# Thread: graphing function

1. ## graphing function

i am having trouble graphing functions i missed the lesson an i'm lost for example:

$y=\frac{x}{x^2-4}$

i know it has V.A at x=-2,x=2 and H.A. at y=0 because the degree of the denominator > degree of numerator

but where do i go from here?(figuring out odd and even asymptotes, x and y intercepts) etc....

2. Originally Posted by euclid2
i am having trouble graphing functions i missed the lesson an i'm lost for example:

$y=\frac{x}{x^2-4}$

i know it has V.A at x=-2,x=2 and H.A. at y=0 because the degree of the denominator > degree of numerator

but where do i go from here?(figuring out odd and even asymptotes, x and y intercepts) etc....
You can get the first derivative to find out on which intervals it is decresing/increasing. You can use the info to know on which side of each vertical asymptote does the function go to (positive or negative infinity).

You can use the second derivative to get the concavity. that might help.

To get the x intercept(s), set y = 0 and solve for x.

To get the y intercept(s), set x = 0 and solve for y.

Don't forget that as the function goes to positive or negative infinity, it must converge to y = 0 (horizontal asymptote, as you've mentioned)..

Cheers!

3. Hello, euclid2!

I use a very primitive approach.
But it has always worked well for me.

Graph: . $f(x) \:=\:\frac{x}{x^2-4}$

i know it has V.A. at $x = \pm2$ and H.A. at $y = 0$
. . because (deg. of denom'r) > (deg of num'r) . . . . Right!

Sketch the asymptotes . . .
Code:
                      |
:     |     :
:     |     :
:     |     :
:     |     :
- - - - - - + - - o - - + - - - - - - -
-2     |     2
:     |     :
:     |     :
:     |     :
|

The only intercept is at (0,0).

The $x$-axis is partitioned into four intervals.
Evaluate a point in each interval.

. . $\begin{array}{ccccc}f(\text{-}3) &=& -\frac{3}{5} & \Rightarrow & (-3,\:-\frac{3}{5}) \\ \\[-3mm]
f(\text{-}1) &=& +\frac{1}{5} & \Rightarrow & (-1,\:\frac{1}{5})\\ \\[-3mm]
f(1) &=& -\frac{1}{3}& \Rightarrow & (1,\:-\frac{1}{3}) \\ \\[-3mm]
f(3) &=& +\frac{3}{5} & \Rightarrow & (3,\:\frac{3}{5})\end{array}$

Plot these points.

Using your knowledge of asymptotes, sketch the graph.

Code:
                :     |     :
:*    |     :*
:     |     :
:     |     :
: *   |     : *
:     |     :  o
:  o  |     :    *
:   * |     :           *
- - - - - - - : - - o - - : - - - - - - -
*          -2     | *   2
*    :     |  o  :
o  :     |     :
* :     |   * :
:     |     :
*:     |     :
:     |    *:
:     |

4. Originally Posted by Soroban
Hello, euclid2!

I use a very primitive approach.
But it has always worked well for me.

Sketch the asymptotes . . .
Code:
                      |
:     |     :
:     |     :
:     |     :
:     |     :
- - - - - - + - - o - - + - - - - - - -
-2     |     2
:     |     :
:     |     :
:     |     :
|

The only intercept is at (0,0).

The $x$-axis is partitioned into four intervals.
Evaluate a point in each interval.

. . $\begin{array}{ccccc}f(\text{-}3) &=& -\frac{3}{5} & \Rightarrow & (-3,\:-\frac{3}{5}) \\ \\[-3mm]
f(\text{-}1) &=& +\frac{1}{5} & \Rightarrow & (-1,\:\frac{1}{5})\\ \\[-3mm]
f(1) &=& -\frac{1}{3}& \Rightarrow & (1,\:-\frac{1}{3}) \\ \\[-3mm]
f(3) &=& +\frac{3}{5} & \Rightarrow & (3,\:\frac{3}{5})\end{array}$

Plot these points.

Using your knowledge of asymptotes, sketch the graph.

Code:
                :     |     :
:*    |     :*
:     |     :
:     |     :
: *   |     : *
:     |     :  o
:  o  |     :    *
:   * |     :           *
- - - - - - - : - - o - - : - - - - - - -
*          -2     | *   2
*    :     |  o  :
o  :     |     :
* :     |   * :
:     |     :
*:     |     :
:     |    *:
:     |
this is great. does it matter which numerical values i plug in to obtain other points on the graph, or in other words why did you use f(-1,1,-3,3) besides the fact that these are the values that the function approaches around the asymptotes?