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For these questions, find the domain and determine if there are any vertical asymptote(s) and/or holes.

 

a) \(g(x)=\frac{4x^2+6}{x+1}\)

 

b) \(h(x)=\frac{5x^2+2}{x^2-9}\)

 Oct 13, 2017
 #1
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g(x)   =  [ 4x^2 + 6 ]  /  [ x + 1]

 

Domain  =   all  reals except  for  x  = -1    [ this makes the denominator  = 0 ]

There will be something known as a "slant" asymptote and a vertical asymptote  at x = -1

See the graph, here :  https://www.desmos.com/calculator/orq4zkm2xz

 

 

h(x)  = [ 5x^2 + 2] / [ x^2  -  9 ]

 

Domain = all reals except x = -3   and x  = 3   [ these make the denominator  = 0 ]

Vertical asymptotes will occur at  x = - 3  and x  =3

See the graph, here : https://www.desmos.com/calculator/6flwyghtuw

 

 

cool cool cool

 Oct 13, 2017

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