How do you find the vertical, horizontal and slant asymptotes of: #(x^2-25)/(x^2+5x)#?

Question

Ariana Alvarez · Accepted Answer

vertical asymptote: #x=0#
horizontal asymptote: #f(x)=1#
slant asymptote: does not exist

Finding the Vertical Asymptote Given, #f(x)=(x^2-25)/(x^2+5x)# Factor the numerator. #f(x)=((x+5)(x-5))/(x(x+5)# Cancel out any factors that appear in the numerator and denominator. #f(x)=(x-5)/x# Set the denominator equal to #0# and solve for #x#. #color(green)(|bar(ul(color(white)(a/a)color(black)(x=0)color(white)(a/a)|)))# Finding the Horizontal Asymptote Given, #f(x)=(color(darkorange)1x^2-25)/(color(purple)1x^2+5x)# Divide the #color(darkorange)("leading coefficient")# of the leading term in the numerator by the #color(purple)("leading coefficient")# of the leading term in the denominator. #f(x)=color(darkorange)1/color(purple)1# #color(green)(|bar(ul(color(white)(a/a)color(black)(f(x)=1)color(white)(a/a)|)))# Finding the Slant Asymptote Given, #f(x)=(x^2-25)/(x^2+5x)# There would be a slant asymptote if the degree of the leading term in the numerator was #1# value larger than the degree of the leading term in the denominator. In your case, we see that the degree in both the numerator and denominator are equal. #:.#, the slant asymptote does not exist.

Lincoln Anderson · Answer

undefined To find the vertical, horizontal, and slant asymptotes of the function $ \frac{x^2 - 25}{x^2 + 5x} $, follow these steps:

**Vertical Asymptotes:**

Vertical asymptotes occur where the denominator of a rational function is equal to zero, provided the numerator is not zero at the same point, as this would result in a hole rather than a vertical asymptote.

Set the denominator $ x^2 + 5x $ equal to zero and solve for $ x $:

$$ x^2 + 5x = 0 $$
$$ x(x + 5) = 0 $$

From this equation, $ x = 0 $ and $ x = -5 $ are the values where the denominator is zero.

**Horizontal Asymptotes:**

To find the horizontal asymptote, compare the degrees of the numerator and the denominator:

- If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $ y = 0 $.
- If the degrees are equal, divide the leading coefficients.
- If the degree of the numerator is greater than the degree of the denominator by one, there is a slant (oblique) asymptote.
- If the degree of the numerator is greater than the degree of the denominator by more than one, there is no horizontal asymptote.

Here, both the numerator and denominator have the same degree of 2. So, divide the leading coefficients:

$$ \text{Horizontal asymptote} = \frac{1}{1} = 1 $$

So, $ y = 1 $ is the horizontal asymptote.

**Slant Asymptote:**

Since the degrees of the numerator and denominator are the same, there is no slant asymptote.

In summary:
- Vertical asymptotes: $ x = 0 $ and $ x = -5 $
- Horizontal asymptote: $ y = 1 $
- No slant asymptote.