Question

Help anyone????

find the asymptotes, if any, of the graph of the rational function f(x) = x/x^2 + 1

Answer 1

if the curve is

\[y = \frac{x}{x^2 + 1}\]

well as for the vertical asymptotes: there are no real numbers that make the denominator zero.

so the curve is continuous

you can find the 1st and 2nd derivatives and as a result can find a maximum value, a minimum value and point of inflection.

the only thing of not is that as x approaches infinity the curve approaches zero from above.

similar for negative infinity, the curve approaches zero from below.

To get a good understanding graph the curve using

https://www.desmos.com/calculator

if the curve is

\[y = \frac{x}{x^2} + 1\]

then there will be a vertical and horizontal asymptote...

and to see them, just graph the curves

Answer 2

if the graph is
(x/x^2) + 1 ==> (x+1)/x
You have a horizontal asymptote at y = 1
because your top degree, of numerator is 1 and degree of denominator is 1, too

thus, if your top and bottom degrees are the same, Horizontal asymptote is the ratio of the co-efficients in this case 1 over 1

for the V.asymptote, equal the denominator to zero
x=0
thus V.A at x = 0

Answer 3

Thanks y'all. Sorry it took so long to reply, but it helped. I got it right!