Question 1000931
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If I understand correctly, to find the vertical asymptote you set the numerator to zero and solve.
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It is incorrect.


To find the vertical asymptote you set the &nbsp;<U>denominator</U>&nbsp; to zero and solve.


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{{{f(x)}}} = {{{5 + e^(-x^2) }}}.


No vertical asymptotes.
Horizontal asymptote &nbsp;y = 5, &nbsp;because &nbsp;{{{e^(-x^2)}}} &nbsp;tends to zero &nbsp;when &nbsp;x --> {{{infinity}}}.

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{{{f(x)}}} = {{{(x^3-9x+4)/(2x^3+6x+7) }}}


Horizontal asymptotes &nbsp;{{{1/2}}} &nbsp;when x ---> {{{infinity}}},

and &nbsp;{{{1/2}}} &nbsp;when x ---> {{{-infinity}}}.


Vertical asymptotes are there where the denominator is zero: &nbsp;{{{2x^3 + 6x + 7}}} = {{{0}}}. 


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{{{f(x)}}} = {{{(1)/(ln(2x+4)) }}}


No horizontal asymptotes.


Vertical asymptote at &nbsp;x = -1.5 &nbsp;&nbsp;(where &nbsp;2x+4 = 1 &nbsp;and &nbsp;ln(2x+4) = ln(1) = 0).


The function is defined at &nbsp;x > -2 &nbsp;&nbsp;(where &nbsp;2x+4 > 0).


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{{{f(x)}}} = {{{(x^2-3x-10)/(sqrt(x+2))}}}.


No horizontal asymptotes.


Vertical asymptote &nbsp;x = -2 &nbsp;&nbsp;(where x+2 = 0).


The function is defined at &nbsp;x > -2.