Given J(x,-8) and K(-1,-5) and the graph of line l, find the value of x so that line JK is parallel to l
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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,-∞)
