Question

State the horizontal asymptote of the rational function. f(x) = x+9÷x^2+2x+3

Answer 1

It will look like this

There is no horizontal aseymtope, because it passes through all of possible y

Answer 2

Assuming you know the asymptote rules. We want to use the one that states if numerator's degree = 1 + denominators degree, the asymptote is a slant symptote in the form of y = mx + b.
$\lim_{n \to \infty} \frac{f(x)}{x} \ \textgreater \ \lim_{n \to \infty} [tex] \lim_{n \to \infty} ( \frac{x + 9}{x^2} + 2x + 3 - 2x) \ \textgreater \ refine \ \textgreater \ \lim_{n \to \infty} ( \frac{x + 9}{x^2} + 3)$ \ \textgreater \ 2 [/tex]
Now write the exception of indeterminate form.
$\lim_{n \to \infty} ( \frac{x + 9}{x^2} \lim_{n \to \infty}3$
For  $\lim_{n \to \infty} ( \frac{x + 9}{x^2})$ we want to divide by the highest denominator power.
$( \frac{x}{x^2} + \frac{9}{x^2})/ \frac{x^2}{x^2} \ \textgreater \ refine \ \textgreater \ \frac{1}{x} + \frac{9}{x^2}$
Write the exception again.
$\lim_{n \to \infty} \frac{1}{x} + \lim_{n \to \infty} \frac{9}{x^2}$
Apply the infinite property to both! $\lim_{n \to \infty} ( \frac{c}{x^a} ) = 0 \ \textgreater \ \lim_{n \to \infty} \frac{1}{x} = 0 \ \textgreater \ \lim_{n \to \infty} \frac{9}{x^2} = 0 \ \textgreater \ 0 + 0 = 0$
For 3 apply $\lim_{x \to a^c}= c \ \textgreater \ 3 \ \textgreater \ 0 + 3 = 3$
Now combine our terms and we get y = 2x + 3