at what value of x does the graph of the following function f(x) have a vertical asymptote? check all that apply. f(x)=1/(x-4)(x+1)

a. 4
b. 0
c. -1
d. 1
e. -4
f. 2

Answer :

Answer:4,-1

Step-by-step explanation:

A function [tex]f(x)[/tex] is said to have vertical asymptote at a point [tex]x_{0}[/tex] if [tex]f(x_{0})[/tex] tends to infinity or negative infinity.

Given function is [tex]f(x)=\frac{1}{(x-4)(x+1)}[/tex]

When [tex]x=4[/tex],

[tex]f(x)=\frac{1}{(4-4)(4+1)}=\frac{1}{0}[/tex]

[tex]f(4)[/tex]→∞

So,[tex]f(x)[/tex] has a vertical asymptote at [tex]4[/tex].

When [tex]x=-1[/tex],

[tex]f(x)=\frac{1}{(4+1)(-1+1)}=\frac{1}{0}[/tex]

[tex]f(-1)[/tex]→∞

So,[tex]f(x)[/tex] has a vertical asymptote at [tex]-1[/tex].

So,[tex]4,-1[/tex] are the vertical asymptotes of [tex]f(x)[/tex]

At values 4 and -1, the function F(x) has a vertical asymptote

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