# Question #04999

Oct 15, 2016

$p > \frac{1}{2} \left(6 + \sqrt{36 + {f}^{2}}\right)$

#### Explanation:

If $y > 0 \forall x$ then $p > 0$ because $p$ is the coeficient of ${x}^{2}$ and $y \left(x\right)$ cannot have $x$ crossings so $y \left(0\right) = p - 6 > 0$

Anyway the $x$ crossings are given at

$x = \frac{- f \pm \sqrt{{f}^{2} + 24 p - 4 {p}^{2}}}{2 p}$ so if

${f}^{2} + 24 p - 4 {p}^{2} < 0$ no feasible crossings

or solving for $p$

$p < \frac{1}{2} \left(6 - \sqrt{{6}^{2} + {f}^{2}}\right)$ and $p > \frac{1}{2} \left(6 + \sqrt{{6}^{2} + {f}^{2}}\right)$

Concluding

$p > 6$ and $p > \frac{1}{2} \left(6 + \sqrt{{6}^{2} + {f}^{2}}\right)$ or finally

$p > \frac{1}{2} \left(6 + \sqrt{{6}^{2} + {f}^{2}}\right)$