$y=x^{\frac{2}{3}}$ why the function of plot in wolframlpha is like this?

Question

why the plot looks like this? when x < 0, y should be bigger than 0 ? imaginary part is up of 0 meanwhile real part is in the below.

wolfram|alpha

The right plot I think it looks like the one below from mathway.

so,what's wrong with my plot?

Answer 1

Wolfram products use the notation that $$(-1)^{\frac{n}{m}}$$ is a the $n^{th}$ numbered root of the equation $x^m=-1$.

E.g.

$$(-1)^{\frac{0}{3}}=(-1)^0=1$$

$$(-1)^{\frac{1}{3}}=\frac{1}{2}+\frac{\sqrt{3}i}{2}$$

$$(-1)^{\frac{2}{3}}=-\frac{1}{2}+\frac{\sqrt{3}i}{2}$$

$$(-1)^{\frac{3}{3}}=(-1)^1=-1$$

$$(-1)^{\frac{4}{3}}=-\frac{1}{2}-\frac{\sqrt{3}i}{2}$$

$$(-1)^{\frac{5}{3}}=\frac{1}{2}-\frac{\sqrt{3}i}{2}$$

It then treats evaluating fractional powers of negative numbers as the product of a positive number and minus one. So any negative value on your graph will have a real and imaginary component.

The pathway graph appears to be doing $$(x^2)^{\frac{1}{3}}$$

If you put this into Wolfram then you'll get the same graph.