finding a tangent line to a parabola












1












$begingroup$


I am practicing for a math contest and I encountered the following problem that I don't know how to solve:



For which value of $b$ is there only one intersection between the line $y = x + b$ and the parabola $y = x^2 + 5x + 3$?



How do I solve it?






algebra-precalculus contest-math







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  • 1




    $begingroup$
    Do you know how to find a derivative?
    $endgroup$
    – R. Burton
    2 hours ago










  • $begingroup$
    Simply $x^2+5x+3=x+b$ must have only one solution, i.e. $x^2+4x+(3-b)=0$ should have zero discriminant.
    $endgroup$
    – the_fox
    2 hours ago










  • $begingroup$
    A Parabola could have many tangent lines...However, it has only 1 min. value, may be this one is the one you are after. More here:desmos.com/calculator/zz7e1nwgf9 and here:info:math.stackexchange.com/questions/92165/…
    $endgroup$
    – NoChance
    2 hours ago








  • 1




    $begingroup$
    @NoChance Your observation would be relevant if the line’s equation were of the form $y=b$. The line in the problem is not horizontal, though.
    $endgroup$
    – amd
    1 hour ago










  • $begingroup$
    @amd, thanks for your comment. I am not sure which tangent line one should pick then?
    $endgroup$
    – NoChance
    30 mins ago
















1












$begingroup$


I am practicing for a math contest and I encountered the following problem that I don't know how to solve:



For which value of $b$ is there only one intersection between the line $y = x + b$ and the parabola $y = x^2 + 5x + 3$?



How do I solve it?






algebra-precalculus contest-math







share|cite|improve this question









New contributor




swagbutton8 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    Do you know how to find a derivative?
    $endgroup$
    – R. Burton
    2 hours ago










  • $begingroup$
    Simply $x^2+5x+3=x+b$ must have only one solution, i.e. $x^2+4x+(3-b)=0$ should have zero discriminant.
    $endgroup$
    – the_fox
    2 hours ago










  • $begingroup$
    A Parabola could have many tangent lines...However, it has only 1 min. value, may be this one is the one you are after. More here:desmos.com/calculator/zz7e1nwgf9 and here:info:math.stackexchange.com/questions/92165/…
    $endgroup$
    – NoChance
    2 hours ago








  • 1




    $begingroup$
    @NoChance Your observation would be relevant if the line’s equation were of the form $y=b$. The line in the problem is not horizontal, though.
    $endgroup$
    – amd
    1 hour ago










  • $begingroup$
    @amd, thanks for your comment. I am not sure which tangent line one should pick then?
    $endgroup$
    – NoChance
    30 mins ago














1












1








1





$begingroup$


I am practicing for a math contest and I encountered the following problem that I don't know how to solve:



For which value of $b$ is there only one intersection between the line $y = x + b$ and the parabola $y = x^2 + 5x + 3$?



How do I solve it?






algebra-precalculus contest-math







share|cite|improve this question









New contributor




swagbutton8 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I am practicing for a math contest and I encountered the following problem that I don't know how to solve:



For which value of $b$ is there only one intersection between the line $y = x + b$ and the parabola $y = x^2 + 5x + 3$?



How do I solve it?






algebra-precalculus contest-math




algebra-precalculus contest-math






share|cite|improve this question









New contributor




swagbutton8 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




swagbutton8 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 2 hours ago









NoChance

3,79221321




3,79221321






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asked 2 hours ago









swagbutton8swagbutton8

61




61




New contributor




swagbutton8 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





swagbutton8 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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swagbutton8 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    Do you know how to find a derivative?
    $endgroup$
    – R. Burton
    2 hours ago










  • $begingroup$
    Simply $x^2+5x+3=x+b$ must have only one solution, i.e. $x^2+4x+(3-b)=0$ should have zero discriminant.
    $endgroup$
    – the_fox
    2 hours ago










  • $begingroup$
    A Parabola could have many tangent lines...However, it has only 1 min. value, may be this one is the one you are after. More here:desmos.com/calculator/zz7e1nwgf9 and here:info:math.stackexchange.com/questions/92165/…
    $endgroup$
    – NoChance
    2 hours ago








  • 1




    $begingroup$
    @NoChance Your observation would be relevant if the line’s equation were of the form $y=b$. The line in the problem is not horizontal, though.
    $endgroup$
    – amd
    1 hour ago










  • $begingroup$
    @amd, thanks for your comment. I am not sure which tangent line one should pick then?
    $endgroup$
    – NoChance
    30 mins ago














  • 1




    $begingroup$
    Do you know how to find a derivative?
    $endgroup$
    – R. Burton
    2 hours ago










  • $begingroup$
    Simply $x^2+5x+3=x+b$ must have only one solution, i.e. $x^2+4x+(3-b)=0$ should have zero discriminant.
    $endgroup$
    – the_fox
    2 hours ago










  • $begingroup$
    A Parabola could have many tangent lines...However, it has only 1 min. value, may be this one is the one you are after. More here:desmos.com/calculator/zz7e1nwgf9 and here:info:math.stackexchange.com/questions/92165/…
    $endgroup$
    – NoChance
    2 hours ago








  • 1




    $begingroup$
    @NoChance Your observation would be relevant if the line’s equation were of the form $y=b$. The line in the problem is not horizontal, though.
    $endgroup$
    – amd
    1 hour ago










  • $begingroup$
    @amd, thanks for your comment. I am not sure which tangent line one should pick then?
    $endgroup$
    – NoChance
    30 mins ago








1




1




$begingroup$
Do you know how to find a derivative?
$endgroup$
– R. Burton
2 hours ago




$begingroup$
Do you know how to find a derivative?
$endgroup$
– R. Burton
2 hours ago












$begingroup$
Simply $x^2+5x+3=x+b$ must have only one solution, i.e. $x^2+4x+(3-b)=0$ should have zero discriminant.
$endgroup$
– the_fox
2 hours ago




$begingroup$
Simply $x^2+5x+3=x+b$ must have only one solution, i.e. $x^2+4x+(3-b)=0$ should have zero discriminant.
$endgroup$
– the_fox
2 hours ago












$begingroup$
A Parabola could have many tangent lines...However, it has only 1 min. value, may be this one is the one you are after. More here:desmos.com/calculator/zz7e1nwgf9 and here:info:math.stackexchange.com/questions/92165/…
$endgroup$
– NoChance
2 hours ago






$begingroup$
A Parabola could have many tangent lines...However, it has only 1 min. value, may be this one is the one you are after. More here:desmos.com/calculator/zz7e1nwgf9 and here:info:math.stackexchange.com/questions/92165/…
$endgroup$
– NoChance
2 hours ago






1




1




$begingroup$
@NoChance Your observation would be relevant if the line’s equation were of the form $y=b$. The line in the problem is not horizontal, though.
$endgroup$
– amd
1 hour ago




$begingroup$
@NoChance Your observation would be relevant if the line’s equation were of the form $y=b$. The line in the problem is not horizontal, though.
$endgroup$
– amd
1 hour ago












$begingroup$
@amd, thanks for your comment. I am not sure which tangent line one should pick then?
$endgroup$
– NoChance
30 mins ago




$begingroup$
@amd, thanks for your comment. I am not sure which tangent line one should pick then?
$endgroup$
– NoChance
30 mins ago










3 Answers
3






active

oldest

votes


















3












$begingroup$

Hint:
$$x^2+5x+3=x+biff x^2+4x+3-b=0,$$ this quadratic equation should have exactly one solution therefore the discriminant $Delta=?$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Note that you like the line $y=x+b$ to be tangent to the parabola. The slope of this line is $ m=1.$ Thus the derivative of your parabola should be the same as the slope of the tangent line. Find the point of the tangency and find the $b$ value so that the line passes through that point.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      The tangent to parabola has slope $y' = 2x+5$ which should be slope of line =1 at the point of contact. So point of contact is at x=-2. Substituting in parabola,we get y=-3. Now this point lies on line as well.






      share|cite|improve this answer









      $endgroup$














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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        Hint:
        $$x^2+5x+3=x+biff x^2+4x+3-b=0,$$ this quadratic equation should have exactly one solution therefore the discriminant $Delta=?$






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          Hint:
          $$x^2+5x+3=x+biff x^2+4x+3-b=0,$$ this quadratic equation should have exactly one solution therefore the discriminant $Delta=?$






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            Hint:
            $$x^2+5x+3=x+biff x^2+4x+3-b=0,$$ this quadratic equation should have exactly one solution therefore the discriminant $Delta=?$






            share|cite|improve this answer









            $endgroup$



            Hint:
            $$x^2+5x+3=x+biff x^2+4x+3-b=0,$$ this quadratic equation should have exactly one solution therefore the discriminant $Delta=?$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 hours ago









            HAMIDINE SOUMAREHAMIDINE SOUMARE

            3,1471422




            3,1471422























                1












                $begingroup$

                Note that you like the line $y=x+b$ to be tangent to the parabola. The slope of this line is $ m=1.$ Thus the derivative of your parabola should be the same as the slope of the tangent line. Find the point of the tangency and find the $b$ value so that the line passes through that point.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Note that you like the line $y=x+b$ to be tangent to the parabola. The slope of this line is $ m=1.$ Thus the derivative of your parabola should be the same as the slope of the tangent line. Find the point of the tangency and find the $b$ value so that the line passes through that point.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Note that you like the line $y=x+b$ to be tangent to the parabola. The slope of this line is $ m=1.$ Thus the derivative of your parabola should be the same as the slope of the tangent line. Find the point of the tangency and find the $b$ value so that the line passes through that point.






                    share|cite|improve this answer









                    $endgroup$



                    Note that you like the line $y=x+b$ to be tangent to the parabola. The slope of this line is $ m=1.$ Thus the derivative of your parabola should be the same as the slope of the tangent line. Find the point of the tangency and find the $b$ value so that the line passes through that point.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 hours ago









                    Mohammad Riazi-KermaniMohammad Riazi-Kermani

                    42.2k42061




                    42.2k42061























                        0












                        $begingroup$

                        The tangent to parabola has slope $y' = 2x+5$ which should be slope of line =1 at the point of contact. So point of contact is at x=-2. Substituting in parabola,we get y=-3. Now this point lies on line as well.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          The tangent to parabola has slope $y' = 2x+5$ which should be slope of line =1 at the point of contact. So point of contact is at x=-2. Substituting in parabola,we get y=-3. Now this point lies on line as well.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            The tangent to parabola has slope $y' = 2x+5$ which should be slope of line =1 at the point of contact. So point of contact is at x=-2. Substituting in parabola,we get y=-3. Now this point lies on line as well.






                            share|cite|improve this answer









                            $endgroup$



                            The tangent to parabola has slope $y' = 2x+5$ which should be slope of line =1 at the point of contact. So point of contact is at x=-2. Substituting in parabola,we get y=-3. Now this point lies on line as well.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 hours ago









                            TojrahTojrah

                            5968




                            5968






















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