+0  
 
0
1575
5
avatar+209 

f(x)= sqrt(x-4)                      as x<=8

        

        (x^3-100)/ x-10            as x>8

 

1.)I need the domain, horizontal and vertical asymptotes, and holes.

2.)Is there supposed to be two VA's an HA's since there's two functions?

 

So far I got-

domain: {x| all real numbers except for x cannot equal 10}

VA: x=10

HA= 0

 Jan 28, 2016

Best Answer 

 #1
avatar+129852 
+15

For the first function....the domain  will be [4, 8]    since, in this case,  the number in a square root can never be < 0 and we have a restricted interval 

 

For the second function....the domain is (8,10)  U (10, inf)

 

There will be a vertical asymptote at x = 10 for the second  function

 

 

 

cool cool cool

 Jan 28, 2016
edited by CPhill  Jan 28, 2016
 #1
avatar+129852 
+15
Best Answer

For the first function....the domain  will be [4, 8]    since, in this case,  the number in a square root can never be < 0 and we have a restricted interval 

 

For the second function....the domain is (8,10)  U (10, inf)

 

There will be a vertical asymptote at x = 10 for the second  function

 

 

 

cool cool cool

CPhill Jan 28, 2016
edited by CPhill  Jan 28, 2016
 #2
avatar+8581 
0

5 stars from me :)

 Jan 28, 2016
 #3
avatar+209 
0

Thank you! There is no HA or holes then?

 Jan 28, 2016
 #4
avatar+129852 
0

I suppose, in some odd way, that the frist function could be considered to be "bounded" by the horizontal asymptote  y= sqrt(8-4) = 2  ....but.....this is only because of the restricted domain......this function would not have an upper bound, normally.....

 

 

 

cool cool cool

 Jan 28, 2016
 #5
avatar+209 
0

Oh okay, thank you, I understand it much better now!

 Jan 28, 2016

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