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!
I believe that they typo meant 300 fossils of which 21% are snail shells.
Since % means per one hundred...
300(21/100)=63
Elmer has 63 fossilized snail shells.
Two circles<span> of </span>radius<span> 4 are </span>tangent<span> to the </span>graph<span> of y^</span>2<span> = </span>4x<span> at the </span>point<span> (</span>1<span>, </span>2<span>). ... I know how to </span>find<span> the </span>tangent<span> line from a circle and a given </span>point<span>, but ... </span>2a2=42. a2=8. a=±2√2. Then1−xc=±2√2<span> and </span>2−yc=±2√2. ... 4 from (1,2<span>), so you could </span>find these<span> centers, and from there the</span>equations<span> of the circle
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