙; y(6)=1/(3√9)=1/9.
y(-4)=1/(3√-1)=1/(3i)=-⅓i (complex number), where i is the imaginary square root of -1.