Use the properties of real numbers to rewrite the expression. 2/5 * (-7) * 5/2
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When taking square roots, you can't take square roots of negative roots of negative numbers. So, what will work for the domain of u(x) is what makes u(x) zero or more. We can make an inequality for that.
u(x) ≥ 0.
9x + 27 ≥ 0 by squaring both sides
9x ≥ -27
x ≥ -3
So the domain of the function is when x ≥ -3 is true.
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