find all horizontal asymptotes of the following function.\n\n$f(x)=\\frac{10x - 6}{5x^{2}-13x + 6}$

Explanation:

Step1: Determine the degrees of numerator and denominator

The degree of the numerator (10x - 6) (highest power of (x)) is (n = 1). The degree of the denominator (5x^{2}-13x + 6) is (m=2).

Step2: Apply the rule for horizontal asymptotes

When (n<m) (where (n) is the degree of the numerator and (m) is the degree of the denominator), the horizontal asymptote is (y = 0). We can also use the limit formula. We know that (\lim_{x\rightarrow\pm\infty}f(x)=\lim_{x\rightarrow\pm\infty}\frac{10x - 6}{5x^{2}-13x + 6}). Divide numerator and denominator by (x^{2}): [ \begin{align*} \lim_{x\rightarrow\pm\infty}\frac{\frac{10x}{x^{2}}-\frac{6}{x^{2}}}{\frac{5x^{2}}{x^{2}}-\frac{13x}{x^{2}}+\frac{6}{x^{2}}}&=\lim_{x\rightarrow\pm\infty}\frac{\frac{10}{x}-\frac{6}{x^{2}}}{5-\frac{13}{x}+\frac{6}{x^{2}}}\ \end{align*} ] As (x\rightarrow\pm\infty), (\frac{10}{x}\rightarrow0), (\frac{6}{x^{2}}\rightarrow0), (\frac{13}{x}\rightarrow0) and (\frac{6}{x^{2}}\rightarrow0). So (\lim_{x\rightarrow\pm\infty}\frac{\frac{10}{x}-\frac{6}{x^{2}}}{5-\frac{13}{x}+\frac{6}{x^{2}}}=0)

Answer:

The horizontal asymptote is (y = 0)