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

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SUMMARY

The discussion focuses on finding the vertex and axis of symmetry for the parabola represented by the equation (y-2)² = 4(x-3). The vertex is determined to be at the point (3, 2), which is derived from the standard form of a parabola. The axis of symmetry is the vertical line x = 3. Additionally, the distance p, which is 1, indicates the distance from the vertex to the focus, confirming that this is a horizontal parabola opening to the right.

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  • Familiarity with the relationship between p, the vertex, and the focus
  • Basic algebra skills for manipulating equations
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Alain12345
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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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