If u (x) = -2x^2 and v (x) = 1 over x what is the range of (uov) (x)

u (x) = - 2x², v (x) = 1/x

(u o v) (x) = u (v (x)) = - 2 (1/x) ² = - 2/x²

We can see that domain for x is going to bee all real numbers except 0. From the equation above we can see that graph of the function (u o v) (x) = - 2/x² has horizontal asymptote y=0, because degree of numerator is less than degree of denominator ((u o v) (x) = - 2x⁰/x²).

x² is always going to be positive, so range is going to be all negative numbers.

y<0, or y∈ (0,-∞)