Determining the possible values of y for the graph y=x+2/3-x^2

  • Thread starter TFGordon
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  • #1
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Homework Statement


Determine the possible values that y can take for the graph y=x+2/3-x^2


Homework Equations


N/A


The Attempt at a Solution


y=k
k(3-x^2)=x+2
3k-kx^2=x+2
-kx^2+3k-x-2=0

I've tried factorising my k terms and much more faffing about to no avail...help please? :s
 

Answers and Replies

  • #2
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Hi TFGordon :smile:

I suppose you need to find the range of this function. This is very easy, you'll just need to solve the equation for x, and see where it is undefined.

Example, y=x2. We need to solve this for x. We get [tex]x=\sqrt{y}[/tex] or [tex]x=-\sqrt{y}[/tex]. In either case, this equation is undefined for y<0. Thus if we have [tex]y\geq 0[/tex], then there exists a corresponding y-value. Otherwise, such a y-value does not exist. Thus [tex]\mathbb{R}^+[/tex] is our range...
 
  • #3
hunt_mat
Homework Helper
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Complete the square:

[tex]
y=-\left( x+\frac{1}{2}\right)^{2}+\frac{11}{12}
[/tex]

What happens when x=1/2? Does this graph have a maximum/minimum?
 
  • #4
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I'm a further mathematics A level student, micromass. I appreciate your help but do you really think I would be attempting to to solve this equation if I was unaware that x^2=y works out as x=y^1/2? I know how graphical inequalities work pal, my problem is with the algebra, not the concept of an inequality.

Thankyou hunt mat! Just the ticket :) As it happens, the next part of the question is 'Find the co-ordinates of the stationary points of the curve' so yes, I imagine it does. Thankyou very much, I'd completely neglected to consider that approach :)
 
  • #5
22,129
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I'm a further mathematics A level student, micromass.

So? No need to get an attitude here...

I appreciate your help but do you really think I would be attempting to to solve this equation if I was unaware that x^2=y works out as x=y^1/2?

If you want better help, then you should have written more information in your attempt. You think it's easy to identify somebody's problems? I thought that you had problems with the general method. If you had problems with the algebra, then you should have written that.

I know how graphical inequalities work pal, my problem is with the algebra, not the concept of an inequality.

Very nice, but I don't quite see how graphical inequalities come into play here..

Anyway, let me give you another example: [tex]y=x^2+x[/tex]. This corresponds to the quadratic equality [tex]x^2+x-y=0[/tex]. So, the discriminant is [tex]D=1+4y[/tex] This is postive if [tex]y\geq -1/4[/tex]. So the range is [tex][-1/4,+\infty[ [/tex].
With this example, you can easily calculate the range in your problem. It's the same thing really. So further mathematics A level student should have no problems with it...
 
  • #6
SammyS
Staff Emeritus
Science Advisor
Homework Helper
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Do you mean: y=(x+2)/(3-x2) ?

Find the local min & max.
 

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