1)2

x^2-2x-3

Find) Domain

Range

Is This Continuous?

Local min/max

Vertical Asymptote

Horizontal Asymptote

X Intercepts

Y Intercepts

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- Oct 25th 2008, 10:48 AMgreenpumpkinsHow 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 - Oct 25th 2008, 02:29 PMShyamReply
for function $\displaystyle f(x) = \frac{2}{x^2-2x-3}$

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

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

**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 $\displaystyle = \{y \in \mathbb{R} | \;y\ne0\}$

**Horizontal Asymptote (HA)**

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

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

divide with highest degree, $\displaystyle x^2$,

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

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

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

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

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,

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

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.