Answer:
Step-by-step explanation:
Let f(x) = p(x)/q(x), where p and q are polynomials and reduced to lowest terms. (If p and q have a common factor, then they contribute removable discontinuities ('holes').)
Write this in cases:
(i) If deg p(x) ≤ deg q(x), then f(x) is a proper rational function, and lim(x→ ±∞) f(x) = constant.
If deg p(x) < deg q(x), then these limits equal 0, thus yielding the horizontal asymptote y = 0.
If deg p(x) = deg q(x), then these limits equal a/b, where a and b are the leading coefficients of p(x) and q(x), respectively. Hence, we have the horizontal asymptote y = a/b.
Note that there are no obliques asymptotes in this case. ------------- (ii) If deg p(x) > deg q(x), then f(x) is an improper rational function.
By long division, we can write f(x) = g(x) + r(x)/q(x), where g(x) and r(x) are polynomials and deg r(x) < deg q(x).
As in (i), note that lim(x→ ±∞) [f(x) - g(x)] = lim(x→ ±∞) r(x)/q(x) = 0. Hence, y = g(x) is an asymptote. (In particular, if deg g(x) = 1, then this is an oblique asymptote.)
This time, note that there are no horizontal asymptotes. ------------------ In summary, the degrees of p(x) and q(x) control which kind of asymptote we have.
I hope this helps!
Answer:
6
Step-by-step explanation:
Answer: Choice C) -11
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Explanation:
The first equation given is y = 3 - 1/2x
In other words, y is the same as 3 - 1/2x.
We can replace y in the second equation with 3 - 1/2x
This is known as substitution (think of a substitute teacher who is a temporary replacement for your teacher)
Doing this leads to...
3x+4y = 1
3x+4*y = 1
3x+4*( y ) = 1
3x+4*( 3 - 1/2x ) = 1 <<--- y has been replaced with 3-1/2x
3x+4*(3) +4*(-1/2x) = 1
3x+12-2x = 1
3x-2x+12 = 1
x+12 = 1
x+12-12 = 1-12 <<-- subtracting 12 from both sides
x = -11
Which is why the answer is choice C) -11
