The domain is simply the range of x values.
-4≤x<1, x=1, x>1. From this set we see that x is continuously defined for x≥-4, but is not defined for x<-4. A) The domain is therefore [-4,∞).
B) As x→∞, f(x)→-∞, so this the minimum value. The minimum value of the range→-∞, because for -4≤x<1, the minimum value of f(x) is f(-4)=1; for x=1, the minimum value is 9; for x>1, the minimum value is negatively unbounded. The minimum of these three values is negatively unbounded (-∞). The maximum values for these three regions are: f(1-)=6- (slightly less than 6 for x slightly less than 1); f(1)=9; f(1+)=2- (slightly less than 2 for x slightly more than 1). The maximum of these values is 9. The range is therefore (-∞,9].
C) The x-intercepts are found by solving f(x)=0 (y-axis). x+5=0 when x=-5, but this value of x is outside the domain, so we reject this. f(1)=9 and 9≠0, so there can be no x-intercept here because (1,9) is just a point. f(x)=-x+3=0 when x=3 and 3>1 so (3,0) is the x-intercept.
The y-intercepts are found by plugging x=0 into the piecewise function. The only part that's valid for x=0 is -4≤x<1, so this means f(0) when f(x)=x+5, that is, f(0)=5 is the y-intercept (0,5).
D) The graph is discontinuous. (1-,6-), (1,9), (1+,2-) are the discontinuities. Around x=1, the graph jumps from f(x)~6 to f(x)=9 to f(x)~2.