How to find the asymptotes of #f(x) = (x+3) /( x^2 + 8x + 15)# ?

Answer 1

Simplify #f(x)# to see that there is a horizontal asymptote #y=0#, vertical asymptote #x=-5# and hole (removable singularity) at #(-3, 1/2)#

#f(x) = (x+3)/(x^2+8x+15)#

#=(x+3)/((x+3)(x+5))#

#=1/(x+5)#

with exclusion #x != -3#

This function has horizontal asymptote #y=0# since #1/(x+5)->0# as x->+-oo#

It has a vertical asymptote #x=-5#, where the denominator is zero, but the numerator non-zero.

It has a hole (removable singularity) at #(-3, 1/2)# since the left and right limits exist at #(-3, 1/2)# and are equal.

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Answer 2

Greyson Brown

To find the asymptotes of $f(x) = \frac{x+3}{x^2 + 8x + 15}$ , follow these steps:

Factor the denominator, $x^2 + 8x + 15$ , to determine any potential vertical asymptotes.
Identify any horizontal or slant asymptotes by examining the degrees of the numerator and denominator polynomials.

Applying these steps to the given function, $f(x) = \frac{x+3}{x^2 + 8x + 15}$ :

Factor the denominator: $x^2 + 8x + 15 = (x + 3)(x + 5)$ .
Determine if there are any vertical asymptotes by setting the denominator equal to zero: $x^2 + 8x + 15 = 0$ .
- Solving this equation yields no real solutions, indicating no vertical asymptotes.
Identify horizontal or slant asymptotes by comparing the degrees of the numerator and denominator.
- Since the degree of the numerator (1) is less than the degree of the denominator (2), there is a horizontal asymptote at $y = 0$ .

Therefore, the function $f(x) = \frac{x+3}{x^2 + 8x + 15}$ has no vertical asymptotes and a horizontal asymptote at $y = 0$ .

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.