In the question the graph is given and I am to determine the Laplace transform of the function. I managed to get this function which produces the right graph but I can't help thinking that I've overcomplicated it as it seems a little tricky determining the transform. Also, I lost track while I was forming this and I can't work out where the +1 comes from in the heaviside(-t/T+1) part.
Bonus question - I had to set T= to a constant for it to work, how do I tell wolfram to treat the value as a constant rather than a variable? I tried adding 'for t' at the end but it didn't work.
You can write the function for the ramp between $1$ and $5$: $$y(t)=\frac{x}{5} $$ $$Y(s)=\frac{1}{5s^2}$$
You can do this with the formula: $y-y_1=(\frac{y_2-y_1}{x_2-x_1})(x-x_1)$, between the point: $A(x_1=0 ; y_1=0)$ and $B(x_2=5 ; y_2=1)$.
Now you have to consider the ramp just in the interval $[0,5]$, and to do so you use the frequency shift property: $$\mathscr{L}^{-1}\{F(s)e^{-as}\}=f(x-a)$$ In this way: $$Y(s)=\frac{1}{5s^2}-\frac{1}{5s^2}e^{-5s}$$ That, if $f(t)=\frac{x}{5}$, comes from: $$y(t)=f(t)-f(t-5)$$ With, of course, $\mathscr{L}\{x\}=\frac{1}{s^2}$
Because the step function goes to infinity, the result $Y(s)$ is your solution. In fact, if you perform the difference between the two functions, you obtain a straight constant line equal to $5$ that starts from $x=5$.