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

5 pm

step-by-step explanation:

, and

for

step-by-step explanation:

we can analyze the pattern at which the terms increase. let us denote , , , so that we can call

note the following:

the pattern: to get the next term value, we just add the next term number (e.g. 1st, 2nd, to the previous term value (term values are the given numbers in the question, 1, 3, 6, 10, 15,

from , to get to , we add

from , to get to , we add

etc.

we need to form a recursive formula, a formula that has the next term as a function of the previous term. so a possible recursive formula can be

, and for