graph with no aysmptotes?

[rachael724]

Which of the following functions has a graph with no aysmptotes?

a. f(x) = 1/(x2 + 1)
b. g(x) = 1/(x - 1)
c. h(x) = 1/(x2 - 1)
d. p(x) = 1/[(x2 - 1)(x - 1)]
e. q(x) = x3/[(x2 - 3)(x2 - 5)]
f. r(x) = x3/(x2 + 1)
g. none of these

I am not too good at these. My friend was helping me and said the answer is either b or f, but i just dont know.

 

[soroban]

Hello, rachael724!

I had to read the question twice . . .

Which of the following functions has a graph with no asymptotes?

\(\displaystyle [a]\;f(x)\ =\ \frac{1}{x^2 + 1}\qquad\;\;\;g(x)\ =\ \frac{1}{x - 1}\qquad\;\;[c]\;h(x)\ =\ \frac{1}{x^2 - 1}\qquad\;\;[d]\;p(x)\ =\ \frac{1}{(x^2 - 1)(x - 1)}\)

\(\displaystyle [e]\;q(x)\ =\ \frac{x^3}{(x^2 - 3)(x^2 - 5)]}\qquad\;\;[f]\;r(x)\ =\ \frac{x^3}{x^2 + 1}\qquad\;\;[g]\;\text{none of these}\)

Vertical asymptotes occur where the denominator equal zero.

Horizontal asymptotes occur when \(\displaystyle \lim_{x\to\infty} f(x)\) has a finite limit.

\(\displaystyle [a]\;f(x)\ =\ \frac{1}{x^2+1}\)
. . . The denominator cannot equal 0; there are no verical asymptotes.
. . . \(\displaystyle \lim_{x\to\infty}\ \frac{1}{x^2 + 1}\ =\ 0\). . Horizontal asymptote: \(\displaystyle y = 0\)

\(\displaystyle \;g(x)\ =\ \frac{1}{x - 1}\)
. . . Vertical asymptote at \(\displaystyle x = 1\)

\(\displaystyle [c]\;g(x)\ =\ \frac{1}{x^2 - 1}\)
. . . Vertical asymptotes at \(\displaystyle x = \pm1\)

\(\displaystyle [d]\;[(x)\ =\ \frac{1}{(x^2-1)(x - 1)}\)
. . . Vertical asmptotes at \(\displaystyle x = \pm1\)

\(\displaystyle [e]\;q(x)\ =\ \frac{x^3}{(x^2-3)(x^2-5)}\)
. . . Vertical asymptotes at \(\displaystyle x = \pm\sqrt{3},\ \pm\sqrt{5}\)

\(\displaystyle [f]\;r(x)\ =\ \frac{x^3}{x^2 + 1}\)
. . . The denominator cannot equal zero; no vertical asymptotes.
. . . \(\displaystyle \lim_{x\to\infty}\ \frac{x^3}{x^2+1}\ =\ \infty\) * . .No horizontal asymtptote.

. . . . . This one!

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

* . We have: . \(\displaystyle \L\frac{x^3}{x^2 + 1}\)
Divide top and bottom by \(\displaystyle x^2:\;\;\L\frac{\frac{x^3}{x^2}}{\frac{x^2}{x^2} + \frac{1}{x^2}} \;= \;\frac{x}{1 + \frac{1}{x^2}}\)

Take the limit: . \(\displaystyle \lim_{x\to\infty}\left[\frac{x}{1 + \frac{1}{x^2}}\right] \;= \;\frac{\infty}{1 + 0} \;= \;\infty\)

 

[tkhunny]

Horizontal?

y = x looks interesting.

I'm going with g) None of These

 

[stapel]

The answer may depend on whether slant ("oblique") asymptotes are under consideration. If only horizontals and verticals have been covered, then Soroban's answer is correct. Otherwise, tkhunny's is correct.

Eliz.