Write the equation of the image of the parabola $y=x^2-2x+1$ under a translation $vec{p}=hat{i}+3hat{j}$












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$begingroup$


Write the equation of the image of the parabola $y=x^2-2x+1$ under a translation $vec{p}=hat{i}+3hat{j}$




$y=(x-1)^2$ is a parabola whose vertex is $(1,0)$ and focus is $(1,frac{1}{4})$.I dont know how to solve further.






vectors conic-sections







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    $begingroup$


    Write the equation of the image of the parabola $y=x^2-2x+1$ under a translation $vec{p}=hat{i}+3hat{j}$




    $y=(x-1)^2$ is a parabola whose vertex is $(1,0)$ and focus is $(1,frac{1}{4})$.I dont know how to solve further.






    vectors conic-sections







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Write the equation of the image of the parabola $y=x^2-2x+1$ under a translation $vec{p}=hat{i}+3hat{j}$




      $y=(x-1)^2$ is a parabola whose vertex is $(1,0)$ and focus is $(1,frac{1}{4})$.I dont know how to solve further.






      vectors conic-sections







      share|cite|improve this question









      $endgroup$




      Write the equation of the image of the parabola $y=x^2-2x+1$ under a translation $vec{p}=hat{i}+3hat{j}$




      $y=(x-1)^2$ is a parabola whose vertex is $(1,0)$ and focus is $(1,frac{1}{4})$.I dont know how to solve further.






      vectors conic-sections




      vectors conic-sections






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      asked Dec 17 '18 at 14:13









      user984325user984325

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          $begingroup$


          Write the equation of the image of the parabola $y=x^2-2x+1$ under a translation $vec{p}=hat{i}+3hat{j}$




          Translating $y=f(x)$ over $vec{p}=color{blue}{a}hat{i}+color{red}{b}hat{j}$ corresponds to the equation $y-color{red}{b}=f(x-color{blue}{a})$ in the same coordinate system; so with $y=x^2-2x+1$ and $vec{p}=color{blue}{1}hat{i}+color{red}{3}hat{j}$, you get:
          $$y-color{red}{3}=(x-color{blue}{1})^2-2(x-color{blue}{1})+1 iff y=x^2 - 4 x + 7 tag{$*$}$$




          $y=(x-1)^2$ is a parabola whose vertex is $(1,0)$




          After rewriting the parabola in this standard form, you could also simply shift the vertex towards $(1+color{blue}{1},0+color{red}{3})=color{purple}{(2,3)}$, so the equation becomes:
          $$y=(x-color{purple}{2})^2+color{purple}{3}iff y=x^2 - 4 x + 7 tag{$star$}$$



          Note that $(*)$ and $(star)$ match.






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            0












            $begingroup$

            $p(x,y)=(x,y)+(1,3)=(x+1,y+3), quad p(x,y)^t=(x+1,y+3)^t$ so $p$ can be seen as a matrix



            $pmatrix{1 & 0 & 1\ 0 & 1 & 3 \ 0 & 0 & 1}$ which is $pmatrix{1 & 0 & 1\ 0 & 1 & 3 \ 0 & 0 & 1}pmatrix{x \ y \ 1}=pmatrix{x+1 \ y+3 \ 1}$



            so for $y=(x-1)^2$ it is



            $pmatrix{1 & 0 & 1\ 0 & 1 & 3 \ 0 & 0 & 1}pmatrix{x \ (x-1)^2 \ 1}=pmatrix{x+1 \ (x-1)^2+3 \ 1}$



            so $y'=(x-1)^2+3=x^2-2x+4=(x'-1)^2-2(x'-1)+4=x'^2-4x'+7$






            share|cite|improve this answer









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              1












              $begingroup$


              Write the equation of the image of the parabola $y=x^2-2x+1$ under a translation $vec{p}=hat{i}+3hat{j}$




              Translating $y=f(x)$ over $vec{p}=color{blue}{a}hat{i}+color{red}{b}hat{j}$ corresponds to the equation $y-color{red}{b}=f(x-color{blue}{a})$ in the same coordinate system; so with $y=x^2-2x+1$ and $vec{p}=color{blue}{1}hat{i}+color{red}{3}hat{j}$, you get:
              $$y-color{red}{3}=(x-color{blue}{1})^2-2(x-color{blue}{1})+1 iff y=x^2 - 4 x + 7 tag{$*$}$$




              $y=(x-1)^2$ is a parabola whose vertex is $(1,0)$




              After rewriting the parabola in this standard form, you could also simply shift the vertex towards $(1+color{blue}{1},0+color{red}{3})=color{purple}{(2,3)}$, so the equation becomes:
              $$y=(x-color{purple}{2})^2+color{purple}{3}iff y=x^2 - 4 x + 7 tag{$star$}$$



              Note that $(*)$ and $(star)$ match.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$


                Write the equation of the image of the parabola $y=x^2-2x+1$ under a translation $vec{p}=hat{i}+3hat{j}$




                Translating $y=f(x)$ over $vec{p}=color{blue}{a}hat{i}+color{red}{b}hat{j}$ corresponds to the equation $y-color{red}{b}=f(x-color{blue}{a})$ in the same coordinate system; so with $y=x^2-2x+1$ and $vec{p}=color{blue}{1}hat{i}+color{red}{3}hat{j}$, you get:
                $$y-color{red}{3}=(x-color{blue}{1})^2-2(x-color{blue}{1})+1 iff y=x^2 - 4 x + 7 tag{$*$}$$




                $y=(x-1)^2$ is a parabola whose vertex is $(1,0)$




                After rewriting the parabola in this standard form, you could also simply shift the vertex towards $(1+color{blue}{1},0+color{red}{3})=color{purple}{(2,3)}$, so the equation becomes:
                $$y=(x-color{purple}{2})^2+color{purple}{3}iff y=x^2 - 4 x + 7 tag{$star$}$$



                Note that $(*)$ and $(star)$ match.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$


                  Write the equation of the image of the parabola $y=x^2-2x+1$ under a translation $vec{p}=hat{i}+3hat{j}$




                  Translating $y=f(x)$ over $vec{p}=color{blue}{a}hat{i}+color{red}{b}hat{j}$ corresponds to the equation $y-color{red}{b}=f(x-color{blue}{a})$ in the same coordinate system; so with $y=x^2-2x+1$ and $vec{p}=color{blue}{1}hat{i}+color{red}{3}hat{j}$, you get:
                  $$y-color{red}{3}=(x-color{blue}{1})^2-2(x-color{blue}{1})+1 iff y=x^2 - 4 x + 7 tag{$*$}$$




                  $y=(x-1)^2$ is a parabola whose vertex is $(1,0)$




                  After rewriting the parabola in this standard form, you could also simply shift the vertex towards $(1+color{blue}{1},0+color{red}{3})=color{purple}{(2,3)}$, so the equation becomes:
                  $$y=(x-color{purple}{2})^2+color{purple}{3}iff y=x^2 - 4 x + 7 tag{$star$}$$



                  Note that $(*)$ and $(star)$ match.






                  share|cite|improve this answer









                  $endgroup$




                  Write the equation of the image of the parabola $y=x^2-2x+1$ under a translation $vec{p}=hat{i}+3hat{j}$




                  Translating $y=f(x)$ over $vec{p}=color{blue}{a}hat{i}+color{red}{b}hat{j}$ corresponds to the equation $y-color{red}{b}=f(x-color{blue}{a})$ in the same coordinate system; so with $y=x^2-2x+1$ and $vec{p}=color{blue}{1}hat{i}+color{red}{3}hat{j}$, you get:
                  $$y-color{red}{3}=(x-color{blue}{1})^2-2(x-color{blue}{1})+1 iff y=x^2 - 4 x + 7 tag{$*$}$$




                  $y=(x-1)^2$ is a parabola whose vertex is $(1,0)$




                  After rewriting the parabola in this standard form, you could also simply shift the vertex towards $(1+color{blue}{1},0+color{red}{3})=color{purple}{(2,3)}$, so the equation becomes:
                  $$y=(x-color{purple}{2})^2+color{purple}{3}iff y=x^2 - 4 x + 7 tag{$star$}$$



                  Note that $(*)$ and $(star)$ match.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 17 '18 at 14:56









                  StackTDStackTD

                  22.8k2051




                  22.8k2051























                      0












                      $begingroup$

                      $p(x,y)=(x,y)+(1,3)=(x+1,y+3), quad p(x,y)^t=(x+1,y+3)^t$ so $p$ can be seen as a matrix



                      $pmatrix{1 & 0 & 1\ 0 & 1 & 3 \ 0 & 0 & 1}$ which is $pmatrix{1 & 0 & 1\ 0 & 1 & 3 \ 0 & 0 & 1}pmatrix{x \ y \ 1}=pmatrix{x+1 \ y+3 \ 1}$



                      so for $y=(x-1)^2$ it is



                      $pmatrix{1 & 0 & 1\ 0 & 1 & 3 \ 0 & 0 & 1}pmatrix{x \ (x-1)^2 \ 1}=pmatrix{x+1 \ (x-1)^2+3 \ 1}$



                      so $y'=(x-1)^2+3=x^2-2x+4=(x'-1)^2-2(x'-1)+4=x'^2-4x'+7$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        $p(x,y)=(x,y)+(1,3)=(x+1,y+3), quad p(x,y)^t=(x+1,y+3)^t$ so $p$ can be seen as a matrix



                        $pmatrix{1 & 0 & 1\ 0 & 1 & 3 \ 0 & 0 & 1}$ which is $pmatrix{1 & 0 & 1\ 0 & 1 & 3 \ 0 & 0 & 1}pmatrix{x \ y \ 1}=pmatrix{x+1 \ y+3 \ 1}$



                        so for $y=(x-1)^2$ it is



                        $pmatrix{1 & 0 & 1\ 0 & 1 & 3 \ 0 & 0 & 1}pmatrix{x \ (x-1)^2 \ 1}=pmatrix{x+1 \ (x-1)^2+3 \ 1}$



                        so $y'=(x-1)^2+3=x^2-2x+4=(x'-1)^2-2(x'-1)+4=x'^2-4x'+7$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          $p(x,y)=(x,y)+(1,3)=(x+1,y+3), quad p(x,y)^t=(x+1,y+3)^t$ so $p$ can be seen as a matrix



                          $pmatrix{1 & 0 & 1\ 0 & 1 & 3 \ 0 & 0 & 1}$ which is $pmatrix{1 & 0 & 1\ 0 & 1 & 3 \ 0 & 0 & 1}pmatrix{x \ y \ 1}=pmatrix{x+1 \ y+3 \ 1}$



                          so for $y=(x-1)^2$ it is



                          $pmatrix{1 & 0 & 1\ 0 & 1 & 3 \ 0 & 0 & 1}pmatrix{x \ (x-1)^2 \ 1}=pmatrix{x+1 \ (x-1)^2+3 \ 1}$



                          so $y'=(x-1)^2+3=x^2-2x+4=(x'-1)^2-2(x'-1)+4=x'^2-4x'+7$






                          share|cite|improve this answer









                          $endgroup$



                          $p(x,y)=(x,y)+(1,3)=(x+1,y+3), quad p(x,y)^t=(x+1,y+3)^t$ so $p$ can be seen as a matrix



                          $pmatrix{1 & 0 & 1\ 0 & 1 & 3 \ 0 & 0 & 1}$ which is $pmatrix{1 & 0 & 1\ 0 & 1 & 3 \ 0 & 0 & 1}pmatrix{x \ y \ 1}=pmatrix{x+1 \ y+3 \ 1}$



                          so for $y=(x-1)^2$ it is



                          $pmatrix{1 & 0 & 1\ 0 & 1 & 3 \ 0 & 0 & 1}pmatrix{x \ (x-1)^2 \ 1}=pmatrix{x+1 \ (x-1)^2+3 \ 1}$



                          so $y'=(x-1)^2+3=x^2-2x+4=(x'-1)^2-2(x'-1)+4=x'^2-4x'+7$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 17 '18 at 15:02









                          giannispapavgiannispapav

                          1,534324




                          1,534324






























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