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6. The horizontal and vertical asymptotes are y=1, x=4, and there is a "hole" at x=-2. The appropriate choice is ...
F. Oblique asymptote y = 0
8. The denominator has degree 2 more than the numerator, so the end behavior will tend toward 0 as x gets large. The appropriate choice is ...
J. As x → ±∞, f(x) → 0
6. The horizontal and vertical asymptotes are y=1, x=4, and there is a "hole" at x=-2. The appropriate choice is ...
F. Oblique asymptote y = 0
8. The denominator has degree 2 more than the numerator, so the end behavior will tend toward 0 as x gets large. The appropriate choice is ...
J. As x → ±∞, f(x) → 0
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Answer:
Step-by-step explanation:
Assuming the problem is asking you to factor out the formula, it can be done easily by finding a common number with the two givens.
4 goes into both 12 and 16 evenly, so we will use this to factor. But that's not all, we have the n-variable to worry about.
Look at the n-variable exponents in the problem, take the highest power that can go be subtracted from both and use that. We have a and
. Since 5 can't go into 3, we must use 3 to factor.
Our factoring variable will therefore be . Now, use this to simply divide and get the answer:
So how did we get this? For the whole numbers, we have to divide: (12/4, 16/4). Next, we have to minus the exponents since we are dividing.
When dividing the 12 and 4, minus the and
to get
. Get rid of the
term for the 16 and leave the 4. And there's your answer!