How Do You Find the Vertex and Axis of Symmetry in a Parabola Equation?

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To find the vertex of the parabola given by the equation (y-2)² = 4(x-3), identify the vertex at the point (3, 2). The axis of symmetry is the line x = 3, indicating that the parabola opens horizontally. The positive right-hand side of the equation confirms that the parabola opens to the right. The distance p, which is 1, represents the distance from the vertex to the focus. Understanding these elements allows for a clear analysis of the parabola's properties.
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If I am given (y-2)2= 4 (x-3), how would I find the vertex, equation of the axis of symmetry, and the direction of the opening? I'm guessing that I have to use PF=PD, but it's confusing because it looks different from other things that I have done involving PF=PD.

Thanks.
 
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the right hand side of your equation is positive right?, what is the minimum value of x for which this is posible? and the corresponding value of y? this is the vertex. Now rebember that p=1 is the distance from the vertex to the focus and that you have and horizontal parabola.
 

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