Why does the example solution to this problem assume a specific value for a coefficient?












0












$begingroup$


A given problem is




If the graph of the parabola $y=x^2$ is translated to a position such that its $x$ intercepts are $−d$ and $e$, and its $y$ intercept is $−f$, (where $d,e,f>0$), show that $de=f$.




The example solution is




Since the $x$ intercepts are $-d$ and $e$ the parabola must be of the form $y=a(x+d)(x-e)$. Also since we only translated $y=x^2$ it follows that $a=1$. Now setting $x=0$ we have $-f=-de$ and the results [sic] follows.




Ok, I checked the solution but still do not understand. Can some please explain in the most succinct way? Why do we automatically assume that $a = 1$ when it could have been $2,3,4,...$?










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    $begingroup$


    A given problem is




    If the graph of the parabola $y=x^2$ is translated to a position such that its $x$ intercepts are $−d$ and $e$, and its $y$ intercept is $−f$, (where $d,e,f>0$), show that $de=f$.




    The example solution is




    Since the $x$ intercepts are $-d$ and $e$ the parabola must be of the form $y=a(x+d)(x-e)$. Also since we only translated $y=x^2$ it follows that $a=1$. Now setting $x=0$ we have $-f=-de$ and the results [sic] follows.




    Ok, I checked the solution but still do not understand. Can some please explain in the most succinct way? Why do we automatically assume that $a = 1$ when it could have been $2,3,4,...$?










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      A given problem is




      If the graph of the parabola $y=x^2$ is translated to a position such that its $x$ intercepts are $−d$ and $e$, and its $y$ intercept is $−f$, (where $d,e,f>0$), show that $de=f$.




      The example solution is




      Since the $x$ intercepts are $-d$ and $e$ the parabola must be of the form $y=a(x+d)(x-e)$. Also since we only translated $y=x^2$ it follows that $a=1$. Now setting $x=0$ we have $-f=-de$ and the results [sic] follows.




      Ok, I checked the solution but still do not understand. Can some please explain in the most succinct way? Why do we automatically assume that $a = 1$ when it could have been $2,3,4,...$?










      share|cite|improve this question











      $endgroup$




      A given problem is




      If the graph of the parabola $y=x^2$ is translated to a position such that its $x$ intercepts are $−d$ and $e$, and its $y$ intercept is $−f$, (where $d,e,f>0$), show that $de=f$.




      The example solution is




      Since the $x$ intercepts are $-d$ and $e$ the parabola must be of the form $y=a(x+d)(x-e)$. Also since we only translated $y=x^2$ it follows that $a=1$. Now setting $x=0$ we have $-f=-de$ and the results [sic] follows.




      Ok, I checked the solution but still do not understand. Can some please explain in the most succinct way? Why do we automatically assume that $a = 1$ when it could have been $2,3,4,...$?







      algebra-precalculus functions






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      edited Dec 12 '18 at 6:32









      Nij

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      2,03111223










      asked Dec 12 '18 at 5:15









      Ke KeKe Ke

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          $begingroup$

          This comes from the fact that we are only translating the parabola $y = x^2$. This means that the only operations to get to the new parabola are horizontal or vertical shifts (these would be replacing $x$ with $x+h$ or $y$ with $y+k$, respectively). Neither of these changes the coefficient on $x^2$.



          Intuitively, changing $a$ will affect the curvature of the parabola — how quickly it gets steep. But translation means that we don’t want to alter this at all, just “slide” the parabola around rigidly.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            We assume that $a=1$ because the coefficient of the $x^2$ term in $y=x^2$ is $1$.



            Let me elaborate.



            The solution makes the most general assumption about the parabola: we assume that $y$ denotes a parabola with $x$-intercepts $-d$ and $e$. Then we know that



            $$y = a(x - (-d))(x - e)$$



            We also know that we are considering $y = x^2$. Translation of $y$ does not change the coefficient of the $x^2$ term. (Play around with a graphing calculator. You'll notice that this term's coefficient only affects how "skinny" or "wide" the parabola is. That is, translating the graph affects the linear/constant terms, but not the $x^2$ term.)



            Thus,



            $$x^2 + ?x + ?= a(x - (-d))(x - e)$$



            We multiply out the right side, and thus,



            $$1x^2 + ?x + ? = ax^2 + a(d-e)x - ade$$



            Notice why $a$ must be $1$? It's because the coefficients of the quadratic terms must be equal, i.e. $a=1$. (Similarly, we can also see that the linear terms and constant terms must be equal but because of the translation we don't know these, which is why I denoted them with $?$'s.)



            Our translation of the graph didn't affect the term.





            To elaborate further, or offer an alternate view, there is an alternate equation for a parabola,



            $$y = a(x-h)^2 + k = ax^2 - 2hax + ah^2 + k$$



            where $(h,k)$ is the vertex. You could make the argument then that translating the graph only affects the location of the vertex, since it's not been stretched or widened. Notice again: once multiplied out, we see that the quadratic term of $y$ has neither a $h$ or $y$ factor, i.e. the quadratic term is completely indifferent to the vertex of the parabola! Thus, translation does not at all affect the $ax^2$ term!



            Thus, since we are only translating the parabola $y =x^2$, we have no problem and letting $a=1$!






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              the three expressions of a parabola are:
              1) y=a(x-h)^2+k;
              2) y=a(x+d)(x-e);
              3) y=ax^2+bx-f;
              But since no Dilation is mentioned just Translation so a=1



              Equations 2) and 3) leads to de=f.






              share|cite|improve this answer









              $endgroup$













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                3 Answers
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                3 Answers
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                $begingroup$

                This comes from the fact that we are only translating the parabola $y = x^2$. This means that the only operations to get to the new parabola are horizontal or vertical shifts (these would be replacing $x$ with $x+h$ or $y$ with $y+k$, respectively). Neither of these changes the coefficient on $x^2$.



                Intuitively, changing $a$ will affect the curvature of the parabola — how quickly it gets steep. But translation means that we don’t want to alter this at all, just “slide” the parabola around rigidly.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  This comes from the fact that we are only translating the parabola $y = x^2$. This means that the only operations to get to the new parabola are horizontal or vertical shifts (these would be replacing $x$ with $x+h$ or $y$ with $y+k$, respectively). Neither of these changes the coefficient on $x^2$.



                  Intuitively, changing $a$ will affect the curvature of the parabola — how quickly it gets steep. But translation means that we don’t want to alter this at all, just “slide” the parabola around rigidly.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    This comes from the fact that we are only translating the parabola $y = x^2$. This means that the only operations to get to the new parabola are horizontal or vertical shifts (these would be replacing $x$ with $x+h$ or $y$ with $y+k$, respectively). Neither of these changes the coefficient on $x^2$.



                    Intuitively, changing $a$ will affect the curvature of the parabola — how quickly it gets steep. But translation means that we don’t want to alter this at all, just “slide” the parabola around rigidly.






                    share|cite|improve this answer









                    $endgroup$



                    This comes from the fact that we are only translating the parabola $y = x^2$. This means that the only operations to get to the new parabola are horizontal or vertical shifts (these would be replacing $x$ with $x+h$ or $y$ with $y+k$, respectively). Neither of these changes the coefficient on $x^2$.



                    Intuitively, changing $a$ will affect the curvature of the parabola — how quickly it gets steep. But translation means that we don’t want to alter this at all, just “slide” the parabola around rigidly.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 12 '18 at 6:06









                    plattyplatty

                    3,370320




                    3,370320























                        1












                        $begingroup$

                        We assume that $a=1$ because the coefficient of the $x^2$ term in $y=x^2$ is $1$.



                        Let me elaborate.



                        The solution makes the most general assumption about the parabola: we assume that $y$ denotes a parabola with $x$-intercepts $-d$ and $e$. Then we know that



                        $$y = a(x - (-d))(x - e)$$



                        We also know that we are considering $y = x^2$. Translation of $y$ does not change the coefficient of the $x^2$ term. (Play around with a graphing calculator. You'll notice that this term's coefficient only affects how "skinny" or "wide" the parabola is. That is, translating the graph affects the linear/constant terms, but not the $x^2$ term.)



                        Thus,



                        $$x^2 + ?x + ?= a(x - (-d))(x - e)$$



                        We multiply out the right side, and thus,



                        $$1x^2 + ?x + ? = ax^2 + a(d-e)x - ade$$



                        Notice why $a$ must be $1$? It's because the coefficients of the quadratic terms must be equal, i.e. $a=1$. (Similarly, we can also see that the linear terms and constant terms must be equal but because of the translation we don't know these, which is why I denoted them with $?$'s.)



                        Our translation of the graph didn't affect the term.





                        To elaborate further, or offer an alternate view, there is an alternate equation for a parabola,



                        $$y = a(x-h)^2 + k = ax^2 - 2hax + ah^2 + k$$



                        where $(h,k)$ is the vertex. You could make the argument then that translating the graph only affects the location of the vertex, since it's not been stretched or widened. Notice again: once multiplied out, we see that the quadratic term of $y$ has neither a $h$ or $y$ factor, i.e. the quadratic term is completely indifferent to the vertex of the parabola! Thus, translation does not at all affect the $ax^2$ term!



                        Thus, since we are only translating the parabola $y =x^2$, we have no problem and letting $a=1$!






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          We assume that $a=1$ because the coefficient of the $x^2$ term in $y=x^2$ is $1$.



                          Let me elaborate.



                          The solution makes the most general assumption about the parabola: we assume that $y$ denotes a parabola with $x$-intercepts $-d$ and $e$. Then we know that



                          $$y = a(x - (-d))(x - e)$$



                          We also know that we are considering $y = x^2$. Translation of $y$ does not change the coefficient of the $x^2$ term. (Play around with a graphing calculator. You'll notice that this term's coefficient only affects how "skinny" or "wide" the parabola is. That is, translating the graph affects the linear/constant terms, but not the $x^2$ term.)



                          Thus,



                          $$x^2 + ?x + ?= a(x - (-d))(x - e)$$



                          We multiply out the right side, and thus,



                          $$1x^2 + ?x + ? = ax^2 + a(d-e)x - ade$$



                          Notice why $a$ must be $1$? It's because the coefficients of the quadratic terms must be equal, i.e. $a=1$. (Similarly, we can also see that the linear terms and constant terms must be equal but because of the translation we don't know these, which is why I denoted them with $?$'s.)



                          Our translation of the graph didn't affect the term.





                          To elaborate further, or offer an alternate view, there is an alternate equation for a parabola,



                          $$y = a(x-h)^2 + k = ax^2 - 2hax + ah^2 + k$$



                          where $(h,k)$ is the vertex. You could make the argument then that translating the graph only affects the location of the vertex, since it's not been stretched or widened. Notice again: once multiplied out, we see that the quadratic term of $y$ has neither a $h$ or $y$ factor, i.e. the quadratic term is completely indifferent to the vertex of the parabola! Thus, translation does not at all affect the $ax^2$ term!



                          Thus, since we are only translating the parabola $y =x^2$, we have no problem and letting $a=1$!






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            We assume that $a=1$ because the coefficient of the $x^2$ term in $y=x^2$ is $1$.



                            Let me elaborate.



                            The solution makes the most general assumption about the parabola: we assume that $y$ denotes a parabola with $x$-intercepts $-d$ and $e$. Then we know that



                            $$y = a(x - (-d))(x - e)$$



                            We also know that we are considering $y = x^2$. Translation of $y$ does not change the coefficient of the $x^2$ term. (Play around with a graphing calculator. You'll notice that this term's coefficient only affects how "skinny" or "wide" the parabola is. That is, translating the graph affects the linear/constant terms, but not the $x^2$ term.)



                            Thus,



                            $$x^2 + ?x + ?= a(x - (-d))(x - e)$$



                            We multiply out the right side, and thus,



                            $$1x^2 + ?x + ? = ax^2 + a(d-e)x - ade$$



                            Notice why $a$ must be $1$? It's because the coefficients of the quadratic terms must be equal, i.e. $a=1$. (Similarly, we can also see that the linear terms and constant terms must be equal but because of the translation we don't know these, which is why I denoted them with $?$'s.)



                            Our translation of the graph didn't affect the term.





                            To elaborate further, or offer an alternate view, there is an alternate equation for a parabola,



                            $$y = a(x-h)^2 + k = ax^2 - 2hax + ah^2 + k$$



                            where $(h,k)$ is the vertex. You could make the argument then that translating the graph only affects the location of the vertex, since it's not been stretched or widened. Notice again: once multiplied out, we see that the quadratic term of $y$ has neither a $h$ or $y$ factor, i.e. the quadratic term is completely indifferent to the vertex of the parabola! Thus, translation does not at all affect the $ax^2$ term!



                            Thus, since we are only translating the parabola $y =x^2$, we have no problem and letting $a=1$!






                            share|cite|improve this answer









                            $endgroup$



                            We assume that $a=1$ because the coefficient of the $x^2$ term in $y=x^2$ is $1$.



                            Let me elaborate.



                            The solution makes the most general assumption about the parabola: we assume that $y$ denotes a parabola with $x$-intercepts $-d$ and $e$. Then we know that



                            $$y = a(x - (-d))(x - e)$$



                            We also know that we are considering $y = x^2$. Translation of $y$ does not change the coefficient of the $x^2$ term. (Play around with a graphing calculator. You'll notice that this term's coefficient only affects how "skinny" or "wide" the parabola is. That is, translating the graph affects the linear/constant terms, but not the $x^2$ term.)



                            Thus,



                            $$x^2 + ?x + ?= a(x - (-d))(x - e)$$



                            We multiply out the right side, and thus,



                            $$1x^2 + ?x + ? = ax^2 + a(d-e)x - ade$$



                            Notice why $a$ must be $1$? It's because the coefficients of the quadratic terms must be equal, i.e. $a=1$. (Similarly, we can also see that the linear terms and constant terms must be equal but because of the translation we don't know these, which is why I denoted them with $?$'s.)



                            Our translation of the graph didn't affect the term.





                            To elaborate further, or offer an alternate view, there is an alternate equation for a parabola,



                            $$y = a(x-h)^2 + k = ax^2 - 2hax + ah^2 + k$$



                            where $(h,k)$ is the vertex. You could make the argument then that translating the graph only affects the location of the vertex, since it's not been stretched or widened. Notice again: once multiplied out, we see that the quadratic term of $y$ has neither a $h$ or $y$ factor, i.e. the quadratic term is completely indifferent to the vertex of the parabola! Thus, translation does not at all affect the $ax^2$ term!



                            Thus, since we are only translating the parabola $y =x^2$, we have no problem and letting $a=1$!







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 12 '18 at 6:14









                            Eevee TrainerEevee Trainer

                            5,7571936




                            5,7571936























                                0












                                $begingroup$

                                the three expressions of a parabola are:
                                1) y=a(x-h)^2+k;
                                2) y=a(x+d)(x-e);
                                3) y=ax^2+bx-f;
                                But since no Dilation is mentioned just Translation so a=1



                                Equations 2) and 3) leads to de=f.






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  the three expressions of a parabola are:
                                  1) y=a(x-h)^2+k;
                                  2) y=a(x+d)(x-e);
                                  3) y=ax^2+bx-f;
                                  But since no Dilation is mentioned just Translation so a=1



                                  Equations 2) and 3) leads to de=f.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    the three expressions of a parabola are:
                                    1) y=a(x-h)^2+k;
                                    2) y=a(x+d)(x-e);
                                    3) y=ax^2+bx-f;
                                    But since no Dilation is mentioned just Translation so a=1



                                    Equations 2) and 3) leads to de=f.






                                    share|cite|improve this answer









                                    $endgroup$



                                    the three expressions of a parabola are:
                                    1) y=a(x-h)^2+k;
                                    2) y=a(x+d)(x-e);
                                    3) y=ax^2+bx-f;
                                    But since no Dilation is mentioned just Translation so a=1



                                    Equations 2) and 3) leads to de=f.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Dec 12 '18 at 8:08









                                    Monah DebianMonah Debian

                                    1




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