Vertex Equation of an inverse quadratic function.I'm working on a graphing web tool using JSXGraph, The user should be able to draw different functions. I was able to allow the user to draw quadratic functions by creating the vertex of the function where the user clicks, and then creating another point up-right of the the vertex. Then I construct the function by obtaining a through the vertex equation y = a ( x − h ) 2 + kI isolate a, which gives a = y − k x 2 − 2 x h + h 2 As I have the vertex and another point, I obtain a, and then isolating y f ( x ) = a x 2 − 2 a h x + a h 2 + kwith that, I obtain b and then c.The problem is I also need to allow the user to draw inverted parabolas, however I can't find a way to obtain the function given the vertex and a point. I believe i would be able to, with the vertex equation of the inverse quadratic, following a similar procedure as the one above, but I'm not even sure if that exists given the dual nature of the inverse quadratic.So is there a vertex equation for the inverse of a quadratic function that would allow me to accomplish what I did above?

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Vertex Equation of an inverse quadratic function.I&#039;m working on a graphing web tool using JSXGraph, The user should be able to draw different functions. I was able to allow the user to draw quadratic functions by creating the vertex of the function where the user clicks, and then creating another point up-right of the the vertex. Then I construct the function by obtaining a through the vertex equation  y  =  a  (  x  −  h      )    2    +  kI isolate a, which gives  a  =            y      −      k                      x        2            −      2      x      h      +              h        2            As I have the vertex and another point, I obtain a, and then isolating y  f  (  x  )  =  a      x    2    −  2  a  h  x  +  a      h    2    +  kwith that, I obtain b and then c.The problem is I also need to allow the user to draw inverted parabolas, however I can&#039;t find a way to obtain the function given the vertex and a point. I believe i would be able to, with the vertex equation of the inverse quadratic, following a similar procedure as the one above, but I&#039;m not even sure if that exists given the dual nature of the inverse quadratic.So is there a vertex equation for the inverse of a quadratic function that would allow me to accomplish what I did above?

h2a2l1i2morz · Accepted Answer

Step 1As you explained above, a parabola can be uniquely defined by its vertex   V  =  (      v    x    ,      v    y    ) and one more point   P  =  (      p    x    ,      p    y    ). The function term of the parabola then has the form  y  =  a  (  x  −      v    x        )    2    +      v    y    .Then, a can be determined by solving      p    y    =  a  (      p    x    −      v    x        )    2    +      v    y  for a which gives  a  =  (      p    y    −      v    y    )      /    (      p    x    −      v    x        )    2    .Step 2Conversely, also the inverse quadratic function can be uniquely defined by its vertex   V  =  (      v    x    ,      v    y    ) and one more point   P  =  (      p    x    ,      p    y    ). The function term of the inverse function has the form  y  =                    x        −                  v          x                    a        +      v    y    .Again, a can be determined by solving      p    y    =                              p          x                −                  v          x                    a        +      v    y  for a which gives  a  =  (      p    x    −      v    x    )      /    (      p    y    −      v    y        )    2    .

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