[tex]x^2+9x+14>0 \\ \\first\ we \ solve: \\ x^2 + 9x+14 =0 \\ x^2 + 7x+2x+14 =0\\x(x + 7 )+2(x+7) =0 \\ (x+7) (x+2)=0 \\x+7=0 \ \ or \ \x+2=0\\x=-7\ \ or \ \ x=-2 \\\\ "a" \ is \ positive, \ so \ it \ is \ an \ "upwards" \ graph[/tex]
So we know that the curve is u-shaped and it crosses
the t-axis at x=-2 and x= - 7
The curve is more than zero : [tex]x \in (-\infty ,-7)\cup (-2,+\infty )[/tex]
