91Ó°ÊÓ

We have:

$x^2-x-6>0$

$\mathrm{The} \mathrm{vertex} \mathrm{of} \mathrm{an} \mathrm{up}-\mathrm{down} \mathrm{facing} \mathrm{parabola} \mathrm{of} \mathrm{the} \mathrm{form} y=a{x}^2+bx+c \mathrm{is} x_v=-\frac{b}{2a}$

$$\begin{gathered} \mathrm{The} \mathrm{parabola} \mathrm{params} \mathrm{are}: \\ a=1, b=-1, c=-6 \\ {x}_v=-\frac{b}{2a} \\ {x}_v=-\frac{\left(-1\right)}{2· 1} \\ {x}_v=\frac{1}{2} \end{gathered}$$

Plug in $x_v=\frac{1}{2}$in the original equation to get role="math" localid="1645295400095" $y_v:$

$$\begin{gathered} y_v=-\frac{25}{4} \\ \mathrm{Therefore} \mathrm{the} \mathrm{parabola} \mathrm{vertex} \mathrm{is} \\ \left(\frac{1}{2}, -\frac{25}{4}\right) \end{gathered}$$

Substitute $y=f\left(x\right)=0$:

$$\begin{gathered} x^2-x-6=0 \\ \mathrm{For} \mathrm{a} \mathrm{quadratic} \mathrm{equation} \mathrm{of} \mathrm{the} \mathrm{form} a{x}^2+bx+c=0 \mathrm{the} \mathrm{solutions} \mathrm{are}  \\ {x}_{1, 2}=\frac{-bÂ\pm \sqrt{b^2-4ac}}{2a} \\ \mathrm{For} a=1, b=-1, c=-6 \\ {x}_{1, 2}=\frac{-\left(-1\right)Â\pm \sqrt{{\left(-1\right)}^2-4· 1·\left(-6\right)}}{2· 1} \\ {x}_{1, 2}=\frac{-\left(-1\right)Â\pm  5}{2· 1} \\ x=3, x=-2 \end{gathered}$$

$\mathrm{X} \mathrm{Intercepts}: \left(3, 0\right), \left(-2, 0\right)$

Plot the vertex of the function at $\left(\frac{1}{2}, -\frac{25}{4}\right)$with intercepts then draw the curve passes through the plotted points:

Since the solution of the function should be greater than $0,$then solution is$\left(-∞,-2\right)∪\left(3,∞\right)$.