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
