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wombat
 2 years ago
Best ResponseYou've already chosen the best response.0Ok. Well automatically, you can tell that it is a parabola (quadratic) right? Quadratics are more easily graphed when you have the equation in quadratic form: y=a(xh)^2 +k So you will want to convert the above equation to this form. They you can take the vertex and y intercept from the form

keenaH
 2 years ago
Best ResponseYou've already chosen the best response.0i forgot the step after finding the vertex and the formula you use to do it

wombat
 2 years ago
Best ResponseYou've already chosen the best response.0The vertex is the point (h,k)

keenaH
 2 years ago
Best ResponseYou've already chosen the best response.0im pretty sure the vertex is(6,6)

keenaH
 2 years ago
Best ResponseYou've already chosen the best response.0i used the formula ...x=(b)/2(a)

wombat
 2 years ago
Best ResponseYou've already chosen the best response.0That formula works too :) That will give you the x value that is the axis of symmetry. In the case you are doing, the vertex will be ( [b/2a], f([b/2a]))

keenaH
 2 years ago
Best ResponseYou've already chosen the best response.0i dont know what you just wrote, its like a foriegn language

wombat
 2 years ago
Best ResponseYou've already chosen the best response.0OH! sorry! um what I wrote represents the coordinate pair of the vertex. b/2a is the x value. f(b/2a) is the y value that coresponds with the above x value

jhonyy9
 2 years ago
Best ResponseYou've already chosen the best response.0 for you can solve it you need make it equal zero  so 1/3 x^2 4x +6 =0 multiply both sides by 3 3x^2  12x +18 =0 3 x^2 12x +18 =0 D =144  4*1*18 =144 72 =72 x_1,2= (12+/ sqrt72)/2 =(12+/ sqrt(9*4*2))/2 =(12+/ 6sqrt2)/2 = =2(6+/3sqrt2)/2 =6+/3sqrt2 x_1,2=6+/3sqrt2
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