Adding Dirac Delta with Unit Step What does the graph of the function $x(t)=-\delta(t)+u(t)$ look like?
$$\delta(t)\ldots\text{ Dirac delta impulse}\\
u(t)\ldots\text{unit     step function}$$

Will the impulse at the origin start from 0 downwards or from 1?
I'm having trouble because the unit step is discontinuous at t=0, so I don't know how it will shift the impulse.
I am talking about continuous time.
Note that $\delta(t)$ is no function but a distribution. This means that it doesn't have function values but it is only defined by its integral. Consequently, the (generalized) function $x(t)=-\delta(t)+u(t)$ has no value at $t=0$ (the only useful value would be $-\infty$ but that's kind of hard to sketch). For sketching the function graph, you can define function values for $t\neq 0$ in the usual way. At $t=0$ you simply sketch a negative $\delta$-impulse, for $t&lt;0$ the function is zero, and for $t&gt;0$ you assign it the value $1$.