How do you find all the asymptotes for function #f(x)= (17x)/( 2x^2 + 3)#?

1 Answer
Sep 24, 2015

Answer:

Its horizontal asymptote is the #x#-axis (#y=0#) as #x-> pm infty# and it has no vertical asymptotes.

Explanation:

The line #y=0# is a horizontal asymptote of the rational function #f(x)=(17x)/(2x^2+3)# since the degree of the bottom (denominator) is greater than the degree of the top (numerator). Furthermore, the denominator #2x^2+3# has no real roots (the roots are the imaginary numbers #pm i sqrt{3/2}#). Hence, the function has no vertical asymptotes.

To confirm the horizontal asymptote, you can do the following limit calculation, which I'm not rigorously justifying:

#lim_{x-> pm infty}(17x)/(2x^2+3)=lim_{x-> pm infty}(17x)/(2x^2+3) * (1/x^2)/(1/x^2)#

#=lim_{x-> pm infty}(17/x)/(2+3/x^2)=(lim_{x->pm infty}(17/x))/(lim_{x->pm infty} 2+lim_{x->pm infty}(3/x^2))=0/(2+0)=0#