Narayah
consider a sequence whose first five terms are: 0, -2,-8, -18, -32
which function (with domain all integers n 1) could be used to define and continue this sequence?
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f(n)= - 2n2
f(n) = -(n)-1
f(n) = -2(n-1)
1
=
-1
fo
f(n) = -2(n-1)?

None of the given functions can be used to define this sequence.

Step-by-step explanation:

Here the given sequence is 0, -2, -8 , -18 , -32, ....

The given function are

Now, check for each functions:

1)

for  n = 1 , ≠ 0

Hence, the given function DO NOT satisfy the sequence.

2)

for n = 1, f(0) =  -1 -1  = -2 ≠ 0

Hence, the given function DO NOT satisfies the sequence.

3)

for  n = 1 , f(1) = -2(1 -1)  = -2 x 0   = 0

for n = 2 , f(2) = -2(2 -1)  = -2 x 1   = -2

for n = 3,  f(2) = -2(3 -1)  = -2 x 2    =  -4   ≠ -8

Hence, the given function DO NOT satisfies the sequence.

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.