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Please help w/ this!!

 

Question 1:

 

Question 2:

 Mar 4, 2019
 #1
avatar+81 
+1

Part 1:

When shifting an equation, if the value shifting is inside the equation, you go the opposite direction. So, for example, if it was:

\(f(x)=|x+7|\), then you would shift 7 to the left.

So, for your problem of \(f(x)= |x-3|+2\), when you replace x with x+2, the graph will shift 2 units to the left.

 

Part 2:

\(f(x)=\frac{1}{2}\sqrt{x+5}\)

So, square roots have specific domain restrictions, such that the value inside can't be negative because no number, when squared, can be negative, so when doing the opposite it can't be negative. 

So, with that in mind, x can be -5 but no lower.

So the domain would be \([-5,\infty)\)

 

The range is all possible y values. Since the square root can put out either positive or negative numbers, the range is \((-\infty,\infty)\).

 Mar 4, 2019
 #2
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Thanks a lot!

 

So for this question, how would I graph this, and what graph would match the function?

function:

 

Guest Mar 4, 2019
 #3
avatar+81 
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I'm not sure how to explain how to graph it, but the correct graph is the bottom left one. My best suggestion for graphing would be to test it out on www.desmos.com/calculator.

MemeLord  Mar 4, 2019
 #4
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That's ok! I appreciate you doing your best to help out :)

Guest Mar 4, 2019
 #5
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Question 2

 

The domain  is  [-5, inf )

 

However.....from a positive square root we can only get 0  or a positive 

 

So.....the range is  [0, inf)

 

here's a graph to show this : https://www.desmos.com/calculator/quxajtdn9b

 

 

cool cool cool

 Mar 4, 2019
 #6
avatar+81 
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@CPhill But what about the requirement of there being a \(\pm\) outside of every square root?

MemeLord  Mar 4, 2019

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