Question

Can anyone give me refresher of horozonal and verticle asymptote?

Answer 1

@satellite73

Answer 2

Vertical asymptotes are vertical lines which correspond to the zeroes of the denominator of a rational function. (They can also arise in other contexts, such as logarithms, but you'll almost certainly first encounter asymptotes in the context of rationals.)

Let's consider the following equation:

y = [x^2 + 2x - 3] / [x^2 - 5x - 6]

This is a rational function. More to the point, this is a fraction. Can you have a zero in the denominator of a fraction? No. So if I set the denominator of the above fraction equal to zero and solve, this will tell me the values that x cannot be:

x2 – 5x – 6 = 0
(x – 6)(x + 1) = 0
x = 6 or –1

So x cannot be 6 or –1, because then I'd be dividing by zero.



Now look at the graph:



graph of y = [x^2 + 2x - 3] / [x^2 - 5x - 6]

You can see how the graph avoided the vertical lines x = 6 and x = –1. This avoidance occurred because x cannot be –1 or 6. In other words, the fact that the function's domain is restricted is reflected in the function's graph. More usefully, you can use the domain to help you graph, because whichever values are not allowed in the domain will be vertical asymptotes on the graph.



You can draw the vertical asymptote as a dashed line to remind you not to graph there, like this:



Copyright © Elizabeth Stapel 2003-2011 All Rights Reserved
(It's alright that the graph appears to climb right up the sides of the asymptote on the left. This is common. As long as you don't draw the graph crossing the vertical asymptote, you'll be fine.)



graph of y = [x^2 + 2x - 3] / [x^2 - 5x - 6] with asymptotes dashed in

Let's review this relationship between the domain and the vertical asymptotes.

Find the domain and vertical asymptotes(s), if any, of the following function:
y = [x + 2] / [x^2 + 2x - 8]









The domain is the set of all x-values that I'm allowed to use. The only values that could be disallowed are those that give me a zero in the denominator. So I'll set the denominator equal to zero and solve.

x2 + 2x – 8 = 0
(x + 4)(x – 2) = 0
x = –4 or x = 2

Since I can't have a zero in the denominator, then I can't have x = –4 or x = 2 in the domain. This tells me that the vertical asymptotes (which tell me where the graph can not go) will be at the values x = –4 or x = 2.

domain: all x not equal to -4 or 2
vertical asymptotes: x = –4, 2

Note that the domain and vertical asymptotes are "opposites". The vertical asymptotes are at –4 and 2, and the domain is everywhere but –4 and 2. This is always true.

Find the domain and vertical asymptote(s), if any, of the following function:
y = [x^3 - 8] / [x^2 + 9]

To find the domain and vertical asymptotes, I'll set the denominator equal to zero and solve. The solutions will be the values that are not allowed in the domain, and will also be the vertical asymptotes.

x2 + 9 = 0
x2 = –9
Oops! That doesn't solve! So there are no zeroes in the denominator. Since there are no zeroes in the denominator, then there are no forbidden x-values, and the domain is "all x". Also, since there are no values forbidden to the domain, there are no vertical asymptotes.

domain: all x
vertical asymptotes: none

Note again how the domain and vertical asymptotes were "opposites" of each other.

Find the domain and vertical asymptote(s), if any, of the following function:
y = [x^3 - 8] / [x^2 + 5x + 6]

I'll check the zeroes of the denominator:

x2 + 5x + 6 = 0
(x + 3)(x + 2) = 0
x = –3 or x = –2

Since I can't divide by zero, then I have vertical asymptotes at x = –3 and x = –2, and the domain is all other x-values.

domain: for x not equal to -3 or -2
vertical asymptotes: x = –3 and x = –2

When graphing, remember that vertical asymptotes stand for x-values that are not allowed. Vertical asymptotes are sacred ground. Never, on pain of death, can you cross a vertical asymptote. Don't even try!

Top | 1 | 2 | 3

Answer 3

a question i have for example

Answer 4

And if that dosent help heres this



















I'll start by showing you the traditional method, but then I'll explain what's really going on and show you how you can do it in your head. It'll be easy!

Given some polynomial guy

f( x ) = a( x )^n + ... / b( x )^m + ... ... the top is an nth degree polynomial ... the bottom is an mth degree polynomial

1
If
n < m
, then the x-axis is the horizontal asymptote.
2

If

n = m

, then the horizontal asymptote is the line
y = a / b
3
If
n > m
, then there is no horizontal asymptote. (There is a slant diagonal or oblique asymptote.)

Yeah, yeah, you COULD just memorize these things... but it's way better to KNOW what's going on. Then you can just do it.

What we're really doing is some quick long division to divide the denominator into the numerator. The cool thing is that we only need to do the first part -- no remainder crud! And we can do it in our heads!

Find the horizontal asymptote for f( x ) = ( 4x^2 - 3x + 1 ) / ( 2x^2 - 1 )
If we write out the long division...

( 2x^2 - 1 ) / ( 4x^2 - 3x + 1 ) ... which gives 4x^2 - 2 ... subtracting gives -3x yadda

We got this 2 by looking at 4x^2 / 2x^2 = 2
So, the horizontal asymptote is the line

y = 2


We could have just taken a quick look at

4x^2 / 2x^2


and have been done with it!


So, let's do this one the quick way:
Find the horizontal asymptote of f( x ) = ( x^2 - x - 6 ) / ( x^2 - 1 )


Look at x^2 / x^2 = 1
The horizontal asymptote is the line

y = 1


graph with a horizontal asymptote at y = 1



Now, try another one in your head:



Find the horizontal asymptote of

f( x ) = ( x + 3x^2 ) / ( 5x^2 - 6x - 1 ) ... the x^2 is the king of the top

Look at 3x^2 / 5x^2

The horizontal asymptote is the line

y = ( 3 / 5 )

RY IT:

Find the horizontal asymptote of

f( x ) = ( 2x^2 - x + 6 ) / ( 9 - 2x - x^2 )

What about this one?

Find the horizontal asymptote of f( x ) = 2x / ( x^2 - 5x - 3 )

Look at 2x / x^2 ... The denominator goes into the numerator 0 times!

So, the horizontal asymptote is the line

y = 0
(which is the x-axis)

This one falls under part 1 on our list.


YOUR TURN:

Find the horizontal asymptote of

f( x ) = ( x - 3 ) / ( x^2 - 3x - 10 )

OK, so what about this one?
f( x ) = ( 3x^3 + 2 ) / ( x^2 - x - 7 )

If we look at 3x^3 / x^2 , we find that the x^2 WILL divide in...
But, there's going to be some x stuff left over to deal with. This is when you need to start in with some long division... and we get to ignore the remainder!












( x^2 - x - 7 ) / ( 3x^3 + 0x^2 + 0x + 2 ) ... which gives 3x^3 - 3x^2 - 21x ... subtracting gives 3x^2 + yadda ... The 3x + 3 is the slant asymptote. It's the line y = 3x + 3

You can stop here since the rest will be remainder stuff.


a graph with the slant asymptote y = 3x + 3



TRY IT:

Find the slant asymptote of

f( x ) = ( 5x^2 - 3x + 1 ) / ( x + 2 )

Answer 5

you can find the second one on http://www.coolmath.com/precalculus-review-calculus-intro/precalculus-algebra/18-rational-functions-finding-horizontal-slant-asymptotes-05

Answer 6

first one u find http://www.purplemath.com/modules/asymtote.htm