A complete Nyquist diagram can tell if the closed loop system is stable. Namely when going from negative infinite imaginary frequency to positive infinite imaginary frequency and back again to negative infinite imaginary frequency over the D contour you encircle the entire right half plane on the input.
When drawing the Nyquist diagram of a system, then each unstable zero and pole will cause a clockwise- and anticlockwise-encirclement of the origin respectively.
If you write the open loop as a quotient of two polynomials
$$
L(s) = C(s) H(s) = \frac{N(s)}{D(s)},
$$
then the closed loop becomes
$$
\frac{L(s)}{1 + L(s)} = \frac{N(s)}{D(s) + N(s)}.
$$
If you draw the Nyquist diagram of $L(s)$ and look at the encirclements of the minus one point you are basically looking at the unstable zeros and poles of
$$
1 + L(s) = \frac{D(s) + N(s)}{D(s)}.
$$
The number of clockwise encirclements of the minus one point will be equal to the number of unstable roots of $D(s)+N(s)$ minus the number of unstable roots of $D(s)$. The first are also the number of unstable poles of the closed loop system and the second are the number of unstable poles of the open loop system.
For every physical system the part of Nyquist diagram of the negative frequencies is a mirror image of the positive frequencies. So it is not required to draw them both, but it can help identify the total number of encirclements.
PS: Also be aware of encirclements you get infinity. For example $1/s^3$ does have one clockwise encirclement of the minus one point.