You have a function...
$\displaystyle f(x)= a_{0} + a_{1}\ x$ (1)
... and is $\displaystyle f(4)=3$ and $\displaystyle f(10)=-3$... then You can find $\displaystyle a_{0}$ and $\displaystyle a_{1}$ in 'standard way'...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
to make equation of line through two points :
$\displaystyle (y-y_1)=\displasyle \frac{y_2-y_1}{x_2-x_1} \cdot (x-x_1) $
now your two points $\displaystyle M_1 (4,3) \; M_2(10,-3) $
so you will have
$\displaystyle \displaystyle (y-3)=\frac{-3-3}{10-4} \cdot ( x-10) $
$\displaystyle \displaystyle (y-3)=\frac{-6}{6} \cdot ( x-10) $
$\displaystyle \displaystyle y-3= 10-x $
$\displaystyle \displaystyle y=7- x $
if you don't trust me just put any x from 4 to 10 and you will get exact value of y...
for example x=7 -> y=0 (as you see on graphic)
if u need to represent your function over step functions it would be like this
$\displaystyle x(t) = \frac{3t}{4}[u(t)-u(t-4)] + (7-t)[u(t-4) - u(t-10)] $
P.S. you can take any two points from that line and put in formula and you will get same result (doesn't have to be "start" and "end")