Question

where are the asymptotes for the following function located? f(x)=14/(x-5)(x+1)
a. x=-1 and x=5
b. x=-1 and x=14
c. x=1 and x=-5
d. x=14 and x=5

Answer 1

Given:
f(x)=14/(x-5)(x+1)

The asymptotes for the following function is located in a. x=-1 and x=5.

x = -1 and x = 5 is the vetical asymptotes. 
the horizontal aymptotes is located at y = 0.
there is no oblique asymptotes.

Pls. see attachment for addtl explanation. 

Answer 2

Answer:

Option a is correct.

x=-1 and x=5

Step-by-step explanation:

To find the asymptotes for the rational function:

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

The vertical asymptotes for this function is to set the denominator equals to 0.

The horizontal asymptotes for this given fucntion is 0.

Denominator of the given function is: $(x-5)(x+1)$

By definition of asymptotes;

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

By zero product property:

$x-5 = 0$ and $x+1 = 0$

⇒x = 5 and x= -1

Therefore, the asymptotes for the following function located are: x = -1 and x =5