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







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  • 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

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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























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      2 Answers
      2






      active

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      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






























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