# How to find intercepts, asymptotes, end behaviors, etc.

• October 25th 2008, 11:48 AM
greenpumpkins
How to find intercepts, asymptotes, end behaviors, etc.
1) 2
x^2-2x-3
Find) Domain
Range
Is This Continuous?
Local min/max
Vertical Asymptote
Horizontal Asymptote
X Intercepts
Y Intercepts
• October 25th 2008, 03:29 PM
Shyam
for function $f(x) = \frac{2}{x^2-2x-3}$

$f(x) = \frac{2}{(x+1)(x-3)}$

Domain
Domain is the set of values of x for which function is defined.
The given function is not defined at x = -1 and x = 3

So, domain $= \{x \in \mathbb{R} | \;x\ne-1,3 \}$

Range
Range is f(x) values (or y values) you get, when you put these domain x values in function f(x). Since the function has a horizontal Asymptote at y = 0, so, y = 0 is excluded from Range.

So, Range $= \{y \in \mathbb{R} | \;y\ne0\}$

Horizontal Asymptote (HA)

To find HA, divide Numerator and denominator of the function with highest degree (power) of x. Then take limit $x \rightarrow \infty$

$f(x) = \frac{2}{x^2-2x-3}$

divide with highest degree, $x^2$,

$f(x) = \frac{\frac{2}{x^2}}{\frac{x^2}{x^2}-\frac{2x}{x^2}-\frac{3}{x^2}}$

$f(x) = \frac{\frac{2}{x^2}}{1-\frac{2}{x}-\frac{3}{x^2}}$

Now take $\lim{x\to \infty}$

$= \frac{0}{1-0-0}=0$

HA is y = 0

Vertical Asymptote (VA)

For VA, Denominator = 0

(x+1)(x-3) = 0

x = -1 and x = 3

The graph is discontinuous at x = -1 and x = 3

y-intercepts

For y-intercept, put, x = 0 in equation,

we get, y = -2/3

y-intercept = -2/3, the graph cuts y-axis at -2/3.

x-intercepts

For x-intercept, put, y = 0 in equation,

$0 = \frac{2}{(x+1)(x-3)}$

we do not get any answer from this, so there is no, x-intercept. i.e., the graph does not cuts x-axis

Please see attached graph, the green lines show the asymptotes.