Question 1026206
Vertical asymptotes : Find where the denominator equals zero.
{{{4x^2-1=0}}}
{{{4x^2=1}}}
{{{x^2=1/4}}}
There are 2 of them ({{{x=1/2}}}) and ({{{x=-1/2}}}).
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Horizontal asymptotes : Look at the limit of the quotient as x goes to {{{infinity}}} and {{{-infinity}}}
{{{lim(x->infinity,((2x^2-5x-1))/((4x^2-1)))=lim(x->infinity,((2-5/x-1/x^2))/((4-1/x^2)))=(2-0-1)/(4-0)=2/4=1/2}}}
There is 1 horizontal asymptote ({{{y=1/2}}}).
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