What are the equations of the asymptotes of the graph of the function f(x)=3x^2-2x-1/x^2+3x-10

Answer:Vertical asymptotes [tex]x = -5\\\\x = 2[/tex] Horizontal asymptote[tex]y=3[/tex]Step-by-step explanation:We have the function [tex]f(x) = \frac{3x ^ 2-2x-1}{x ^ 2 + 3x-10}[/tex] and we want to find its asymptotes. First we find their vertical asymptotes. To do this we must factor the denominator of the expression. [tex]x ^ 2 + 3x-10[/tex]We must look for two numbers that when adding them obtain 3, and when multiplying those same numbers, obtain -10. The searched numbers are 5 and -2 Then the factors are: [tex](x + 5)(x-2)\\\\x ^ 2 + 3x-10 = (x + 5)(x-2)[/tex]Since the division between zero is not defined then when x tends to -5 the function tends to infinity and when x tends to 2 the function tends to infinity. So the vertical asymptotes will be the straight lines: [tex]x = -5\\\\x = 2[/tex] The horizontal asymptote is calculated as:[tex]\lim_{x \to\infty}f(x)\\\\= \lim_{x \to\infty}\frac{3x ^ 2-2x-1}{x ^ 2 + 3x-10}[/tex]The highest exponent of the function is 2. Then the terms with exponents less than 2 tend to zero[tex]\lim_{x \to\infty}\frac{3x ^ 2-2x-1}{x ^ 2 + 3x-10} =\lim_{x \to\infty}\frac{3x ^ 2}{x ^ 2}\\\\ = \lim_{x \to\infty}\frac{3}{1} = 3\\\\[/tex]Then [tex]\lim_{x \to\infty}f(x)=3[/tex]Therefore the horizontal asymptote is:[tex]y=3[/tex]Observe the attached image