Answer:1st problem: Find all asymptotes of [tex]f(x)=x^2+\frac{1}{x}[/tex].Horizontal: noneVertical: x=0Slant: none2nd problem: Find all asymptotes of [tex]f(x)=\frac{x^2+1}{x}[/tex].Horizontal: noneVertical: x=0Slant: y=xStep-by-step explanation:I'm going to do two problems:1st problem: Find all asymptotes of [tex]f(x)=x^2+\frac{1}{x}[/tex].2nd problem: Find all asymptotes of [tex]f(x)=\frac{x^2+1}{x}[/tex].Problem 1:[tex]f(x)=x^2+\frac{1}{x}[/tex]I'm going to combine the terms; find a common denominator.This means I will need to multiply first term by x/x:[tex]f(x)=\frac{x^3}{x}+\frac{1}{x}[/tex][tex]f(x)=\frac{x^3+1}{x}[/tex]Horizontal asymptotes will not exist if the degree of the numerator is larger than the degree of the denominator. This is the case here because the degree of the numerator is 3 while the degree of the denominator is 1.Slant asymptotes will only exist when the degree of the numerator is one more than the degree of the denominator. Since 3 is not one more than 1, then we don't have any slant aymptotes.The vertical asymptotes will cause the denominator to be 0 while the numerator will be another number other than 0.x is 0 when x=0. We plug into top, 0^3+1=1.x=0 gives us 1/0 so you definitely have a vertical asymptote at x=0.Problem 2:There are no horizontal asymptotes again because degree of top is more than degree of bottom.Slant asymptote will exist when the degree of the top is 1 more than the degree of the bottom which is here.The degree of the top is 2 while the degree of the bottom is 1.Let's divide to find the slant aysmptote:[tex]\frac{x^2+1}{x}=\frac{x^2}{x}+\frac{1}{x}=x+\frac{1}{x}[/tex]So the slant aysmptote will be the quotient part of this which is x.So the slant asymptote is y=x.The vertical asymptote will happen when denominator is 0 and the top is something that is not 0. (It should remain this way after simplifying as well).So again we are dividing just by x.So x is 0 when x=0.Since 0^2+1=1 and not 0 then you have a vertical asymptote at x=0.