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 parabola has a specific geometrical definition, not just a shape definition. – Andrew Li May 8 at 11:28
• A parabola should be a conic cut, which is not here the case. – dmtri May 8 at 11:34

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.

• Excellent answer! Thanks – FoitGuy May 8 at 11:55

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".

• The fact that the OP refered as $x^4$ as a parabola hints me that maybe he actually wants to know if $x\to |x^3|$ is still a polynomial, i.e if there is a polynomial that coincides with it for all x – Harmonic Sun May 8 at 11:33
• The first sentence in my answer was supposed to (implicitly) address that too, namely that $x^4$ isn't a parabola but I didn't consider the possibility that OP mistakenly used "parabola" and possibly meant a "polynomial function" - perhaps OP can clarify if this is indeed the case. – StackTD May 8 at 11:34
• I was under the impression that all even-ordered polynomials were considered parabolas. However, I would also be interested in the answer to this question. Obviously, you've answered the question that I asked: Only polynomials of order 2 are technically parabolas, so |x^3| is not technically a parabola. – FoitGuy May 8 at 11:53
• If you meant "polynomials", then no: $x^n$ is a polynomial for all $n$ but $|x^n|$ is only a polynomial for even $n$ (since obviously it then reduces back to $x^n$), not for odd $n$. – StackTD May 8 at 11:54