Is $f(x)= x^2-13\sqrt{x}+30$ a quadratic?

Question

Is this function a quadratic? $$f(x)= x^2-13\sqrt{x}+30$$

I understand that only looking at $x^2$ would mean that it is quadratic, however, I am really confused about the $\sqrt{x}$ part. I tried plotting it on Desmos and it does not show up.

How would it not be considered as a quadratic function, if not, why it would be considered a quadratic function?

Answer 1

In general, the term "quadratic" function is reserved for functions that look like $f(x)=ax^2+bx+c$ where $a\neq 0$ and $a,b,c$ are constants. Sometimes we have things that don't look a quadratic at first but can be re-cast as a quadratic in another variable. For instance, $$x^4+2x^2+7$$ can be re-written as a quadratic where the variable is $x^2$ :$$(x^2)^2+2(x^2)+7$$.

So in general, a quadratic is something that the variable shows up at most twice in, and if it shows up twice, one time it is the square of the second. So for yours, no it is not.

Answer 2

Quadratics are polynomials of $2^{nd}$ degree.

Polynomials cannot contain terms with fractional powers, such as $\,x^{1/2}=\sqrt{x}\,$.

Therefore $\,x^2-13\sqrt{x}+30\,$ is not a polynomial, and consequently not a quadratic.

Answer 3

The other answers are somewhat incorrect in their justification, in my view.

The fact that $x^2 - 13\sqrt{x} + 30$ doesn't "look" like a quadratic isn't sufficient to conclude that it isn't a quadratic. For instance, $(e^x+1)x^2 - e^xx^2 - 13 \log x + 13 \log x + 15$ doesn't "look" quadratic either, but if we simplify it, it is.

Proving that $x^2 - 13\sqrt{x} + 30$ is not a quadratic polynomial requires a proof. In this case, the proof is quite simple. The third derivative of a quadratic is identically zero. This is not the case here.