Given that $H(x)$ is a Heaviside function, how would I graph $H(x)+2H(x-3)-3H(x-5)$ Question

Given that $H(x)$ is a Heaviside function, how would I graph $H(x)+2H(x-3)-3H(x-5)$?

I was thinking about shifting the function, but then realized that there are multiple $H$'s
Or do I need to plot points?

It helps to write the function like this:
$$f(x)=\begin{align}&\;\;\;\;\;\color{#eb6235}{0}\\&+(\color{#8778b3}{1}-\color{#eb6235}{0})H(x-\color{#e19c24}{0})\\&+(\color{#5d9ec7}{3}-\color{#8778b3}{1})H(x-\color{#e19c24}{3})\\&+(\color{#996633}0-\color{#5d9ec7}{3})H(x-\color{#e19c24}{5})\end{align}$$
where the numbers being subtracted from $x$ in the unit step function represent the breakpoints, and the matching constants correspond to the pieces that make up your piecewise function. As an example, note that $\color{#5d9ec7}{3}$ shows up in the $\color{#5d9ec7}{3}-\color{#8778b3}{1}$ and $\color{#996633}0-\color{#5d9ec7}{3}$ factors corresponding to $H(x-\color{#e19c24}{3})$ and $H(x-\color{#e19c24}{5})$, so your function takes the value $\color{#5d9ec7}{3}$ in the interval $[\color{#e19c24}{3},\color{#e19c24}{5})$.
In this form, you can then actually make a direct conversion:
$$f(x)=\begin{cases}\color{#eb6235}0&\text{if }x<\color{#e19c24}0\\\color{#8778b3}1&\text{if }\color{#e19c24}0\le x<\color{#e19c24}3\\\color{#5d9ec7}3&\text{if }\color{#e19c24}3\le x<\color{#e19c24}5\\\color{#996633}0&\text{if }\color{#e19c24}5\le x\end{cases}$$
