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,-∞)

false

step-by-step explanation:

4y - 6 + 4y^2 - 9

before simplifying this contains 4 terms

after simplifying

4y -15 + 4y^2

it contains 3 terms

attachment?

step-by-step explanation: