Answers

2014-08-07T21:16:59+08:00
Let \ y \ be \ a \ quadratic \ function \ with \ standard \ form : \\\\ y=a(x-h)^2 + k \\\\vertex(h, k) \\ The \ maximum \ or \ minimum \ value \ of \ y \ occurs \ at \ x=h\\\\If \ a>0 , \ then \ the \ minimum \ value \ of \ y \ is \ y(h)=k \\If \ a< 0 , \ then \ the \ maximu \ value \ of \ y \ is \ y(h)=k

y=x^2-6x+29 =\\\\=(x^2-6x +3^2-3^2)+29=\\\\=(x^2-6x +3^2)-3^2+29=\\\\=(x^2-6x + 9)-9+29=\\\\=(x-3)^2+20

The \ standard \ form \ is: \\\\ y=(x-3)^2+20

The graph is a parabola that has its vertex at (h,k)= (3,20) and a> 0 opens  upward

Since \ the \ coefficient \ of \ x^2 \ is \ positive, \ y \ has \ a \ minimum \ value \\\\ The \ minimum \ value \ is : \ \ y(3) =20



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