SOLUTION: Here is my "domain" problem if it makes any sense: Find domain of the function y=2sqrt of 4x+10 Teresa M.

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Question 142579: Here is my "domain" problem if it makes any sense:

Find domain of the function y=2sqrt of 4x+10
Teresa M.

Found 2 solutions by jim_thompson5910, scott8148:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

2%2Asqrt%284x%2B10%29 Start with the given expression

Remember you cannot take the square root of a negative value. So that means the argument 4x%2B10 must be greater than or equal to zero


4x%2B10%3E=0 Set the inner expression greater than or equal to zero

4x%3E=0-10Subtract 10 from both sides


4x%3E=-10 Combine like terms on the right side


x%3E=%28-10%29%2F%284%29 Divide both sides by 4 to isolate x



x%3E=-5+%2F+2 Reduce


So that means x must be greater than or equal to -5+%2F+2 in order for x to be in the domain

So the domain in set-builder notation is


So here is the domain in interval notation: [-5+%2F+2,)



Notice if we graph y=2%2Asqrt%284x%2B10%29 , we get
+graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+2%2Asqrt%284x%2B10%29%29+ notice how the graph never crosses the line x=-5+%2F+2. So this graphically verifies our answer.

and we can see that x must be greater than or equal to -5+%2F+2 in order to lie on the graph

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
the domain means allowable values of x __ values that don't result in undefined or imaginary quantities

in this case, the argument of the square root (4x+10) must not be negative

4x+10>=0 __ 4x>=-10 __ x>=-2.5

the domain is all real numbers greater than or equal to -2.5