Ranges of simple quadratic functions

AI Thread Summary
The discussion revolves around determining the range of a simple quadratic function represented as a fraction of two quadratic equations. The user presents the equations and derives a discriminant condition, indicating that it can be positive or zero. The critical inequality derived is (b-qy)² - 4(a-py)(c-ry) ≥ 0, which is essential for finding the range of y. Clarification is sought on how this inequality can yield a range for y and the next steps in the analysis. The user also notes a potential misunderstanding regarding the number of equations involved, emphasizing the complexity of working with a quadratic fraction.
Kartik.
Messages
55
Reaction score
1
Taking a general function with two quadratic eqs,
Y = (ax^2+bx+c / px^2+qx+r)
ax+bx+c = pyx^2+qxy+ry
x^2(a-py) + x(b-qy)+(c-ry)=0
The discriminant turns out to be a+ve or a 0.
So,
(b-qy)^2 - 4(a-py)(c-ry) >or= 0
Now how does this equation yield a range for y?or can it?what are the next steps?
 
Physics news on Phys.org
You say you have "two quadratic equations" but I see three equations, none of which is quadratic.
 
that is probably because of me using the m.physicsforum because of which i lack a "^2”(at the 2nd equation, the way you put it) and a couple of signs of implications after the first equation(which is the attempt towards the solution) and by saying i have two quadratic equations, i mean to say one quadratic eq over the other = quadratic fraction
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top