Question 246701
{{{y=sqrt(3x-5) + 6}}}
The parent function is {{{y = sqrt(x)}}}

>>Here's how I solved it: {{{y = 6 + sqrt(3(x-5/3))}}}

This is exactly what you should have done! But you should have stopped here. There is nothing useful to be gained by factoring out {{{sqrt(3)}}}.

>> x - (5/3) > 0
>> x > (5/3)

>> D: x>(5/3)

All correct.<br>
>> the graph is 5 units down, 6 units to  the left of the parent function. 
>> R:  y<= 0
These are not correct.<br>
The {{{(x - 5/3)}}} means the graph is shifted to the right by 5/3.
The 6 added to the square root means the graph is shifted 6 units up.<br>
As for the range, the equation says that y equals 6 plus a square root. Square roots are never negative. They range from zero to infinitely large positive numbers. After we add six to the square root we will end up with numbers from 6 (when the square root is zero) to infinitely large positive values. So
R: {{{y >= 6}}}<br>
(BTW, the factored out 3 inside the square root means that the graph is compressed horizontally by a factor of 3.)<br>
Here are the graphs so you can see all this "in action". The green is the parent. The orange is your function. (The horizontal compression by a factor of 3 explains the different curvature of the two graphs.)
{{{graph(600, 600, -1, 15, -1, 15, 6+sqrt(3x-5), sqrt(x))}}}