Question

What are the equations of the asymptotes of the graph of the function f (x) = StartFraction 3 x squared minus 2 x minus 1 Over x squared + 3 x minus 10 EndFraction?
x = –5, x = 2 and y = 3
x = –2, x = 5 and y = 3
x = 3, y = –5, and y = 2
x = 3, y = –2, and y = 5

Answer 1

I believe the answer is a.

Answer 2

Answer:

As x = -5, x = 2 and y = 3 are the equations of the asymptotes of the graph of the function $f(x)=\frac{3x^{2} -2x - 1}{x^{2}+3x-10 }$.

Therefore, x = -5, x = 2 and y = 3 is the right option.

Step-by-step explanation:

As the given function is

$f(x)=\frac{3x^{2} -2x - 1}{x^{2}+3x-10 }$

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

$x^{2} +3x -10 = 0$

$(x-2)(x-5)=0$

$x=2$  $or$  $x=-5$

x=−5 , check:

$\lim_{x \to -5^{+}}({3x^{2}-2x-11 \over x^{2}+3x-10 }) = -\infty$

Since, the limit is infinite, then x = -5 is a vertical asymptote.

x = 2, check:

$\lim_{x \to 2^{+}}({3x^{2}-2x-11 \over x^{2}+3x-10 }) = -\infty$

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

$\lim_{x \to \infty^{}}{f(x)=L}$ or $\lim_{x \to -\infty^{}}{f(x)=L},$ and L is finite.

Calculating the limits:

$\lim_{x \to \infty^{}}({3x^{2}-2x-11 \over x^{2}+3x-10 }) = 3$

$\lim_{x \to -\infty^{}}({3x^{2}-2x-11 \over x^{2}+3x-10 }) = 3$

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 $f(x)=\frac{3x^{2} -2x - 1}{x^{2}+3x-10 }$.

Therefore, x = -5, x = 2 and y = 3 is the right option.

Keywords: asymptote, vertical asymptote, horizontal asymptote, equation

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