Here's the question you clicked on:
keenaH
y=1/3x^2-4x+6
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(x-h)^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
i forgot the step after finding the vertex and the formula you use to do it
The vertex is the point (h,k)
im pretty sure the vertex is(6,-6)
i used the formula ...x=-(b)/2(a)
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]))
i dont know what you just wrote, its like a foriegn language
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
- 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