Question

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

Answer 1

Answer:

Horizontal asymptote : y = 0

Vertical asymptote :

2x+1 = 0, x-5 = 0

⇒$x = \frac{-1}{2} , 5$

Step-by-step explanation:

Given : $g(x) = \frac{x-1}{(2x+1)(x-5)}$

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 = $\frac{a_{n} }{b_{m}}$

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

⇒$x = \frac{-1}{2} , 5$

Answer 2

I'm not too sure, hopefully the image helps you in some way