Graph problem (1 Viewer)

[Aerlinn]

In a question, I came across this graph: y= sq. root of-3(4-x)
I graphed it on my calculator, and it didn't come up as I expected. It appears to have been dilated by sq. root of 3 parallel to the y-axis, then translated 4 units to the right. What happened to the two reflections? Reflection in the x and y-axis, since there are two negative signs? Since the graphs not symmetrical there's no reason for the reflections to disappear...
Can someone please explain it to me?
~~~
Also, a little sport of confusion. If you have: 3/4= (2x+3)^4, and you want to take the 4th root of both sides, you would get a positive and negative answer wouldn't you? How would you write the next step out? (I get confused with taking roots sometimes)

Thanks a heap ^^
:wave:

 

[bos1234]

2nd qn
-----------------

4th root, you can only get a positive answer because 4th root of a negative s undefined

similarly 6th root, 8th root, 10th root etc all positive answer

 

[ssglain]

I'm not quite sure which two reflections you're referring to.

For y = sqrt{-3(4-x)} to be defined, we need -3(4-x)>0, which leads to the domain being x>4 only. The range is y>0 since we're only graphing the positive square root.

If we rewrite this as y = sqrt(3)*sqrt{-(4-x)}, i.e. y = sqrt(3)*sqrt(x-4) - it can be seen that indeed this is a transformation of the basic y = sqrt(x) graph, obtained by shifting 4 units right and dilating by a factor of sqrt(3) parallel to the y-axis.

 

[ssglain]

Oh and for Qn 2 - bos1234 I think you misread the question (it took me a second take, too). Now, it's true that, as long as you don't want to get stuck in the complex number system, you can only 4th root positive numbers. But if you do 4th root a positive number, you produce both positive and negative numbers just as sqrt(4) = +-2. I think the formal definition runs along the lines of nth-root(x^n) = |x|, given n is (positive?) even number.