Is $y=|x^3| $a parabola? I'm just curious. It seems to have the same shape and a similar form as parabolas such as $ x^2$ and $x^4 $. The odd exponent would normally give negative outputs for negative inputs, but the absolute value function changes that. So would $|x^3|$ technically be considered a parabola?
 A: No, just as we wouldn't call $x^4$'s graph a parabola, although the shape is very similar.
A parabola as a purely geometrical shape, with a specific definition and/or construction and certain properties, is not necessarily the graph of a function: you could e.g. rotate $y=x^2$ in the plane: it would still be a parabola. However, if you're considering functions of the form $y=f(x)$, then a parabola is the graph of any second degree polynomial: $y=ax^2+bx+c$.
Written in a different form, sometimes called "standard form", it is easier to immediately deduce some of its properties: the graph of $y=a(x-p)^2+q$ is a parabola with vertex/top in $(p,q)$ and it opens upwards for $a>0$ and downwards for $a<0$; $|a|$ determines its "width".
A: So as StackID just explained in his comprehensive answer, no $x\to |x^3|$ is not a parabola, and neither is $x\to x^4$, because in terms of polynomials, only second degree polynomials can qualify as a parabola.
But if what you want to know is actually whether $x\to |x^3|$ is still a polynomial, that is to say does there exist a polynomial $P\in \textbf{F}[x]$ such that $\forall x\in\mathbb{R},P(x)=|x^3|$, this question is a bit more interesting.
Indeed, it does look like a polynomial, doesn't it ?
But here again, the answer is no:  let's assume there is such a polynomial $P$. Since $\forall x\geq0,|x^3|=x^3$, we have that $$\forall x\geq0,P(x)=x^3\text{ i.e }P(x)-x^3=0$$
So the polynomial $P(x)-x^3$ has infinitely many roots ; and since a polynomial can have at most as many roots as its degree, we conclude that this can only be the null polynomial (on all $\mathbb{R}$) : $\forall x\in\mathbb{R}, P(x)-x^3=0\text{ }\text{ i.e }\text{ }P(x)=x^3$.
But this is impossible, since $\forall x\in\mathbb{R}, P(x)=|x^3|$, and for $x<0, |x^3|=-x^3$, and we cannot have $\forall x<0,P(x)=x^3=-x^3$, since $\forall x<0, x^3\neq-x^3$.
So by contradiction, such a polynomial $P$ cannot exist, and thus $x\to|x^3|$ isn't even a polynomial.
