How to find range of a Quadratic/Quadratic function easily without plotting its graph?












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Is there any way to find range of a Quadratic/Quadratic function, without plotting its graph?






quadratics







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  • $begingroup$
    Do you know how to complete the square?
    $endgroup$
    – Matthew Leingang
    May 9 '18 at 8:22










  • $begingroup$
    Do you mean a rational function with quadratic numerator or denominator ?
    $endgroup$
    – Yves Daoust
    May 9 '18 at 8:52










  • $begingroup$
    This type of problem was extremely common on tests and in textbooks in the late 1800s and very early 1900s, especially in France. I posted some information about it in this 10 December 2008 ap-calculus post archived at Math Forum.
    $endgroup$
    – Dave L. Renfro
    May 9 '18 at 9:42


















0












$begingroup$


Is there any way to find range of a Quadratic/Quadratic function, without plotting its graph?






quadratics







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you know how to complete the square?
    $endgroup$
    – Matthew Leingang
    May 9 '18 at 8:22










  • $begingroup$
    Do you mean a rational function with quadratic numerator or denominator ?
    $endgroup$
    – Yves Daoust
    May 9 '18 at 8:52










  • $begingroup$
    This type of problem was extremely common on tests and in textbooks in the late 1800s and very early 1900s, especially in France. I posted some information about it in this 10 December 2008 ap-calculus post archived at Math Forum.
    $endgroup$
    – Dave L. Renfro
    May 9 '18 at 9:42
















0












0








0





$begingroup$


Is there any way to find range of a Quadratic/Quadratic function, without plotting its graph?






quadratics







share|cite|improve this question











$endgroup$




Is there any way to find range of a Quadratic/Quadratic function, without plotting its graph?






quadratics




quadratics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 9 '18 at 8:37









José Carlos Santos

176k24137247




176k24137247










asked May 9 '18 at 8:19









ShebSheb

1




1












  • $begingroup$
    Do you know how to complete the square?
    $endgroup$
    – Matthew Leingang
    May 9 '18 at 8:22










  • $begingroup$
    Do you mean a rational function with quadratic numerator or denominator ?
    $endgroup$
    – Yves Daoust
    May 9 '18 at 8:52










  • $begingroup$
    This type of problem was extremely common on tests and in textbooks in the late 1800s and very early 1900s, especially in France. I posted some information about it in this 10 December 2008 ap-calculus post archived at Math Forum.
    $endgroup$
    – Dave L. Renfro
    May 9 '18 at 9:42




















  • $begingroup$
    Do you know how to complete the square?
    $endgroup$
    – Matthew Leingang
    May 9 '18 at 8:22










  • $begingroup$
    Do you mean a rational function with quadratic numerator or denominator ?
    $endgroup$
    – Yves Daoust
    May 9 '18 at 8:52










  • $begingroup$
    This type of problem was extremely common on tests and in textbooks in the late 1800s and very early 1900s, especially in France. I posted some information about it in this 10 December 2008 ap-calculus post archived at Math Forum.
    $endgroup$
    – Dave L. Renfro
    May 9 '18 at 9:42


















$begingroup$
Do you know how to complete the square?
$endgroup$
– Matthew Leingang
May 9 '18 at 8:22




$begingroup$
Do you know how to complete the square?
$endgroup$
– Matthew Leingang
May 9 '18 at 8:22












$begingroup$
Do you mean a rational function with quadratic numerator or denominator ?
$endgroup$
– Yves Daoust
May 9 '18 at 8:52




$begingroup$
Do you mean a rational function with quadratic numerator or denominator ?
$endgroup$
– Yves Daoust
May 9 '18 at 8:52












$begingroup$
This type of problem was extremely common on tests and in textbooks in the late 1800s and very early 1900s, especially in France. I posted some information about it in this 10 December 2008 ap-calculus post archived at Math Forum.
$endgroup$
– Dave L. Renfro
May 9 '18 at 9:42






$begingroup$
This type of problem was extremely common on tests and in textbooks in the late 1800s and very early 1900s, especially in France. I posted some information about it in this 10 December 2008 ap-calculus post archived at Math Forum.
$endgroup$
– Dave L. Renfro
May 9 '18 at 9:42












3 Answers
3






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1












$begingroup$

For $Ane 0$ we have $Ax^2+Bx+C=Ag(x)+D$ where $g(x)=(x+frac {B}{2A})^2$ and $D=C-frac {B^2}{4A}.$ The range of $g(x)$ is $[0,infty).$



If $A>0$ the range of $Ag(x)$ is ${Ay: yin [0,infty)}=[0,infty).$ So the range of $Ax^2+Bx+C$ is ${z+D: zin [0,infty)}=[D,infty).$



If $A<0$ the range of $Ag(x)$ is ${Ay:yin [0,infty)}=(-infty,0].$ So the range of $Ax^2+BX+C$ is ${z+D: zin (-infty,0]}=(-infty,D].$






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    0












    $begingroup$

    Because of the symmetry of the parabola, its stationary point will always be in the middle of its $x$-intercepts (or zeroes). Using the quadratic formula and taking the average of both roots, the $x$-coordinate of the stationary point of any quadratic function $ax^2+bx+c$ (where $aneq0$) is given by $x=-frac{b}{2a}$.



    When $x=frac{-b}{2a}$, $y=c-frac{b^2}{4a}$.



    Therefore the maximum or minimum value of the quadratic is $c-frac{b^2}{4a}$. Whether it is the maximum or minimum can be determined by examining the sign of $a$.



    If $a$ is positive, then the range is $y geq c-frac{b^2}{4a}$.



    If $a$ is negative, then the range is $y leq c-frac{b^2}{4a}$.






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      Hint:



      The rational function $dfrac{p(x)}{q(x)}$ has (one or two) vertical asymptotes if the denominator has real roots. It also has an horizontal asymptote $y=dfrac{p_2}{q_2}$.



      The local extrema are given by



      $$left(frac{p(x)}{q(x)}right)'=0$$ or



      $$p'(x)q(x)=p(x)q'(x).$$



      The two members are two cubic polynomial, with the same leading coefficient, so that this is in fact a quadratic equation (can also be linear or constant).



      After solving for the roots, if any, the limits of the range are given by the value at the extrema, at the horizontal asymptote, and $pminfty$ if there are vertical asymptotes.



      A longer analysis is required to exhaust the cases.






      share|cite|improve this answer









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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        For $Ane 0$ we have $Ax^2+Bx+C=Ag(x)+D$ where $g(x)=(x+frac {B}{2A})^2$ and $D=C-frac {B^2}{4A}.$ The range of $g(x)$ is $[0,infty).$



        If $A>0$ the range of $Ag(x)$ is ${Ay: yin [0,infty)}=[0,infty).$ So the range of $Ax^2+Bx+C$ is ${z+D: zin [0,infty)}=[D,infty).$



        If $A<0$ the range of $Ag(x)$ is ${Ay:yin [0,infty)}=(-infty,0].$ So the range of $Ax^2+BX+C$ is ${z+D: zin (-infty,0]}=(-infty,D].$






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          For $Ane 0$ we have $Ax^2+Bx+C=Ag(x)+D$ where $g(x)=(x+frac {B}{2A})^2$ and $D=C-frac {B^2}{4A}.$ The range of $g(x)$ is $[0,infty).$



          If $A>0$ the range of $Ag(x)$ is ${Ay: yin [0,infty)}=[0,infty).$ So the range of $Ax^2+Bx+C$ is ${z+D: zin [0,infty)}=[D,infty).$



          If $A<0$ the range of $Ag(x)$ is ${Ay:yin [0,infty)}=(-infty,0].$ So the range of $Ax^2+BX+C$ is ${z+D: zin (-infty,0]}=(-infty,D].$






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            For $Ane 0$ we have $Ax^2+Bx+C=Ag(x)+D$ where $g(x)=(x+frac {B}{2A})^2$ and $D=C-frac {B^2}{4A}.$ The range of $g(x)$ is $[0,infty).$



            If $A>0$ the range of $Ag(x)$ is ${Ay: yin [0,infty)}=[0,infty).$ So the range of $Ax^2+Bx+C$ is ${z+D: zin [0,infty)}=[D,infty).$



            If $A<0$ the range of $Ag(x)$ is ${Ay:yin [0,infty)}=(-infty,0].$ So the range of $Ax^2+BX+C$ is ${z+D: zin (-infty,0]}=(-infty,D].$






            share|cite|improve this answer









            $endgroup$



            For $Ane 0$ we have $Ax^2+Bx+C=Ag(x)+D$ where $g(x)=(x+frac {B}{2A})^2$ and $D=C-frac {B^2}{4A}.$ The range of $g(x)$ is $[0,infty).$



            If $A>0$ the range of $Ag(x)$ is ${Ay: yin [0,infty)}=[0,infty).$ So the range of $Ax^2+Bx+C$ is ${z+D: zin [0,infty)}=[D,infty).$



            If $A<0$ the range of $Ag(x)$ is ${Ay:yin [0,infty)}=(-infty,0].$ So the range of $Ax^2+BX+C$ is ${z+D: zin (-infty,0]}=(-infty,D].$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 24 '18 at 7:37









            DanielWainfleetDanielWainfleet

            35.9k31648




            35.9k31648























                0












                $begingroup$

                Because of the symmetry of the parabola, its stationary point will always be in the middle of its $x$-intercepts (or zeroes). Using the quadratic formula and taking the average of both roots, the $x$-coordinate of the stationary point of any quadratic function $ax^2+bx+c$ (where $aneq0$) is given by $x=-frac{b}{2a}$.



                When $x=frac{-b}{2a}$, $y=c-frac{b^2}{4a}$.



                Therefore the maximum or minimum value of the quadratic is $c-frac{b^2}{4a}$. Whether it is the maximum or minimum can be determined by examining the sign of $a$.



                If $a$ is positive, then the range is $y geq c-frac{b^2}{4a}$.



                If $a$ is negative, then the range is $y leq c-frac{b^2}{4a}$.






                share|cite|improve this answer











                $endgroup$


















                  0












                  $begingroup$

                  Because of the symmetry of the parabola, its stationary point will always be in the middle of its $x$-intercepts (or zeroes). Using the quadratic formula and taking the average of both roots, the $x$-coordinate of the stationary point of any quadratic function $ax^2+bx+c$ (where $aneq0$) is given by $x=-frac{b}{2a}$.



                  When $x=frac{-b}{2a}$, $y=c-frac{b^2}{4a}$.



                  Therefore the maximum or minimum value of the quadratic is $c-frac{b^2}{4a}$. Whether it is the maximum or minimum can be determined by examining the sign of $a$.



                  If $a$ is positive, then the range is $y geq c-frac{b^2}{4a}$.



                  If $a$ is negative, then the range is $y leq c-frac{b^2}{4a}$.






                  share|cite|improve this answer











                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Because of the symmetry of the parabola, its stationary point will always be in the middle of its $x$-intercepts (or zeroes). Using the quadratic formula and taking the average of both roots, the $x$-coordinate of the stationary point of any quadratic function $ax^2+bx+c$ (where $aneq0$) is given by $x=-frac{b}{2a}$.



                    When $x=frac{-b}{2a}$, $y=c-frac{b^2}{4a}$.



                    Therefore the maximum or minimum value of the quadratic is $c-frac{b^2}{4a}$. Whether it is the maximum or minimum can be determined by examining the sign of $a$.



                    If $a$ is positive, then the range is $y geq c-frac{b^2}{4a}$.



                    If $a$ is negative, then the range is $y leq c-frac{b^2}{4a}$.






                    share|cite|improve this answer











                    $endgroup$



                    Because of the symmetry of the parabola, its stationary point will always be in the middle of its $x$-intercepts (or zeroes). Using the quadratic formula and taking the average of both roots, the $x$-coordinate of the stationary point of any quadratic function $ax^2+bx+c$ (where $aneq0$) is given by $x=-frac{b}{2a}$.



                    When $x=frac{-b}{2a}$, $y=c-frac{b^2}{4a}$.



                    Therefore the maximum or minimum value of the quadratic is $c-frac{b^2}{4a}$. Whether it is the maximum or minimum can be determined by examining the sign of $a$.



                    If $a$ is positive, then the range is $y geq c-frac{b^2}{4a}$.



                    If $a$ is negative, then the range is $y leq c-frac{b^2}{4a}$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited May 9 '18 at 8:53

























                    answered May 9 '18 at 8:48









                    PKBeamPKBeam

                    31219




                    31219























                        0












                        $begingroup$

                        Hint:



                        The rational function $dfrac{p(x)}{q(x)}$ has (one or two) vertical asymptotes if the denominator has real roots. It also has an horizontal asymptote $y=dfrac{p_2}{q_2}$.



                        The local extrema are given by



                        $$left(frac{p(x)}{q(x)}right)'=0$$ or



                        $$p'(x)q(x)=p(x)q'(x).$$



                        The two members are two cubic polynomial, with the same leading coefficient, so that this is in fact a quadratic equation (can also be linear or constant).



                        After solving for the roots, if any, the limits of the range are given by the value at the extrema, at the horizontal asymptote, and $pminfty$ if there are vertical asymptotes.



                        A longer analysis is required to exhaust the cases.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Hint:



                          The rational function $dfrac{p(x)}{q(x)}$ has (one or two) vertical asymptotes if the denominator has real roots. It also has an horizontal asymptote $y=dfrac{p_2}{q_2}$.



                          The local extrema are given by



                          $$left(frac{p(x)}{q(x)}right)'=0$$ or



                          $$p'(x)q(x)=p(x)q'(x).$$



                          The two members are two cubic polynomial, with the same leading coefficient, so that this is in fact a quadratic equation (can also be linear or constant).



                          After solving for the roots, if any, the limits of the range are given by the value at the extrema, at the horizontal asymptote, and $pminfty$ if there are vertical asymptotes.



                          A longer analysis is required to exhaust the cases.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Hint:



                            The rational function $dfrac{p(x)}{q(x)}$ has (one or two) vertical asymptotes if the denominator has real roots. It also has an horizontal asymptote $y=dfrac{p_2}{q_2}$.



                            The local extrema are given by



                            $$left(frac{p(x)}{q(x)}right)'=0$$ or



                            $$p'(x)q(x)=p(x)q'(x).$$



                            The two members are two cubic polynomial, with the same leading coefficient, so that this is in fact a quadratic equation (can also be linear or constant).



                            After solving for the roots, if any, the limits of the range are given by the value at the extrema, at the horizontal asymptote, and $pminfty$ if there are vertical asymptotes.



                            A longer analysis is required to exhaust the cases.






                            share|cite|improve this answer









                            $endgroup$



                            Hint:



                            The rational function $dfrac{p(x)}{q(x)}$ has (one or two) vertical asymptotes if the denominator has real roots. It also has an horizontal asymptote $y=dfrac{p_2}{q_2}$.



                            The local extrema are given by



                            $$left(frac{p(x)}{q(x)}right)'=0$$ or



                            $$p'(x)q(x)=p(x)q'(x).$$



                            The two members are two cubic polynomial, with the same leading coefficient, so that this is in fact a quadratic equation (can also be linear or constant).



                            After solving for the roots, if any, the limits of the range are given by the value at the extrema, at the horizontal asymptote, and $pminfty$ if there are vertical asymptotes.



                            A longer analysis is required to exhaust the cases.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered May 9 '18 at 9:10









                            Yves DaoustYves Daoust

                            133k676232




                            133k676232






























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