Identify the excluded values of h(x)=x+4x2−16h(x) = \dfrac{x+4}{x^2-16}h(x)=x2−16x+4.
x=4x = 4x=4 and x=−4x = -4x=−4
x=−4x = -4x=−4 only
x=4x = 4x=4 only
x=16x = 16x=16