Recall that the equation for a line can be expressed in the form

#y=mx+b# where #m=(x_2-x_2)/(y_2-y_1)=("vertical rise")/("horizontal run")#

Your #y#-intercept is the value of #y# when #x=0#. When you plug a zero into the above equation you get the following:

#y=mx+b#

#y=m(0)+b#, where we let #x=0#

#y=0+b#, because anything times zero is zero

#y=b#

So the #y#-intercept is the value of #y# when you set #x=0#. But you were given a problem where #x# can *never* be #0#. You were given that #x=-3# and, clearly, #-3 != 0#. Well, if #x != 0# (because it's -3), then the #y#-intercept does not exist.

What about the slope? Remember the slope is #("rise")/("run")#. If you want to know how "steep" a line is, you must divide your vertical distance by your horizontal distance. A typical line (not your question above) looks like this:

In order to turn this into a vertical line, you would have to make the run part really short, and the rise part really big. The graph starts to look like this:

So when you are given #x=-3#, you have a vertical line. A vertical line has zero run because there is simply no left or right steepness. Additionally, the rise becomes infinite because there is only up and down in a vertical line. It's all rise and no run! Thus, the slope, #m=oo#.