Bhavana+V

Equation: f(x)= -x^2 Domanin: (- inifinity, + inifinity) Range: [0, + infinity) Symmetry: - (-x)^2 =-x^2 Yes, symmetric Even: Yes (show work)

End behavior (using limits)

f(x) = -x^2 x-> 0^+ = - infinity f(x)= -x^2 x-> 0^- = - infinity f(x)= - infinity x-> 0

f(x)= - infinity x-> infinity f(x)= - infinity x-> - infinity

Hence f(x) in this case, forever continues to - infinity, therefore it has NO discontinuities.