[tex]y=2(x+ \frac{5}{4})^2-\frac{49}{8}[/tex]
Vertex-form equation for a vertical parabola:
[tex]y = a(x - h)^2 + k \\vertex \ is \ (h, k) \\\\The \ vertex \ is : \ (-\frac{5}{4}, -\frac{49}{8} )[/tex]
a > 0, \ so \ the \ parabola \ opens \ upwards.
The minimum value of y is at the vertex, where [tex]y = -\frac{49}{8}[/tex]
Since the parabola opens upwards and the vertex is \ [tex](-\frac{5}{4} ,-\frac{49}{8}): \\it \ is \ decreasing \ when \ x < -\frac{5}{4} \\it \ is \ increasing \ when \ x > -\frac{5}{4}[/tex]
[tex]the \ roots \ of \ the \ parabola :\\ \\2(x+ \frac{5}{4})^2-\frac{49}{8}=0\\\\2(x^2+\frac{5}{2}x+\frac{25}{16})-\frac{49}{8}=0\\\\2 x^2+5x+\frac{25}{8} -\frac{49}{8}=0\\\\2 x^2+5x-\frac{24}{8} =0[/tex]
[tex]2 x^2+5x-3 =0 \\a=2, \ \ b=5 , \ \ c=-3 \\\\x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-5-\sqrt{5^2-4 \cdot2 \cdot (-3) }}{2 \cdot 2}=\frac{-5-\sqrt{25+24 }}{4}=\\\\=\frac{-5-\sqrt{49 }}{4}=\frac{-5-7}{4}=\frac{-12}{4}=-3\\\\x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{-5+7}{4}= \frac{-2}{4}=-\frac{1}{2}[/tex]
