Respuesta :
Answer:
Horizontal asymptote : y = 0
Vertical asymptote :
2x+1 = 0, x-5 = 0
⇒[tex]x = \frac{-1}{2} , 5[/tex]
Step-by-step explanation:
Given : [tex]g(x) = \frac{x-1}{(2x+1)(x-5)}[/tex]
Solution :
Case 1. When the degree of the denominator is larger than the degree of the numerator), then the graph of y = f(x) will have a horizontal asymptote at y = 0
Case 2 . the degrees of the numerator and denominator are the same then the graph of y = f(x) will have a horizontal asymptote at y = [tex]\frac{a_{n} }{b_{m}}[/tex]
Case 3 . the degree of the numerator is larger than the degree of the denominator, then the graph of y = f(x) will have no horizontal asymptote.
Since in the given function degree of numerator is 1 and degree of denominator is 2 So, Case 1 applies here .
Thus Horizontal asymptote : y = 0
Vertical asymptote : for this we equate denominator = 0
2x+1 = 0, x-5 = 0
⇒[tex]x = \frac{-1}{2} , 5[/tex]
