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One last problem:

Graph
 

f ( x ) = ( x^3 ) / ( x^2 - 9 )
 

2 things
and
2 sentences!
 

1
 

y-intercepts: Find f(0)
 

f ( 0 ) = ( ( 0 )^3 ) / ( ( 0 )^2 - 9 ) = 0

 

( 0 , 0 ) ... nowhere else!
 

2
 

x-intercepts: numerator = 0, solve
 

x^3 = 0

 

( 0 , 0 ) ... nowhere else!

 

3
 

Vertical asymptotes: denominator = 0, solve
 

x^2 - 9 = 0 gives ( x - 3 ) ( x + 3 ) = 0 which gives x = 3 and x = -3

The lines x = -3 and x = 3

4
 

Horizontal asymptote:
 
Look at
 
( x^3 ) / ( x^2 ) = x something
 
Hey -- it's a slant!
 


5
 


Slant asymptote
:
 

( x^3 + 0x^2 + 0x ) / ( x^2 - 9 ) = x ... which gives x^3 - 9x ... subtracting gives 9x ... stop here since x^2 no longer goes in
 

The line y = x

OK, Skippy -- are you ready to graph this bad boy?

Intercepts and asymptotes:

the asymptotes and intercepts for f ( x ) = ( x^3 ) / ( x^2 - 9 )

First, let's think about the left and right neighborhoods:

Here's what we've got -- see if you can figure it out!

Graphs hug asymptotes.

These guys are functions.

He can ONLY cross the x-axis at x = 0.

Can he do this?

one possible function graph ... Buzz! Nope! He'd have to cross the x-axis where he's not allowed to.

So, it must be like this:

the left and right neighborhoods for f ( x ) = ( x^3 ) / (x^2 - 9 )

What about the middle neighborhood?

We'll need to quickie plot two points...  Try it before going on.

f ( x ) = ( x^3 ) / ( ( x - 3 ) ( x + 3 ) )

plugging in -1 for x gives ( - ) / ( ( - ) ( + ) ) = ( + ) ... above for y, and plugging in 1 for x gives ( + ) / ( ( - ) ( + ) ) = ( - ) ... below for y

graph of f ( x ) = ( x^3 ) / ( x^2 - 9 ) ... DONE!

NOTE: A common mistake that students make is to think that a graph cannot cross a slant or horizontal asymptote.  This is not the case!  A graph CAN cross slant and horizontal asymptotes (sometimes more than once).  It's those vertical asymptote critters that a graph cannot cross.  This is because these are the bad spots in the domain.

 

YOUR TURN:

Graph

f ( x ) = ( 4 - 3x - x^2 ) / ( x - 2 )