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Find the vertical asymptote(s) of f(x)=\frac{\left(x^{2}+3x+6\right)}{x^{2}-4}

x = −1, 2
x = −2, 2
x = 1, −2
x = −1, 1

User Haomin
by
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1 Answer

5 votes

Answer:

x = −2, 2

Explanation:


f(x)=(\left(x^(2)+3x+6\right))/(x^(2)-4)

Find the horizontal asymptotes by comparing the degrees of the numerator and denominator.

Vertical Asymptotes:

x=−2,2Horizontal Asymptotes:

y=1

No Oblique Asymptotes

Full Explanation

The line x=L is a vertical asymptote of the function
y=(x^(2) + 3 x + 6)/(x^(2) - 4) 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.

So, find the points where the denominator equals 0 and check them.

x=−2, check:
\lim_(x \to -2^+)\left((x^(2) + 3 x + 6)/(x^(2) - 4)\right)=-\infty

Since the limit is infinite, then x=−2 is a vertical asymptote.

x=2, check:
\lim_(x \to 2^+)\left((x^(2) + 3 x + 6)/(x^(2) - 4)\right)=\infty

Since the limit is infinite, then x=2 is a vertical asymptote.

Vertical asymptotes: x=−2; x=2

User Lko
by
3.6k points