quadratic function positive

Question

Put constraints on a quadratic function. I know that for $x > 0$ then $ax^2 + bx + c > 0$

I read around but I just found positive for all $x$. Thanks a lot

Answer 1

If $\;a\neq0\;$ and that is then an actual quadratic, then it describes an upwards parabola if $\;a>0\;$ and a downwards one if $\;a<0\;$.

So you want $\;a>0\;$ (first constraint), and then one way to succeed here is to make the parabola all the time positive, and that means it has no roots (points where it vanishes), and this is the same as requiring its discriminant to be negative:

$$\Delta=b^2-4ac$$

and then you succeed...and, in fact, $\;ax^2=bx+c>0\;$ for all $\;x\in\Bbb R\;$ , not only for $\;x>0\;$ .

Answer 2

You want $ax^2+bx+c>0$ for $x>0$.

This require $a>0$ as a first condition ( the parabola must be upward)

Then we have three possibilities:

1) the parabola is always positive ( for all $x \in \mathbb{R}$) if $b^2-4ac<0$ .

2)The parabola has value $0=$ for $x=0$ and is positive for all other values of $x$. This means that $b=c=0$.

3) The parabola has two roots, one for a value $x<0$ and the other for $x=0$: if $c=0$ and $b>0$.

Answer 3

Let $f(x)=ax^2+bx+c$

When $x\to \infty$, $\displaystyle \frac{f(x)}{x^2} \to a \implies a>0$

When $x\to 0^{+}$, $f(x)\to c \implies c \geq 0$

Now

$$\left \{ \begin{align*} a &>0 \\ c &\geq 0 \end{align*} \right.$$

Case I: $\Delta=b^2-4ac \geq 0$, the largest root should be non-positive.

$$\frac{-b+\sqrt{b^2-4ac}}{2a} \leq 0$$

$$0 \leq \sqrt{b^2-4ac} \leq b$$

$$\implies b \geq 2\sqrt{ac}$$

Case II: $\Delta=b^2-4ac<0$, $f(x)>0 \; \forall x$

$$b^2 < 4ac$$

$$\implies -2\sqrt{ac} < b < 2\sqrt{ac}$$

where $ac>0$

Combining all together:

\begin{align*} a &>0 \\ c &\geq 0 \\ b & \quad \left \{ \begin{array}{rcll} \geq & 0 & , & c =0 \\ > & -2\sqrt{ac} & , & c>0 \end{array} \right. \end{align*}