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

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SUMMARY

The discussion focuses on determining the possible values of y for the function y=(x+2)/(3-x^2). Participants emphasize the need to find the range of this function by solving for x and identifying where the function is undefined. A key insight is the use of completing the square to analyze the function, leading to the conclusion that the range can be derived from the discriminant of the corresponding quadratic equation. The final form of the function indicates that the maximum value occurs at specific stationary points.

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  • Knowledge of completing the square technique
  • Familiarity with quadratic equations and discriminants
  • Basic concepts of stationary points in calculus
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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
 
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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 x=\sqrt{y} or x=-\sqrt{y}. In either case, this equation is undefined for y<0. Thus if we have y\geq 0, then there exists a corresponding y-value. Otherwise, such a y-value does not exist. Thus \mathbb{R}^+ is our range...
 
Complete the square:

<br /> y=-\left( x+\frac{1}{2}\right)^{2}+\frac{11}{12}<br />

What happens when x=1/2? Does this graph have a maximum/minimum?
 
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 :)
 
TFGordon said:
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: y=x^2+x. This corresponds to the quadratic equality x^2+x-y=0. So, the discriminant is D=1+4y This is positive if y\geq -1/4. So the range is [-1/4,+\infty[.
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...
 
Do you mean: y=(x+2)/(3-x2) ?

Find the local min & max.
 

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