| Author |
 Topic  |
|
|
R.Damian
Average Member
USA
7 Posts |
 Posted - 11/03/2009 : 02:15:56
|
Hi, I'm trying to graph the functions
y=(2^x)+5/(2^x)-7
The power is x so that throws me off a bit. I'm suppose to follow somei n class directions to do these problem so i need to find if there is any symmetry, asymptotes (vertical, horizontal, oblique). I know how the graph looks like and there seems to be one vertical asymptote. I just don't know how to show it on my work. |
 |
|
|
Ultraglide
Advanced Member
Canada
164 Posts |
Posted - 11/03/2009 : 15:19:01
|
I assume that from your comments you have graphed the function. If you don't have a graphing calculator, try something like Graphmatica which is free to download. Now back to your problem. First of all I am assuming you have 2^x in the demoninator of the second term. This is never zero so you can't have any vertical asymptotes.
Now for the horizontal asymptotes:
If x gets very large 2^x also gets large and 5/2^x goes to zero, so you don't have an asymptote as you go to positive infinity. Similarly as x gets very small 2^x goes to zero but 5/2^x will get very large so you don't have a horizontal asymptote as you go to negative infinity.
You also cannot have oblique asymptotes because the degree of the denominator in the second term is not one less than the degree of the numerator.
In fact, the function looks a lot like a parabole, but it is not a parabola. |
|
|
|
uberclay
Advanced Member
Canada
147 Posts |
Posted - 11/03/2009 : 15:19:17
|
The vertical asymptote will appear at x where the value of your function is undefined; as is the case with division by zero.
Assuming your function is y == (2x + 5) / (2x - 7);
this can be determined by finding the value of x for which the divisor (2x - 7) is equal to zero.
|
|
|
|
Ultraglide
Advanced Member
Canada
164 Posts |
Posted - 11/03/2009 : 15:22:04
|
I assumed that it looks like
y=2^x+(5/2^x)-7 |
|
|
|
R.Damian
Average Member
USA
7 Posts |
Posted - 11/03/2009 : 17:02:40
|
I'm sorry, i wrote the function pretty bad. Uberclay has it right. Its y = (2^x + 5) / (2^x - 7)
So there is a Asymptote on this graph and i understand its when X has a restriction on its domain I'm must not sure how to show how (2^x-7) = 0, It also seems to show some horizontal asymptotes, not sure about those though. |
|
|
|
Ultraglide
Advanced Member
Canada
164 Posts |
Posted - 11/03/2009 : 23:46:18
|
Ok, the vertical asymptote occurs when the denominator is zero. So we have to solve 2^x-7=0
This is the same as 2^x=7.
Take the log of both sides gives
xlog2=log7 or x = log7/log2.
The horizontal asymptote can be found by letting x go to infinity, but both numerator and denominator will get exceedingly large so we have to do some manipulation first.
Divide each term in the ratio by 2^x gives
y = (1+5/2^x)/(1-7/2^x). Now when x goes to infinity both the terms with 2^x in the denominator go to zero so the value of the function approaches 1.
Now as x goes to positive infinity the ratio will approach 1 from above (because the ratio is slightly greater than 1).
For x approaching negative infinity, use the original expression noting that 2^x will approach zero, so the ratio will approach -5/7 from below. |
|
|
|
R.Damian
Average Member
USA
7 Posts |
Posted - 11/05/2009 : 08:22:59
|
Thanks, That really helped me out. I just have one more. I'm trying to figure out how to find the oblique asymptote for this one function. I usually use Synthetic division but its not coming out for me on this one.
F(x)= (X^3 + 4) / (2x^2 + x -1)
I usually have only x-1 or x-2 but for this one i get (2x-1)(x+1) for the denominator..so.. do i just pick one of the factors in the denominator and divide it into the numerator or do i use 2x^2+x-1? or am i completely off? |
|
|
|
Ultraglide
Advanced Member
Canada
164 Posts |
Posted - 11/05/2009 : 11:43:03
|
Synthetic division as I recall, only works with linear factors, i.e. ax+b so I would just divide using the denominator as given. All you really want is the quotient (answer).
So just divide using algebraic long division remembering to write the dividend as x^3+0x*2+0x+4.
Note that your numerator is one degree higher than the denominator so your quotient will be linear, i.e. ax+b. If the numerator was 2 degrees higher, your quotient would be quadratic and so on. |
|
|
| |
Topic |
|
Rational Function Graphing
