Does this set of coordinates result in a curve?












0












$begingroup$


Coordinates: $(0,0), (3,3), (6,4.5), (9, 5.25)$



If this is a curve is there a formula for determining the $y$ value for any given $x$ within the range $0$ to $9?$






algebra-precalculus







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$endgroup$








  • 1




    $begingroup$
    What type of curve are you talking about ?
    $endgroup$
    – Vivek Kaushik
    Jul 19 '16 at 3:29






  • 1




    $begingroup$
    Sadly there are many curves that pass through these four points, so it requires more information such as "cubic polynomial" to make a unique determination.
    $endgroup$
    – hardmath
    Jul 19 '16 at 3:39
















0












$begingroup$


Coordinates: $(0,0), (3,3), (6,4.5), (9, 5.25)$



If this is a curve is there a formula for determining the $y$ value for any given $x$ within the range $0$ to $9?$






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What type of curve are you talking about ?
    $endgroup$
    – Vivek Kaushik
    Jul 19 '16 at 3:29






  • 1




    $begingroup$
    Sadly there are many curves that pass through these four points, so it requires more information such as "cubic polynomial" to make a unique determination.
    $endgroup$
    – hardmath
    Jul 19 '16 at 3:39














0












0








0





$begingroup$


Coordinates: $(0,0), (3,3), (6,4.5), (9, 5.25)$



If this is a curve is there a formula for determining the $y$ value for any given $x$ within the range $0$ to $9?$






algebra-precalculus







share|cite|improve this question











$endgroup$




Coordinates: $(0,0), (3,3), (6,4.5), (9, 5.25)$



If this is a curve is there a formula for determining the $y$ value for any given $x$ within the range $0$ to $9?$






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 15:37









user376343

3,9584829




3,9584829










asked Jul 19 '16 at 3:27









TomTom

1




1








  • 1




    $begingroup$
    What type of curve are you talking about ?
    $endgroup$
    – Vivek Kaushik
    Jul 19 '16 at 3:29






  • 1




    $begingroup$
    Sadly there are many curves that pass through these four points, so it requires more information such as "cubic polynomial" to make a unique determination.
    $endgroup$
    – hardmath
    Jul 19 '16 at 3:39














  • 1




    $begingroup$
    What type of curve are you talking about ?
    $endgroup$
    – Vivek Kaushik
    Jul 19 '16 at 3:29






  • 1




    $begingroup$
    Sadly there are many curves that pass through these four points, so it requires more information such as "cubic polynomial" to make a unique determination.
    $endgroup$
    – hardmath
    Jul 19 '16 at 3:39








1




1




$begingroup$
What type of curve are you talking about ?
$endgroup$
– Vivek Kaushik
Jul 19 '16 at 3:29




$begingroup$
What type of curve are you talking about ?
$endgroup$
– Vivek Kaushik
Jul 19 '16 at 3:29




1




1




$begingroup$
Sadly there are many curves that pass through these four points, so it requires more information such as "cubic polynomial" to make a unique determination.
$endgroup$
– hardmath
Jul 19 '16 at 3:39




$begingroup$
Sadly there are many curves that pass through these four points, so it requires more information such as "cubic polynomial" to make a unique determination.
$endgroup$
– hardmath
Jul 19 '16 at 3:39










2 Answers
2






active

oldest

votes


















0












$begingroup$

I recommend looking at Lagrange's Interpolation Formula. Given any $n$ points, they can always be interpolated by a polynomial of degree $n-1$ or less. This means that those points will always lie on the curve of that polynomial. So given your 4 points, there is a polynomial of degree 3 or less such that the points are on the curve of that polynomial. The explicit equation for the polynomial can be found using the formula included in the link.



However, these four points most definitely do not determine uniquely a continuous function (a curve). There are very many curves that could pass through these points.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    I agree with Christian. This Lagrange interpolation formula will give you the polynomial $$f(x)=dfrac{4}{3}x-dfrac{1}{8}x^2+dfrac{1}{216}x^3$$
    $endgroup$
    – Jean Marie
    Jul 19 '16 at 7:03



















0












$begingroup$

This particular curve can be also $f(x)=6(1-2^{-x/3})$. Here is how I guessed this function:




  1. you go in steps of 3, so you might have something $x/3$.


  2. At the first step the function increases by 3, then at the next step by 3/2, then 3/4. From here, I would guess that the formula is $$sum_s3left(frac{1}{2}right)^s,$$ where $s=x/3$


  3. $sum_s(1/2)^s=frac{1-(1/2)^{s+1}}{1-1/2}$







share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    I recommend looking at Lagrange's Interpolation Formula. Given any $n$ points, they can always be interpolated by a polynomial of degree $n-1$ or less. This means that those points will always lie on the curve of that polynomial. So given your 4 points, there is a polynomial of degree 3 or less such that the points are on the curve of that polynomial. The explicit equation for the polynomial can be found using the formula included in the link.



    However, these four points most definitely do not determine uniquely a continuous function (a curve). There are very many curves that could pass through these points.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      I agree with Christian. This Lagrange interpolation formula will give you the polynomial $$f(x)=dfrac{4}{3}x-dfrac{1}{8}x^2+dfrac{1}{216}x^3$$
      $endgroup$
      – Jean Marie
      Jul 19 '16 at 7:03
















    0












    $begingroup$

    I recommend looking at Lagrange's Interpolation Formula. Given any $n$ points, they can always be interpolated by a polynomial of degree $n-1$ or less. This means that those points will always lie on the curve of that polynomial. So given your 4 points, there is a polynomial of degree 3 or less such that the points are on the curve of that polynomial. The explicit equation for the polynomial can be found using the formula included in the link.



    However, these four points most definitely do not determine uniquely a continuous function (a curve). There are very many curves that could pass through these points.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      I agree with Christian. This Lagrange interpolation formula will give you the polynomial $$f(x)=dfrac{4}{3}x-dfrac{1}{8}x^2+dfrac{1}{216}x^3$$
      $endgroup$
      – Jean Marie
      Jul 19 '16 at 7:03














    0












    0








    0





    $begingroup$

    I recommend looking at Lagrange's Interpolation Formula. Given any $n$ points, they can always be interpolated by a polynomial of degree $n-1$ or less. This means that those points will always lie on the curve of that polynomial. So given your 4 points, there is a polynomial of degree 3 or less such that the points are on the curve of that polynomial. The explicit equation for the polynomial can be found using the formula included in the link.



    However, these four points most definitely do not determine uniquely a continuous function (a curve). There are very many curves that could pass through these points.






    share|cite|improve this answer









    $endgroup$



    I recommend looking at Lagrange's Interpolation Formula. Given any $n$ points, they can always be interpolated by a polynomial of degree $n-1$ or less. This means that those points will always lie on the curve of that polynomial. So given your 4 points, there is a polynomial of degree 3 or less such that the points are on the curve of that polynomial. The explicit equation for the polynomial can be found using the formula included in the link.



    However, these four points most definitely do not determine uniquely a continuous function (a curve). There are very many curves that could pass through these points.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jul 19 '16 at 3:41









    ChristianChristian

    2,1111522




    2,1111522








    • 1




      $begingroup$
      I agree with Christian. This Lagrange interpolation formula will give you the polynomial $$f(x)=dfrac{4}{3}x-dfrac{1}{8}x^2+dfrac{1}{216}x^3$$
      $endgroup$
      – Jean Marie
      Jul 19 '16 at 7:03














    • 1




      $begingroup$
      I agree with Christian. This Lagrange interpolation formula will give you the polynomial $$f(x)=dfrac{4}{3}x-dfrac{1}{8}x^2+dfrac{1}{216}x^3$$
      $endgroup$
      – Jean Marie
      Jul 19 '16 at 7:03








    1




    1




    $begingroup$
    I agree with Christian. This Lagrange interpolation formula will give you the polynomial $$f(x)=dfrac{4}{3}x-dfrac{1}{8}x^2+dfrac{1}{216}x^3$$
    $endgroup$
    – Jean Marie
    Jul 19 '16 at 7:03




    $begingroup$
    I agree with Christian. This Lagrange interpolation formula will give you the polynomial $$f(x)=dfrac{4}{3}x-dfrac{1}{8}x^2+dfrac{1}{216}x^3$$
    $endgroup$
    – Jean Marie
    Jul 19 '16 at 7:03











    0












    $begingroup$

    This particular curve can be also $f(x)=6(1-2^{-x/3})$. Here is how I guessed this function:




    1. you go in steps of 3, so you might have something $x/3$.


    2. At the first step the function increases by 3, then at the next step by 3/2, then 3/4. From here, I would guess that the formula is $$sum_s3left(frac{1}{2}right)^s,$$ where $s=x/3$


    3. $sum_s(1/2)^s=frac{1-(1/2)^{s+1}}{1-1/2}$







    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      This particular curve can be also $f(x)=6(1-2^{-x/3})$. Here is how I guessed this function:




      1. you go in steps of 3, so you might have something $x/3$.


      2. At the first step the function increases by 3, then at the next step by 3/2, then 3/4. From here, I would guess that the formula is $$sum_s3left(frac{1}{2}right)^s,$$ where $s=x/3$


      3. $sum_s(1/2)^s=frac{1-(1/2)^{s+1}}{1-1/2}$







      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        This particular curve can be also $f(x)=6(1-2^{-x/3})$. Here is how I guessed this function:




        1. you go in steps of 3, so you might have something $x/3$.


        2. At the first step the function increases by 3, then at the next step by 3/2, then 3/4. From here, I would guess that the formula is $$sum_s3left(frac{1}{2}right)^s,$$ where $s=x/3$


        3. $sum_s(1/2)^s=frac{1-(1/2)^{s+1}}{1-1/2}$







        share|cite|improve this answer









        $endgroup$



        This particular curve can be also $f(x)=6(1-2^{-x/3})$. Here is how I guessed this function:




        1. you go in steps of 3, so you might have something $x/3$.


        2. At the first step the function increases by 3, then at the next step by 3/2, then 3/4. From here, I would guess that the formula is $$sum_s3left(frac{1}{2}right)^s,$$ where $s=x/3$


        3. $sum_s(1/2)^s=frac{1-(1/2)^{s+1}}{1-1/2}$








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jul 19 '16 at 5:14









        AndreiAndrei

        13.2k21230




        13.2k21230






























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