Question

Where are the asymptotes for the following function located? f (x) = StartFraction 14 Over (x minus 5) (x + 1) EndFraction

Answer 1

Answer:

2 vertical asymptotes occurring at x = 5 and x = -1

Step-by-step explanation:

given

$f(x) = \frac{14}{(x-5)(x+1)}$

recall that asymptotic occur at the locations that will make the equation undefined. In this case, the asymptote will occur at x-locations which will cause the denominator to become zero (and hence undefined)

Equating the denominator to zero,

(x-5)(x+1) = 0

(x-5) =0

x = 5 (first asymptote)

or (x+1) = 0

x = -1  (2nd asymptote)