How do you find the asymptotes for #f(x) =(x^2 - 16)/(x^2- 5x + 4)#?

Question

Theodore Amidon · Accepted Answer

Vertical asymptotes are #x=1# and #x=4# and horizontal asymptote is #y=1#

The vertical asymptotes are given by restrictions on the domain and hence come from the zeroes of the denominator.

Putting #x^2-5x+4=0# ad solving this will give domain

#x^2-5x+4=0# i.e. #x^2-x-4x+4=0# or

#x(x-1)-4(x-1)=0# or #(x-1)(x-4)=0#

i.e. #x= 1 or 4#

Hence #x# can take all values other than #1# and #4#

and vertical asymptotes are #x=1# and #x=4#.

Since, numerator and denominator of the function are of same degree, horizontal asymptote is found by dividing the leading terms

i.e. #x^2/x^2# i.e. #1#.

Hence horizontal asymptote is #y=1#

Chloe Alexander · Answer

undefined To find the asymptotes for the function $ f(x) = \frac{x^2 - 16}{x^2 - 5x + 4} $, you need to check for vertical asymptotes and horizontal asymptotes.

Vertical asymptotes occur where the denominator becomes zero but the numerator doesn't. Set the denominator equal to zero and solve for $ x $ to find vertical asymptotes.

$$ x^2 - 5x + 4 = 0 $$

This quadratic equation factors as:

$$ (x - 4)(x - 1) = 0 $$

So, the vertical asymptotes occur at $ x = 4 $ and $ x = 1 $.

Horizontal asymptotes can be found by examining the behavior of the function as $ x $ approaches positive or negative infinity. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $ y = 0 $. If the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is greater, there is no horizontal asymptote.

In this case, the degrees of the numerator and denominator are both 2, and the leading coefficients are both 1. So, the horizontal asymptote is:

$$ y = \frac{1}{1} = 1 $$

Therefore, the asymptotes for the function are:

- Vertical asymptotes: $ x = 4 $ and $ x = 1 $
- Horizontal asymptote: $ y = 1 $

Elias Ackerman · Answer

undefined To find the asymptotes of $f(x) = \frac{x^2 - 16}{x^2 - 5x + 4}$, first, determine if there are any vertical asymptotes by identifying any values of $x$ that would make the denominator zero. Then, analyze the behavior of the function as $x$ approaches these values.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degrees are equal, divide the leading coefficients. If the degree of the denominator is greater, there are no horizontal asymptotes. If the degree of the numerator is greater, there is a horizontal asymptote at $y = 0$.

For this function:

Vertical asymptotes: Set the denominator equal to zero and solve for $x$.

$$x^2 - 5x + 4 = 0$$

Factor or use the quadratic formula to find the roots. These roots are the values of $x$ where the function may have vertical asymptotes.

Horizontal asymptotes: Compare the degrees of the numerator and denominator. If they are equal, divide the leading coefficients. If the degree of the numerator is greater, there is a horizontal asymptote at $y = 0$.

Once you've found the asymptotes, you can plot the graph of the function to confirm their behavior.