Finding a linear function from given functions












3















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
    2 days ago
















3















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
    2 days ago














3












3








3








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 2 days ago









greedoid

37.7k114794




37.7k114794










asked 2 days ago









Evan Kim

998




998












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


















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
















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




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










3 Answers
3






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oldest

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8














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














    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































      0














      ... and now the conceptual version. "$f(t+2) = f(t)+6$" means that the slope is $6$ (the increase in output) over a run of $2$ (the increase in input), so the slope is $frac{6}{2} = 3$. Then you have a point on the line, $(1,10)$ from "$f(1) = 10$", so $f(t) - 10 = 3(t-1)$, by point-slope, and we have $f(t) = 3t + 7$.






      share|cite|improve this answer





















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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        8














        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


























          8














          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
























            8












            8








            8






            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 2 days ago









            greedoid

            37.7k114794




            37.7k114794























                4














                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




























                  4














                  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


























                    4












                    4








                    4






                    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 2 days ago

























                    answered 2 days ago









                    KM101

                    4,758421




                    4,758421























                        0














                        ... and now the conceptual version. "$f(t+2) = f(t)+6$" means that the slope is $6$ (the increase in output) over a run of $2$ (the increase in input), so the slope is $frac{6}{2} = 3$. Then you have a point on the line, $(1,10)$ from "$f(1) = 10$", so $f(t) - 10 = 3(t-1)$, by point-slope, and we have $f(t) = 3t + 7$.






                        share|cite|improve this answer


























                          0














                          ... and now the conceptual version. "$f(t+2) = f(t)+6$" means that the slope is $6$ (the increase in output) over a run of $2$ (the increase in input), so the slope is $frac{6}{2} = 3$. Then you have a point on the line, $(1,10)$ from "$f(1) = 10$", so $f(t) - 10 = 3(t-1)$, by point-slope, and we have $f(t) = 3t + 7$.






                          share|cite|improve this answer
























                            0












                            0








                            0






                            ... and now the conceptual version. "$f(t+2) = f(t)+6$" means that the slope is $6$ (the increase in output) over a run of $2$ (the increase in input), so the slope is $frac{6}{2} = 3$. Then you have a point on the line, $(1,10)$ from "$f(1) = 10$", so $f(t) - 10 = 3(t-1)$, by point-slope, and we have $f(t) = 3t + 7$.






                            share|cite|improve this answer












                            ... and now the conceptual version. "$f(t+2) = f(t)+6$" means that the slope is $6$ (the increase in output) over a run of $2$ (the increase in input), so the slope is $frac{6}{2} = 3$. Then you have a point on the line, $(1,10)$ from "$f(1) = 10$", so $f(t) - 10 = 3(t-1)$, by point-slope, and we have $f(t) = 3t + 7$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 days ago









                            Eric Towers

                            31.8k22265




                            31.8k22265






























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