Finding a linear function from given functions 
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. 
 A: Since $f(1)=10$ we have $f(3)= f(1)+6=16$. Thus you have to solve the system:
$$ v+c=10$$
$$ 3v+c = 16$$ 
A: 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).
$$v = \frac{\Delta f(t)}{\Delta t}$$
A: ... 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 $f(t) = 3t + 7$.
