two points and an equation












0














So I think this is the question, I forget how its worded exactly, so bear with me. Suppose we have an equation $5x^3 + bx^2 + cx + d$ cirve that passes through the points $(0,0)$ and $(2,0)$ and the question asks me to solve for $b$ in the equation. How do I do it?






graphing-functions problem-solving







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  • 1




    Welcome to Maths SX! You only can have a relation between $b$ and $c$.
    – Bernard
    Jan 4 at 21:11






  • 1




    Did you try putting both points into the equation and seeing what you get? (I suppose it's an equation $y=...$, yours isn't an equation, it doesn't equate anything)
    – orion
    Jan 4 at 21:13
















0














So I think this is the question, I forget how its worded exactly, so bear with me. Suppose we have an equation $5x^3 + bx^2 + cx + d$ cirve that passes through the points $(0,0)$ and $(2,0)$ and the question asks me to solve for $b$ in the equation. How do I do it?






graphing-functions problem-solving







share|cite|improve this question









New contributor




Ryan smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1




    Welcome to Maths SX! You only can have a relation between $b$ and $c$.
    – Bernard
    Jan 4 at 21:11






  • 1




    Did you try putting both points into the equation and seeing what you get? (I suppose it's an equation $y=...$, yours isn't an equation, it doesn't equate anything)
    – orion
    Jan 4 at 21:13














0












0








0







So I think this is the question, I forget how its worded exactly, so bear with me. Suppose we have an equation $5x^3 + bx^2 + cx + d$ cirve that passes through the points $(0,0)$ and $(2,0)$ and the question asks me to solve for $b$ in the equation. How do I do it?






graphing-functions problem-solving







share|cite|improve this question









New contributor




Ryan smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











So I think this is the question, I forget how its worded exactly, so bear with me. Suppose we have an equation $5x^3 + bx^2 + cx + d$ cirve that passes through the points $(0,0)$ and $(2,0)$ and the question asks me to solve for $b$ in the equation. How do I do it?






graphing-functions problem-solving




graphing-functions problem-solving






share|cite|improve this question









New contributor




Ryan smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Ryan smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 21:21









amWhy

192k28225439




192k28225439






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asked Jan 4 at 21:08









Ryan smithRyan smith

4




4




New contributor




Ryan smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





Ryan smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Ryan smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    Welcome to Maths SX! You only can have a relation between $b$ and $c$.
    – Bernard
    Jan 4 at 21:11






  • 1




    Did you try putting both points into the equation and seeing what you get? (I suppose it's an equation $y=...$, yours isn't an equation, it doesn't equate anything)
    – orion
    Jan 4 at 21:13














  • 1




    Welcome to Maths SX! You only can have a relation between $b$ and $c$.
    – Bernard
    Jan 4 at 21:11






  • 1




    Did you try putting both points into the equation and seeing what you get? (I suppose it's an equation $y=...$, yours isn't an equation, it doesn't equate anything)
    – orion
    Jan 4 at 21:13








1




1




Welcome to Maths SX! You only can have a relation between $b$ and $c$.
– Bernard
Jan 4 at 21:11




Welcome to Maths SX! You only can have a relation between $b$ and $c$.
– Bernard
Jan 4 at 21:11




1




1




Did you try putting both points into the equation and seeing what you get? (I suppose it's an equation $y=...$, yours isn't an equation, it doesn't equate anything)
– orion
Jan 4 at 21:13




Did you try putting both points into the equation and seeing what you get? (I suppose it's an equation $y=...$, yours isn't an equation, it doesn't equate anything)
– orion
Jan 4 at 21:13










2 Answers
2






active

oldest

votes


















1














You don't have enough information for a unique solution



What you can do is:



$y = 5x^3 + bx^2+ cx + d$



Plug 0 in for x and 0 for y.



$d = 0$



Then plug 2 in for x and 0 for y



$40 + 4b + 2c = 0$



$b = -10 - frac 12 c$



And that is as far as you can go.






share|cite|improve this answer





























    1














    For the point $(0, 0)$, you get



    $$0 = 5(0)^3+b(0)^2+c(0)+d iff d = 0$$



    So the equation is essentially $f(x) = 5x^3+bx^2+cx$.



    For the point $(2, 0)$, you get



    $$0 = 5(2)^3+b(2)^2+c(2) iff 0 = 40+4b+2c iff -40 = 4b+2c iff -20 = 2b+c$$



    There isn’t enough information to solve for $b$. However, you can come up with a relation between $b$ and $c$, as shown.






    share|cite|improve this answer





















      Your Answer





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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1














      You don't have enough information for a unique solution



      What you can do is:



      $y = 5x^3 + bx^2+ cx + d$



      Plug 0 in for x and 0 for y.



      $d = 0$



      Then plug 2 in for x and 0 for y



      $40 + 4b + 2c = 0$



      $b = -10 - frac 12 c$



      And that is as far as you can go.






      share|cite|improve this answer


























        1














        You don't have enough information for a unique solution



        What you can do is:



        $y = 5x^3 + bx^2+ cx + d$



        Plug 0 in for x and 0 for y.



        $d = 0$



        Then plug 2 in for x and 0 for y



        $40 + 4b + 2c = 0$



        $b = -10 - frac 12 c$



        And that is as far as you can go.






        share|cite|improve this answer
























          1












          1








          1






          You don't have enough information for a unique solution



          What you can do is:



          $y = 5x^3 + bx^2+ cx + d$



          Plug 0 in for x and 0 for y.



          $d = 0$



          Then plug 2 in for x and 0 for y



          $40 + 4b + 2c = 0$



          $b = -10 - frac 12 c$



          And that is as far as you can go.






          share|cite|improve this answer












          You don't have enough information for a unique solution



          What you can do is:



          $y = 5x^3 + bx^2+ cx + d$



          Plug 0 in for x and 0 for y.



          $d = 0$



          Then plug 2 in for x and 0 for y



          $40 + 4b + 2c = 0$



          $b = -10 - frac 12 c$



          And that is as far as you can go.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 21:15









          Doug MDoug M

          44.2k31854




          44.2k31854























              1














              For the point $(0, 0)$, you get



              $$0 = 5(0)^3+b(0)^2+c(0)+d iff d = 0$$



              So the equation is essentially $f(x) = 5x^3+bx^2+cx$.



              For the point $(2, 0)$, you get



              $$0 = 5(2)^3+b(2)^2+c(2) iff 0 = 40+4b+2c iff -40 = 4b+2c iff -20 = 2b+c$$



              There isn’t enough information to solve for $b$. However, you can come up with a relation between $b$ and $c$, as shown.






              share|cite|improve this answer


























                1














                For the point $(0, 0)$, you get



                $$0 = 5(0)^3+b(0)^2+c(0)+d iff d = 0$$



                So the equation is essentially $f(x) = 5x^3+bx^2+cx$.



                For the point $(2, 0)$, you get



                $$0 = 5(2)^3+b(2)^2+c(2) iff 0 = 40+4b+2c iff -40 = 4b+2c iff -20 = 2b+c$$



                There isn’t enough information to solve for $b$. However, you can come up with a relation between $b$ and $c$, as shown.






                share|cite|improve this answer
























                  1












                  1








                  1






                  For the point $(0, 0)$, you get



                  $$0 = 5(0)^3+b(0)^2+c(0)+d iff d = 0$$



                  So the equation is essentially $f(x) = 5x^3+bx^2+cx$.



                  For the point $(2, 0)$, you get



                  $$0 = 5(2)^3+b(2)^2+c(2) iff 0 = 40+4b+2c iff -40 = 4b+2c iff -20 = 2b+c$$



                  There isn’t enough information to solve for $b$. However, you can come up with a relation between $b$ and $c$, as shown.






                  share|cite|improve this answer












                  For the point $(0, 0)$, you get



                  $$0 = 5(0)^3+b(0)^2+c(0)+d iff d = 0$$



                  So the equation is essentially $f(x) = 5x^3+bx^2+cx$.



                  For the point $(2, 0)$, you get



                  $$0 = 5(2)^3+b(2)^2+c(2) iff 0 = 40+4b+2c iff -40 = 4b+2c iff -20 = 2b+c$$



                  There isn’t enough information to solve for $b$. However, you can come up with a relation between $b$ and $c$, as shown.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 4 at 21:16









                  KM101KM101

                  5,8611423




                  5,8611423






















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