How do you find the vertical, horizontal and slant asymptotes of: #(7x-2 )/( x^2-3x-4)#?

Question

Anna Allen · Accepted Answer

vertical asymptotes x = -1 , x = 4
horizontal asymptote y = 0

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s set the denominator equal to zero. solve : #x^2-3x-4=0rArr(x-4)(x+1)=0# #rArrx=-1,x=4" are the asymptotes"# Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" (a constant)"# divide terms on numerator/denominator by #x^2# #((7x)/x^2-2/x^2)/(x^2/x^2-(3x)/x^2-4/x^2)=(7/x-2/x^2)/(1-3/x-4/x^2)# as #xto+-oo,f(x)to(0-0)/(1-0-0)# #rArry=0" is the asymptote"# Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 1 , denominator-degree 2 ) Hence there are no slant asymptotes. graph{(7x-2)/(x^2-3x-4) [-10, 10, -5, 5]}

Serenity Jones · Answer

undefined To find the vertical, horizontal, and slant asymptotes of the function $ f(x) = \frac{7x - 2}{x^2 - 3x - 4} $, follow these steps:

1. **Vertical Asymptotes**:
   Vertical asymptotes occur where the denominator of the function becomes zero, as long as the numerator does not also become zero simultaneously. To find vertical asymptotes, solve the equation $ x^2 - 3x - 4 = 0 $ for $ x $. These are the values where the function approaches positive or negative infinity.

Factorizing $ x^2 - 3x - 4 $, we get $ (x - 4)(x + 1) = 0 $. So, $ x = 4 $ and $ x = -1 $ are the vertical asymptotes.

2. **Horizontal Asymptotes**:
   Horizontal asymptotes can be determined by examining the behavior of the function as $ x $ approaches positive or negative infinity. If the degrees of the numerator and denominator are the same, the horizontal asymptote can be found by dividing the leading coefficient of the numerator by the leading coefficient of the denominator.

In this case, both the numerator and denominator have the same degree (degree 2). Therefore, divide the leading coefficient of the numerator (7) by the leading coefficient of the denominator (1). The horizontal asymptote is $ y = 7/1 = 7 $.

3. **Slant Asymptote**:
   If the degree of the numerator is exactly one greater than the degree of the denominator, a slant (or oblique) asymptote may exist. To find it, perform polynomial long division or use other methods to divide the numerator by the denominator. The quotient represents the slant asymptote.

In this case, the degrees of the numerator and denominator differ by one (degree of numerator is 1 and degree of denominator is 2). Perform polynomial long division or use another method to divide $ 7x - 2 $ by $ x^2 - 3x - 4 $ to find the quotient.

Upon division, you'll find that the quotient is $ 7 + \frac{25}{x - 4} $. As $ x $ approaches positive or negative infinity, the $ \frac{25}{x - 4} $ term approaches 0, leaving the slant asymptote as $ y = 7 $.

In summary:
- Vertical asymptotes: $ x = 4 $ and $ x = -1 $
- Horizontal asymptote: $ y = 7 $
- Slant asymptote: $ y = 7 $