How to figure out the formula of axis of simmetry in quadratic functionsWell, this is what I did. Assuming that that quadratic equation or function is given by the general formula f ( x ) = a x 2 + b x + c.So, assuming that there exist an axis of simmetry (eje de simetria en español), I had to find two points ( x 1 , f ( x 1 ) ) and ( x 2 , f ( x 2 ) ) so that x 1 ≠ x 2 but f ( x 1 ) = f ( x 2 ). To make the typing job a bit easier x 1 = x and x 2 = d.So f ( x ) = a x 2 + b x + c and f ( d ) = a d 2 + b d + c and f ( x ) = f ( d ): a x 2 + b x + c = a d 2 + b d + c a x 2 + b x = a d 2 + b d a x 2 − a d 2 = b d − b x a ( x 2 − d 2 ) = b ( d − x ) a ( x + d ) ( x − d ) = − ( x − d ) b a ( x + d ) = − bBecause we are talking about quadratic functions b can be 0. So if we b = 0, we have: a ( x + d ) = 0and again, if it is a quadractic function a can´t be 0, so we´d have that: x + d = 0 and x = − d, which is true. For example in the function f ( x ) = x 2 − 16, f ( 4 ) = f ( − 4 ) = 0.Well, continuing with the calculations we´d have: x + d = − b a d = − b a − x d = − b − x a a Because it´s an axis of simmetry, it´s in the middle, so I´ll have to use this formula: x 1 + x 2 2 So: x 1 + x 2 = x + d and d = − b − x a a , so I would have. x + d = x + ( − b − x a a ) = x a a + ( − b − x a a ) = x a − b − x a a x + d = − b a So: x + d 2 = − b a ÷ 2 = − b 2 a That´s what I´ve done so far, but what I probably want to know is why there exists and axis of symmetry (which I think will be the line x = − b 2 a ). I thought I could prove it using that fact but I got nowhere that´s why I posted this forum. (there might be some grammar mistakes, that´s because I am not an english speaker, you can correct where you think I made a mistake.

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How to figure out the formula of axis of simmetry in quadratic functionsWell, this is what I did. Assuming that that quadratic equation or function is given by the general formula   f  (  x  )  =  a      x    2    +  b  x  +  c.So, assuming that there exist an axis of simmetry (eje de simetria en español), I had to find two points   (      x    1    ,  f  (      x    1    )  ) and   (      x    2    ,  f  (      x    2    )  ) so that       x    1    ≠      x    2   but   f  (      x    1    )  =  f  (      x    2    ). To make the typing job a bit easier       x    1    =  x and       x    2    =  d.So   f  (  x  )  =  a      x    2    +  b  x  +  c and   f  (  d  )  =  a      d    2    +  b  d  +  c and   f  (  x  )  =  f  (  d  ):  a      x    2    +  b  x  +  c  =  a      d    2    +  b  d  +  c  a      x    2    +  b  x  =  a      d    2    +  b  d  a      x    2    −  a      d    2    =  b  d  −  b  x  a  (      x    2    −      d    2    )  =  b  (  d  −  x  )  a  (  x  +  d  )  (  x  −  d  )  =  −  (  x  −  d  )  b  a  (  x  +  d  )  =  −  bBecause we are talking about quadratic functions b can be 0. So if we   b  =  0, we have:  a  (  x  +  d  )  =  0and again, if it is a quadractic function a can´t be 0, so we´d have that:  x  +  d  =  0 and   x  =  −  d, which is true. For example in the function   f  (  x  )  =      x    2    −  16,   f  (  4  )  =  f  (  −  4  )  =  0.Well, continuing with the calculations we´d have:  x  +  d  =            −      b        a    d  =            −      b        a    −  x  d  =            −      b      −      x      a        a  Because it´s an axis of simmetry, it´s in the middle, so I´ll have to use this formula:                    x        1            +              x        2              2  So:       x    1    +      x    2    =  x  +  d and   d  =            −      b      −      x      a        a  , so I would have.  x  +  d  =  x  +  (            −      b      −      x      a        a    )  =            x      a        a    +  (            −      b      −      x      a        a    )  =            x      a      −      b      −      x      a        a    x  +  d  =            −      b        a  So:            x      +      d        2    =            −      b        a    ÷  2  =            −      b              2      a      That´s what I´ve done so far, but what I probably want to know is why there exists and axis of symmetry (which I think will be the line   x  =            −      b              2      a      ). I thought I could prove it using that fact but I got nowhere that´s why I posted this forum. (there might be some grammar mistakes, that´s because I am not an english speaker, you can correct where you think I made a mistake.

AtticaPlotowvi · Accepted Answer

Step 1Let us talk about the two improvements that can be made.1) If you want to prove a fact is true in general, there is no need to discuss the particular case of “Because we are talking about quadratic functions b can be 0. So if we (let)   b  =  0, …..”.2) From “ continuing with the calculations we´d have:   x  +  d  =                    −        b            a      ”, we can arrive at the conclusion of                     x        +        d            2        =            −      b              2      a       immediately by dividing both sides by 2.Step 2Regarding your query of “why there exists and axis of symmetry”, I try to explain it in the following way.Firstly, it is a provable fact that   y  =      x    2   is symmetric (and therefore the said axis exists). [The proof is more or less similar to what you did in the first part of your post.]Next, by completing square,   f  (  x  )  =  a      x    2    +  b  x  +  c can be re-written as   f  (  x  )  =  a  (  x  –  h      )    2    +  k; where h and k are functions of a, b and c.Note that (i) The “a” controls the opening (upward or downward) of the curve and the degree (how wide) of opening of the quadratic curve; (ii) The “h” and “k” are respectively horizontal and vertical shifts of the function. All these are under the topic of transformations.In short,   y  =  f  (  x  ) can be transformed equivalently to   y  =      x    2  .Added:- In other words, any quadratic function is just the transformed image of the basic   y  =      x    2   through (series of translations, reflections, amplifications...).

How to figure out the formula of axis of simmetry in quadratic functions. Well, this is what I did. Assuming that that quadratic equation or function is given by the general formula f(x)=ax^2+bx+c.

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