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!
Sub those points for x
f(x) is the y
(x,y)
so for x=0
f(0)=2(0)^4+3
f(0)=0+3
f(0)=3
(0,3) is a opint
for x=1
f(1)=2(1)^4+3
f(1)=2(1)+3
f(1)=2+3
f(1)=5
(1,5) is another point
for x=2
f(2)=2(2^4)+3
f(2)=2(16)+3
f(2)=32+3
f(2)=35
(2,35) is another point
points are
(0,3)
(1,5)
(2,35)
The answer is a I'm pretty sure
