bpeoples is right ... but here's the answer with actual physics equations ...
Weight of Car (mg, also known as W) points straight down
road inclined at (theta - 0 is as close as I can get) angle
you split W into vectors perpendicular and parallel to slanted road surface: W sin 0 perpendicular to road surface, W cos 0 parallel to road surface
Normal Force from road (to prevent car sinking into ground) = Fn = W sin 0
so, centripetal acceleration (ac) = v^2/r, where v = linear velocity of car (the stuff it says on spedometer or whatever) and r = radius of curve (assuming curve is along a circle, not an ellipse
between wheels and road, there is also force of friction (Ff) = u (latin letter mu - coefficient of friction, which depends on road condition ... snow, rain, etc) * Fn
so as long as the W cos 0 (force making car slide down slope) - Ff = mac (mass of car times centripetal acceleration) car will stay stable in turn. If left of equation too big, car slides down slope, if left side too small, then car skids up the slope
as you can see, diagrams are our friend, but I'm sorry I don't have one. This is a common problem in physics classes (probably AP or college level, although Honors might also have it) and is usually found in the section on "friction" and maybe "splitting vectors into their component vectors" or "Newton's Second Law: F = ma"
hope this helps! ^_^