If the numerator n of the function exponent is an even number and the denominator m is an arbitrary integer,

Then y>0, image first; Quadrant. At this time, the parity of the exponent p of (-1) p is irrelevant.

For example: y = x (2/3);

y=x^(-2/3)(x<； & gt0);

y=x^(2/4),y=x^(-2/4)(x<； & gt0).

If the denominator m of the exponent of the function is an even number and the numerator n is an arbitrary integer, then x >；; 0 (or x & gt=0); Y>0 (or y & gt=0), the image is in the first quadrant. Has little to do with the parity of p,

For example: y = x (1/2) (x >; =0); y=x^(-3/4)(x>； 0).

When both m and n are odd numbers, the image must pass through the third quadrant. For example, y = x * (1/3); y=x^(-3).

So when n is even or m is even, the image does not pass through the third quadrant, which has nothing to do with the parity of p.

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