Answer:
As x = -5, x = 2 and y = 3 are the equations of the asymptotes of the graph of the function [tex]f(x)=\frac{3x^{2} -2x - 1}{x^{2}+3x-10 }[/tex].
Therefore, x = -5, x = 2 and y = 3 is the right option.
Step-by-step explanation:
As the given function is
[tex]f(x)=\frac{3x^{2} -2x - 1}{x^{2}+3x-10 }[/tex]
Determining Vertical Asymptote:
The line x = L is a vertical asymptote of the function if if the limit of the function (one-sided) at this point is infinite.
In other words, it means that possible points are points where the denominator equals 0 or doesn't exist.
Such as
[tex]x^{2} +3x -10 = 0[/tex]
[tex](x-2)(x-5)=0[/tex]
[tex]x=2[/tex] [tex]or[/tex] [tex]x=-5[/tex]
x=−5 , check:
[tex]\lim_{x \to -5^{+}}({3x^{2}-2x-11 \over x^{2}+3x-10 }) = -\infty[/tex]
Since, the limit is infinite, then x = -5 is a vertical asymptote.
x = 2, check:
[tex]\lim_{x \to 2^{+}}({3x^{2}-2x-11 \over x^{2}+3x-10 }) = -\infty[/tex]
Since the limit is infinite, then x = 2 is a vertical asymptote.
Determining Horizontal Asymptote:
Line y=L is a horizontal asymptote of the function y = f(x), if either
[tex]\lim_{x \to \infty^{}}{f(x)=L}[/tex] or [tex]\lim_{x \to -\infty^{}}{f(x)=L},[/tex] and L is finite.
Calculating the limits:
[tex]\lim_{x \to \infty^{}}({3x^{2}-2x-11 \over x^{2}+3x-10 }) = 3[/tex]
[tex]\lim_{x \to -\infty^{}}({3x^{2}-2x-11 \over x^{2}+3x-10 }) = 3[/tex]
Thus, the horizontal asymptote is y=3.
So, x = -5, x = 2 and y = 3 are the equations of the asymptotes of the graph of the function [tex]f(x)=\frac{3x^{2} -2x - 1}{x^{2}+3x-10 }[/tex].
Therefore, x = -5, x = 2 and y = 3 is the right option.
Keywords: asymptote, vertical asymptote, horizontal asymptote, equation
Learn more about asymptotes from brainly.com/question/11268104
#learnwithBrainly
