The question is the following:
The graph is zero between $0$ and $2$, is a straight line from the point $(2,0)$ to $(5,5)$, a straight line down from $(5,5)$ to $(4,0)$ and zero everywhere else.
So far, my attempt is: $\dfrac{5t}{3}[(u(t-2)-u(t-5))-(u(t-5)-u(t-8))]$ where $u(t)$ is the Heaviside function. I have chosen the slope of $\dfrac{5t}{3}$ since $(5,5) - (2,0) = (3,5)$ so the rise over run is $\dfrac{5}{3}.$
Any help would be appreciated on where I went wrong.
EDIT: My slope is incorrect... I have corrected this now.
I am assuming every square is 1 unit in your picture.
The first line is
$$\frac 53 (t-2)$$
Second lin eis
$$-\frac 53 (t-8)$$
Which gives:
$$\frac{1}{3} (5 (t-2)) (\theta (t-2)-\theta (t-5))-\frac{1}{3} (5 (t-8)) (\theta (t-5)-\theta (t-8))$$