The presence of a midpoint will result in what type of segments?
2 answers:
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First, note that 
is always positive (except for x=0), so 
must be always negative.
Thus, the only plausible graphs are 1 and 3 since they are below the x-axis.
Now, and are only defined for x≥0, because only for these x'es we can take the square root.
Note that the third graph has domain (-infinity, 0], so it is not the right one, while 1 is ok.
Answer: first graph
Thus, the only plausible graphs are 1 and 3 since they are below the x-axis.
Now,
Note that the third graph has domain (-infinity, 0], so it is not the right one, while 1 is ok.
Answer: first graph
Find the commomon denominator of 6 and 5.
It's 6 · 5 = 30.
Divide by fraction is the same like multiply by reciprocal of this fraction: