Finding a linear function from given functions












1















The question is asking to find the linear function, $f(t) = vt + C$ for $f(t+2) = f(t) + 6$ and $f(1) = 10$




The answer is $3t + 7$, but I have no idea how the answer is produced.






algebra-precalculus functions polynomials







share|cite|improve this question
























  • Set up the two given conditions as equations to be solved for the coefficients $v$ and $C$.
    – zipirovich
    4 hours ago
















1















The question is asking to find the linear function, $f(t) = vt + C$ for $f(t+2) = f(t) + 6$ and $f(1) = 10$




The answer is $3t + 7$, but I have no idea how the answer is produced.






algebra-precalculus functions polynomials







share|cite|improve this question
























  • Set up the two given conditions as equations to be solved for the coefficients $v$ and $C$.
    – zipirovich
    4 hours ago














1












1








1








The question is asking to find the linear function, $f(t) = vt + C$ for $f(t+2) = f(t) + 6$ and $f(1) = 10$




The answer is $3t + 7$, but I have no idea how the answer is produced.






algebra-precalculus functions polynomials







share|cite|improve this question
















The question is asking to find the linear function, $f(t) = vt + C$ for $f(t+2) = f(t) + 6$ and $f(1) = 10$




The answer is $3t + 7$, but I have no idea how the answer is produced.






algebra-precalculus functions polynomials




algebra-precalculus functions polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 4 hours ago









greedoid

37.6k114794




37.6k114794










asked 4 hours ago









Evan Kim

898




898












  • Set up the two given conditions as equations to be solved for the coefficients $v$ and $C$.
    – zipirovich
    4 hours ago


















  • Set up the two given conditions as equations to be solved for the coefficients $v$ and $C$.
    – zipirovich
    4 hours ago
















Set up the two given conditions as equations to be solved for the coefficients $v$ and $C$.
– zipirovich
4 hours ago




Set up the two given conditions as equations to be solved for the coefficients $v$ and $C$.
– zipirovich
4 hours ago










2 Answers
2






active

oldest

votes


















4














Since $f(1)=10$ we have $f(3)= f(1)+6=16$. Thus you have to solve the system:



$$ v+c=10$$
$$ 3v+c = 16$$






share|cite|improve this answer





























    3














    Hint:



    $$f(t+2) = f(t)+6$$



    $$f(1) = 10 = f(-1)+6 iff f(-1) = 4$$



    You now have the two points $(1, 10)$ and $(-1, 4)$.



    Small Addition: From here, you can either use the use a system of equations for the two points $(1, 10)$ and $(-1, 4)$:



    $$begin{cases}
    v+C = 10\


    -v+C = 4
    end{cases}$$



    or you can refer to the definition of a linear equation $y = vt+C$, in which $v$ is the slope and $C$ is the $y$-intercept, or the $y$-coordinate when $x = 0$ (and it’s easy to find here because the point is the midpoint of the two points already found).



    $$v = frac{Delta f(t)}{Delta t}$$






    share|cite|improve this answer























      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054049%2ffinding-a-linear-function-from-given-functions%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4














      Since $f(1)=10$ we have $f(3)= f(1)+6=16$. Thus you have to solve the system:



      $$ v+c=10$$
      $$ 3v+c = 16$$






      share|cite|improve this answer


























        4














        Since $f(1)=10$ we have $f(3)= f(1)+6=16$. Thus you have to solve the system:



        $$ v+c=10$$
        $$ 3v+c = 16$$






        share|cite|improve this answer
























          4












          4








          4






          Since $f(1)=10$ we have $f(3)= f(1)+6=16$. Thus you have to solve the system:



          $$ v+c=10$$
          $$ 3v+c = 16$$






          share|cite|improve this answer












          Since $f(1)=10$ we have $f(3)= f(1)+6=16$. Thus you have to solve the system:



          $$ v+c=10$$
          $$ 3v+c = 16$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 4 hours ago









          greedoid

          37.6k114794




          37.6k114794























              3














              Hint:



              $$f(t+2) = f(t)+6$$



              $$f(1) = 10 = f(-1)+6 iff f(-1) = 4$$



              You now have the two points $(1, 10)$ and $(-1, 4)$.



              Small Addition: From here, you can either use the use a system of equations for the two points $(1, 10)$ and $(-1, 4)$:



              $$begin{cases}
              v+C = 10\


              -v+C = 4
              end{cases}$$



              or you can refer to the definition of a linear equation $y = vt+C$, in which $v$ is the slope and $C$ is the $y$-intercept, or the $y$-coordinate when $x = 0$ (and it’s easy to find here because the point is the midpoint of the two points already found).



              $$v = frac{Delta f(t)}{Delta t}$$






              share|cite|improve this answer




























                3














                Hint:



                $$f(t+2) = f(t)+6$$



                $$f(1) = 10 = f(-1)+6 iff f(-1) = 4$$



                You now have the two points $(1, 10)$ and $(-1, 4)$.



                Small Addition: From here, you can either use the use a system of equations for the two points $(1, 10)$ and $(-1, 4)$:



                $$begin{cases}
                v+C = 10\


                -v+C = 4
                end{cases}$$



                or you can refer to the definition of a linear equation $y = vt+C$, in which $v$ is the slope and $C$ is the $y$-intercept, or the $y$-coordinate when $x = 0$ (and it’s easy to find here because the point is the midpoint of the two points already found).



                $$v = frac{Delta f(t)}{Delta t}$$






                share|cite|improve this answer


























                  3












                  3








                  3






                  Hint:



                  $$f(t+2) = f(t)+6$$



                  $$f(1) = 10 = f(-1)+6 iff f(-1) = 4$$



                  You now have the two points $(1, 10)$ and $(-1, 4)$.



                  Small Addition: From here, you can either use the use a system of equations for the two points $(1, 10)$ and $(-1, 4)$:



                  $$begin{cases}
                  v+C = 10\


                  -v+C = 4
                  end{cases}$$



                  or you can refer to the definition of a linear equation $y = vt+C$, in which $v$ is the slope and $C$ is the $y$-intercept, or the $y$-coordinate when $x = 0$ (and it’s easy to find here because the point is the midpoint of the two points already found).



                  $$v = frac{Delta f(t)}{Delta t}$$






                  share|cite|improve this answer














                  Hint:



                  $$f(t+2) = f(t)+6$$



                  $$f(1) = 10 = f(-1)+6 iff f(-1) = 4$$



                  You now have the two points $(1, 10)$ and $(-1, 4)$.



                  Small Addition: From here, you can either use the use a system of equations for the two points $(1, 10)$ and $(-1, 4)$:



                  $$begin{cases}
                  v+C = 10\


                  -v+C = 4
                  end{cases}$$



                  or you can refer to the definition of a linear equation $y = vt+C$, in which $v$ is the slope and $C$ is the $y$-intercept, or the $y$-coordinate when $x = 0$ (and it’s easy to find here because the point is the midpoint of the two points already found).



                  $$v = frac{Delta f(t)}{Delta t}$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 3 hours ago

























                  answered 4 hours ago









                  KM101

                  4,528418




                  4,528418






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054049%2ffinding-a-linear-function-from-given-functions%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      サソリ

                      Image
                      この記事は検証可能な参考文献や出典が全く示されていないか、不十分です。 出典を追加して記事の信頼性向上にご協力ください。 （ 2019年1月 ） 「 サソリ 」のその他の用法については「サソリ (曖昧さ回避)」をご覧ください。 「 スコーピオン 」と「 スコルピオン 」はこの項目へ転送されています。その他の用法については「スコーピオン」をご覧ください。 サソリ目 生息年代: 430–0 Ma PreЄ Є O S D C P T J K Pg N シルル紀前期-現世 ダイオウサソリ（ Pandinus Imperator ） 分類 界 : 動物界 Animalia 門 : 節足動物門 Arthropoda 亜門 : 鋏角亜門 Chelicerata 綱 : クモガタ綱 Arachnida 目 : サソリ目 Scorpiones 学名 Scorpiones C. L. Koch, 1837 英名 Scorpion 上科 Pseudochactoidea キョクトウサソリ上科 Buthoidea Chaeriloidea Chactoidea Iuroidea Scorpionoidea サソリ （ サソリ類 、 蠍 、 蝎 、 scorpion ）は、鋏角亜門・クモガタ綱の サソリ目 （ Scorpiones ）に属する節足動物の総称である。体の前端に鋏型の触肢、後端に毒針を有する捕食者である。 1700種以上を含め、最古の化石記録は4億3千万年前のシルル紀まで遡る。有毒生物として広く知られるが、人間の命にかかわる毒を持つものはごく一部の種類に過ぎない。 目次 1 形態 1.1 前体 1.2 後体 1.2.1 中体 1.2.2 終体 2 雌雄 3 生態 3.1 天敵 3.2 生殖 4 分布 5 毒 6 蛍光現象 7 分類
                      Read more

                      広島県道265号伴広島線

                      Image
                      一般県道 広島県道265号伴広島線 路線延長 km 起点 広島県広島市安佐南区伴東7丁目 終点 広島県広島市中区榎町 接続する 主な道路 （記法） 広島県道38号広島豊平線 広島県道71号広島湯来線 山陽自動車道五日市IC 国道183号（ 国道261号重複） ■テンプレート（■ノート ■使い方） ■PJ道路 広島県道265号伴広島線 （ひろしまけんどう265ごう ともひろしません）は広島県広島市安佐南区から広島県広島市中区を結ぶ一般県道である。 目次 1 路線データ 2 通過する自治体 3 見どころ 4 接続道路 4.1 本線 4.2 旧道 5 別名 6 関連項目 路線データ 本線 起点：広島市安佐南区伴東7丁目・大原下橋南詰交差点（広島県道38号広島豊平線（本線・旧道）交点） 終点：広島市中区榎町・十日市交差点（国道183号（国道261号重用）交点） 総延長： 旧道 起点：広島市安佐南区伴東4丁目・新細坂交差点（広島県道38号広島豊平線交点） 終点：広島市安佐南区伴中央4丁目・寺組橋(南)交差点（広島県道265号伴広島線（本線）交点） 総延長： 通過する自治体 広島県 広島市（安佐南区、佐伯区、西区、中区） 見どころ 広島高速交通広島新交通1号線（アストラムライン） 広島広域公園 広島広域公園陸上競技場（広島ビッグアーチ） 己斐峠 己斐橋（太田川放水路） 天満橋（天満川） 接続道路 本線 広島県道38号広島豊平線（本線・旧道）（広島市安佐南区伴東7丁目・大原下橋南詰交差点〔起点〕） 広島県道71号広島湯来線（本線）（広島市安佐南区伴中央2丁目・伴交番入口交差点 - 広島市安佐南区伴中央4丁目・寺組橋(南)交差点間で重用） 広島県道71号広島湯来線（旧道）（広島県道265号伴広島線（旧道）重用）（広島市安佐南区伴中央4丁目・寺組橋(南)交差点） 広島県道71号広島湯来線（広島市安佐南区伴中央4丁目・寺組橋(南)交差点 - 広島市佐伯区五日市町石内・五月が丘(南)交差点間で重用）
                      Read more

                      Setup Asymptote in Texstudio

                      Image
                      0 Good time of the day! I am trying to compile the following file: documentclass[12pt]{article} usepackage[inline]{asymptote} begin{document} Here is a venn diagram produced with Asymptote, drawn to width 4cm: defA{A} defB{V{B}} begin{figure} begin{center} begin{asy} size(4cm,0); pen colour1=red; pen colour2=green; pair z0=(0,0); pair z1=(-1,0); pair z2=(1,0); real r=1.5; path c1=circle(z1,r); path c2=circle(z2,r); fill(c1,colour1); fill(c2,colour2); picture intersection=new picture; fill(intersection,c1,colour1+colour2); clip(intersection,c2); add(intersection); draw(c1); draw(c2); //draw("$A$",box,z1)
                      Read more