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Graphing/function help!!

I cannot seem to get y by itself and make it work in a graph. I need to determine whether the following is a function: y^2=3x^2-4x-8. Can someone help explain how to get y by itself so I am able to graph it.Thank you!
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3 Answers

A function assigns to every x-value in its domain unique y-value.
 
The equation
 
y²=3x²-4x-8
 
is not a function. To see this, we should first determine for which values of x the equation is even defined. Since
 
y²≥0, we must have 3x²-4x-8≥0.
 
You can solve this inequality with the quadratic formula and will find that the allowed values for x are
 
(-∞,-(2/3)√7+(2/3)] ∪ [(2/3)√7+(2/3),∞)
 
Let's take one of those allowed x-values, let's say x=10, and see what happens:
 
y²=3x²-4x-8=3(10)²-4(10)-8=252
 
Now y can be +√252 or -√252, so there is no unique y-value for x=10, so y is not a function of x.
 
More generally, when you solve for y,
 
y=±√(3x²-4x-8),
 
so you always get two y-values - not allowed for a function!
 
You can also see that your equation is not a function when you graph it. You will get a hyperbola with two branches, which does not pass the so-called vertical line test: There are vertical lines that will intersect either branch of the hyperbola twice, which cannot happen for the graph of a function.
 
 
 

Comments

So when graphing that is why its not a complete V and says error in some of the tables? Thank you so much for your help!!! I appreciate it
 
Yes! You need to graph the two halves of the parabola separately, though you should be able to do it in the same diagram.
cannot seem to get y by itself and make it work in a graph. I need to determine whether the following is a function: y^2=3x^2-4x-8. Can someone help explain how to get y by itself so I am able to graph it.Thank you!
 

y^2 = 3x^2 -4x -8

You can take the square root of each side.  Then, use a graphing calculator or x/y chart to graph the equation.

 
I hope this helps.
 

Comments

I tried this and it was not a complete line. I wasn't sure what I was doing wrong and go confused. in the y- value of the calculator I put y= √3x-4x-8 it showed up as error.

You will input:

y= √3x^2 - 4x - 8

You will input:

y= √3x^2 - 4x - 8
 
The graph of a square root funtion looks like a V.
My graph is saying error and my line is a V with the point at the bottom cut off with that. That is why I was stuck and confused. That was how I initially did the problem before coming here.
which is only telling me that it doesn't exist. But why is my graph not a complete V? that was where I thought I was wrong.
 

You can try changing the window size.  Mine uses -40 to +40 and a scale of 4 for x and y. Also, you can look at the table and plot points that way.  Let me know if I can help more.

The "V" does not connect at the bottom!  On the left, it reaches it's lowest value at x = -2,  on the right it is x=3.  These answers came from the table.

Erica,
   One way to visualize the function is to assume values for x and see the values of y that are derived.
The function is y2 = 3x2 -4x -8.
For x = 0, y2 = -8. Here the answer is not a real number. 
If x = 1, y2 = 3-4-8 = -9; again not a real number.
If x = 3, y2 = 27-12-8 = 7; here the answer is a real number.
  Real answers for this function exist when y2 is 0 or positive. If your aim is to determine if the function is real for all values of x, then that is not the case.
  Your graphing calculator will be able to plot real values for the function after you input the function into it. Hope this helps with your understanding of the problem. 
 
 
 
 
 

Comments

OK so to determine this all I have to do is place a 0 in the x value and if its a zero or negative number than its not a function? Is that what you are saying? How can I put it into my calculator with the y value being y^2? That was the part I was confused on. Can I leave it and graph from there? I know how to determine if its a function once I put it in my calculator. The question is what do I do with the y value or does it matter?
 
You will input:
 

y= √3x^2 - 4x - 8

 
My graph is saying error and my line is a V with the point at the bottom cut off with that. That is why I was stuck and confused. That was how I initially did the problem before coming here.

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