# 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.

• Set up the two given conditions as equations to be solved for the coefficients $v$ and $C$. – zipirovich Dec 27 '18 at 15:29

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

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

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).

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

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