The graph of y=|x-4|+3 and the graph of y=|x-4| open the same direction,
because y=|x-4|+3 is y=|x-4| shifted 3 units up, entirely. So if one opens upwards, the other does so as well.
So consider the graph of y=|x-4|. This graph is the graph of y=|x| shifted 4 units right.
To show this, in the function y=|x|, consider the points (-2, 2) (because for x=-2, y=|x|=|-2|=2)
and point (2, 2).
Now consider y=|x-4|, if y = 2, then
i) x-4 = 2, which means x=6 OR
ii) x-4 =-2, which means that x=2
So we have the points (2,2) and (6, 2)
the value 2 was taken at -2 and 2 in y=|x|
the value 2 is taken at x=2 and x=6 in y=|x-4|
this proves that the graph of y=|x-4| is the graph of y=|x| shifted 4 units right.
This means that both graphs open the same way.
Thus consider the graph of y=|x|,
i) for x=0, y=0,
ii) for x>0, for larger values of x, we have larger values of y,
for example: 5>3, and |5|>3, so the graph increases
iii) for x<0, for smaller values of x, we have larger values of y
for example: -5<-3, |-5|=5>3=|-3|, thus the graph is decreasing as we move right, up to 0, then it starts increasing.
All these mean that the graph of y=|x| opens upwards, so the graph of y=|x-4|+2 also opens upwards.
Answer: b
