Horizontal asymptotes can occur on both sides of the y-axis, so don’t forget to look at both sides of your graph.
For example, if your equation is f(x)=x2+x+33x2+5{\displaystyle f(x)={\frac {x^{2}+x+3}{3x^{2}+5}}}, remove all but the leading terms to get x23x2{\displaystyle {\frac {x^{2}}{3x^{2}}}}. As another example, your equation might be f(x)=2+5x2+4x3x4+6{\displaystyle f(x)={\frac {2+5x^{2}+4x^{3}}{x^{4}+6}}}. After you remove all but the leading terms, you’ll have 4x3x4{\displaystyle {\frac {4x^{3}}{x^{4}}}}. To give you another example, if you have f(x)=x3−14+x2{\displaystyle f(x)={\frac {x^{3}-1}{4+x^{2}}}}, ignore the constants to get f(x)=x3x2{\displaystyle f(x)={\frac {x^{3}}{x^{2}}}}.
To go with the previous example of x23x2{\displaystyle {\frac {x^{2}}{3x^{2}}}}, cancel out both of the x2{\displaystyle {x^{2}}} to get 13{\displaystyle {\frac {1}{3}}}, which shows you the horizontal asymptote. For the other previous example of 4x3x4{\displaystyle {\frac {4x^{3}}{x^{4}}}}, cancel out the top x3{\displaystyle {x^{3}}} and take away x3{\displaystyle {x^{3}}} from the denominator to get 4x{\displaystyle {\frac {4}{x}}} which becomes 0{\displaystyle 0}. For the last example of f(x)=x3x2{\displaystyle f(x)={\frac {x^{3}}{x^{2}}}}, remove x2{\displaystyle {x^{2}}} from the numerator and denominator to get x{\displaystyle x}.
Going back to our first example of f(x)=x2+x+33x2+5{\displaystyle f(x)={\frac {x^{2}+x+3}{3x^{2}+5}}}, you ended up with 13{\displaystyle {\frac {1}{3}}}. In this example, the HA is 13{\displaystyle {\frac {1}{3}}}.
For our previous example of f(x)=2+5x2+4x3x4+6{\displaystyle f(x)={\frac {2+5x^{2}+4x^{3}}{x^{4}+6}}}, you ended up with 0{\displaystyle 0}, so the horizontal asymptote is 0{\displaystyle 0}, which is also the x-axis.
In the previous example that started with f(x)=x3−14+x2{\displaystyle f(x)={\frac {x^{3}-1}{4+x^{2}}}}, you were left with x{\displaystyle x}. Since x{\displaystyle x} is larger than a nonexistent denominator, no HA is possible with this equation.
