Answers

2014-08-25T22:45:26+08:00
y=2(x+ \frac{5}{4})^2-\frac{49}{8}

Vertex-form equation for a vertical parabola:

y = a(x - h)^2 + k \\vertex \ is \ (h, k) \\\\The \ vertex \ is : \ (-\frac{5}{4}, -\frac{49}{8} )

a > 0, \ so \ the \ parabola \ opens \ upwards.
 
The minimum  value  of   y is  at the   vertex,   where  y = -\frac{49}{8}

Since  the  parabola  opens upwards and   the vertex   is  \ (-\frac{5}{4} ,-\frac{49}{8}): \\it \ is \ decreasing \ when \ x < -\frac{5}{4} \\it \ is \ increasing \ when \ x > -\frac{5}{4}



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

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}






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