1. ## asymptote

Give an example of a RATIONAL function that has

-a vertical asymptote at x-=2
-a horizontal asymptote at y=1

2. Originally Posted by aymers14
Give an example of a RATIONAL function that has

-a vertical asymptote at x-=2
-a horizontal asymptote at y=1
A function of the form $y = \frac{a}{x - b} + c$ (which is rational, by the way) has a vertical asymptote x = b and a horizontal asymptote y = c .......

3. A vertical asymptote is a dashed vertical line that as the graph approaches the number that makes the function undefined, it approaches this line and f(x) blows up to infinity (or - infinity). A vertical asymptote is y = a, where x is any point that makes the denominator 0. You may also have to check if it's a hole or not.

A horizontal asymptote is a dashed horizontal line that as the graph approaches, it extends to infinity in the x and approaches the horizontal asymptote. A horizontal asymptote is x = b. To find the horizontal asymptote, take the limit of a rational function as it goes to infinity.

Examples:
1.
$f(x) = \frac{x^2 + 2}{x-2}$
When does the vertical asymptote exist? When denominator is 0. When does the denominator become 0? When x = 2. To find out what happens on both sides of the asymptote, take the limit as x goes to 2 from both sides.

$\lim_{x \to 2^{+}} \frac{x^2 + 2}{x-2} = +\infty$

$\lim_{x \to 2^{-}} \frac{x^2 + 2}{x-2} = -\infty$

2.
$f(x) = \frac{x+1}{x^2-1} = \frac{x+1}{(x+1)(x-1)}$

What are the vertical asymptotes here? x = 1 only. Why not x = -1? Because, as you can see (x+1) cancels, and this implies that there is a hole at x= -1, not an asymptote.

3. $f(x) = \frac{x+2}{x-3}$
Finding the horizontal asymptote is simply taking the limit as x approaches infinity. If limit is infinity, no horizontal asymptote.

$\lim_{x \to \pm \infty} \frac{x+2}{x-3} = \lim_{x \to \pm \infty} \frac{x}{x} = 1$

Take the first derivative to find where it is increasing/decreasing, the second derivative to find where it is concave up/down.