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wombatBest ResponseYou've already chosen the best response.0
Ok. 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
 2 years ago

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

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

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

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

wombatBest ResponseYou've already chosen the best response.0
That 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]))
 2 years ago

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

wombatBest ResponseYou've already chosen the best response.0
OH! 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
 2 years ago

jhonyy9Best 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
 2 years ago
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