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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.

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  • $\begingroup$ Set up the two given conditions as equations to be solved for the coefficients $v$ and $C$. $\endgroup$ – zipirovich Dec 27 '18 at 15:29
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Since $f(1)=10$ we have $f(3)= f(1)+6=16$. Thus you have to solve the system:

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

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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).

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

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... 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 $f(t) = 3t + 7$.

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