at which values of x does the graph of f(x)=x^2-x-2/(x^2-1)(x^2-16) have a vertical asymptope? check all that apply

x = -4, 1, 4

Step-by-step explanation:

The function is given as:

f(x) = (x² - x -2)/[(x² - 1)(x² -16)]

We want to find the values of x where the graph will have vertical assymptote.

Vertical asymptotes are defined as vertical lines which correspond to the values of x that will make the denominator of a rational function to be zero.

Thus, to get the values of x that we will have vertical assymptotes, we have to equate the denominator to zero.

We have;

[(x² - 1)(x² -16)] = 0

Thus,

x² - 1 = 0

Or

x² - 16 = 0

So;

x² = 1

x = √1

x = 1

Or

x² = 16

x = ±√16

x = ±4

Values of x that give vertical assymptote are;

x = -4, 1, 4

quit

step-by-step explanation:

i think its a and d

step-by-step explanation: