Determine the equations of the vertical and horizontal asymptote if any for g(x)= x-1/(2x+1)(x-5)

Answer:Horizontal asymptote : y = 0Vertical 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 = 0Vertical asymptote : for this we equate denominator = 0 2x+1 = 0, x-5 = 0⇒[tex]x = \frac{-1}{2} , 5[/tex]