SOLUTION: I have the following problem. I will be honest, it is for homework, so the answer is not in the back of the book. I am just trying to understand how to properly solve it. y = (x

Algebra ->  Graphs -> SOLUTION: I have the following problem. I will be honest, it is for homework, so the answer is not in the back of the book. I am just trying to understand how to properly solve it. y = (x      Log On


   

Question 103413This question is from textbook Algebra and Trigonometry
: I have the following problem. I will be honest, it is for homework, so the answer is not in the back of the book. I am just trying to understand how to properly solve it.
y = (x^2 - 4)/2x^4
I tried to simplify this on the website and it gave me y - (x^2 - 4)/2x^4 = 0
I am suppose to state what the intercept is and what it is symmetric to (y-axis, x-axis and/or origin)
Here is what I did.
y = (x^2 - 4)/2x^4
y = (x^2 - 4)/(x^2)(2x^2) - I broke apart the 2x^4
y = -4/2x^2 - I thought the x^2 on the bottom of the fraction cancelled out the top.
After that, I did not know what else to do. I could multiply both sides by 2x^2, but that did not seem to get me anywhere, so I started by making a list of the possible values of x and y.
Here is what I came up with.
x = 2, y = -1/2
x = -2, y = -1/2
x = 1, y = -2
I plotted these and my answer was the following:
1. I could not find an intercept. I don't see where either can equal 0, unless they both equal 0.
2. They would be symmetric to the y-axis.
Am I correct? My main goal is to find out what else to do with the equation. I'm lost.
Thanks for your help in advance. This question is from textbook Algebra and Trigonometry

Found 3 solutions by stanbon, ankor@dixie-net.com, Fombitz:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
simplify this on the website and it gave me y - (x^2 - 4)/2x^4 = 0
I am suppose to state what the intercept is and what it is symmetric to (y-axis, x-axis and/or origin)
-------------------------------------
y - (x^2 - 4)/2x^4 = 0
graph%28400%2C300%2C-10%2C10%2C-10%2C10%2C%28x%5E2-4%29%2F2x%5E4%29%29
---------------
Intercepts:
Since x cannot be zero, no y-intercepts.
If y=0, x=+2 or x=-2 are the x-intercepts.
-----------------------
Symmetry:
f(x)=f(-x) so the function is even and therefore symmetric to the y-axis.
--------------------------
Cheers,
Stan H.

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
solve it.
y = (x^2 - 4)/2x^4
:
Solving it is not that hard and it will give us the x intercepts:
%28x%5E2-4%29%2F2x%5E4 = 0
:
We can also write it like this:
x%5E2%2F2x%5E4 - 4%2F2x%5E4 = 0
or
x%5E2%2F2x%5E4 = 4%2F2x%5E4 = 0
:
Cancel x^2 and 2 and you have;
1%2F2x%5E2%29 = 2%2Fx%5E4
:
Cross multiply:
x^4 = 2(2x^2)
:
x^4 = 4x^2
:
Divide both sides by x^2 and you have:
x^2 = 4
:
x = +/-Sqrt(4)
x = +2
x = -2
:
These are the two x intercepts,
The axis of symmetry are half way between the two intercepts: x = 0
:
The graph would look like this:
+graph%28+300%2C+200%2C+-6%2C+6%2C+-.1%2C+.1%2C+%28x%5E2-4%29%2F2x%5E4%29+


Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
y+=+%28x%5E2+-+4%29%2F2x%5E4
y+=+%28x%5E2+-+4%29%2F%28x%5E2%29%282x%5E2%29 - I broke apart the 2x^4
y+=+-4%2F2x%5E2 - I thought the x^2 on the bottom of the fraction cancelled out the top.
You're on the right track but you lost some terms.
Let's back up.
y+=+%28x%5E2+-+4%29%2F2x%5E4
y+=+x%5E2%2F2x%5E4+-+4%2F2x%5E4
y+=+1%2F2x%5E2+-+4%2F2x%5E4
I don't think doing the division gives you any more insight. I think it's better to go back the your original form.
y+=+%28x%5E2+-+4%29%2F2x%5E4
First off, where does this equation equal zero. The most obvious place is when the numerator equals zero.
x%5E2+-+4=0
x%5E2+=+4
x=2 and x=-2 are both points where y=0.
To check for symmetry, do as you did before.
Try a number and its negative and see what you get for a result.
y%281%29+=+%281%5E2+-+4%29%2F2%281%29%5E4
y%281%29+=+-3%2F2
y%28-1%29+=+%28%28-1%29%5E2-4%29%2F2%28-1%29%5E4
y%28-1%29+=+-3%2F2
If you look at the equation you are squaring x and squaring the square of x. Working with squares of negatives and positives will give you the same answer so you graph will be symmetric about the y axis.
You can also graph some points and look at the graph.
+graph%28+300%2C+300%2C+-5%2C+5%2C+-5%2C+5%2C+%28x%5E2+-+4%29%2F2x%5E4%29+
As you can see, the graph plummets as x or -x approaches 0 because
lim%28x-%3E0%2C%28y%29%29=-infinity%29
After you pass x=2 in the positive sense and x=-2 in the negative sense, you have a slight increase above zero and then quickly approach zero as x gets larger. There is a maximum value of y=0.0312 at x=+/-2.8 and then quickly head towards zero. Easiest way to see this is to use EXCEL to calculate a bunch of x,y points (use one column for your x values and then the next column use a formula for your y values). Then you can graph that also.
+graph%28+300%2C+300%2C+1%2C+5%2C+-1%2C+1%2C+%28x%5E2+-+4%29%2F2x%5E4%29+
Additionally from your graph and from looking at the values you see that:
lim%28x-%3Einfinity%2C%28y%29%29=0%29 and
lim%28x-%3E-infinity%2C%28y%29%29=0%29
Hope it helps. Good Luck!