grabthebasics.com
home faq wannahelp feedback contact

- solving simultaneous equations using ti 89
- solve quadric equations
- books on the quadratic equation
- quadratic equation format
- two ways to solve simultaneous equations
- simultaneous equations the easy way
- "solving three equations"
- ti 83 quadratic formula program code5c
- how do i factorise -5c-30d
- quadratic equasions
- simulataneous quadratic equation c++
- step by step how to solve quadratic equation
- uses for quadratic equation
- easy to enter simultaneous equations
- solving the quadratic equations
- lineage2.patch box
- write the equation of a quadratic function with vertex at %28-1,1%29
- simultaneous equation calculater
- rules to solve brackets in mathematics
- solving four step equations for sudoku

Factorise X Squared 100


Solving Quadratic Equations

But there was more!:

I'm looking for the equation of a demand curve, which is usually negative sloping, i.e. y = -x. Well, at least that is how it should look. However, I know that y = -x is a linear equation, and that's not exactly what I need. I guess it could be negatively exponential, but I just don't know. Perhaps it might help to give you a few examples of the points on the curve (points are in x:y format)

1) 1,000:400
2) 1,500:300
3) 2,400:200
4) 4,000:100

I should think that the y-intercept would be around 800. It doesn't matter very much whether the curve touches the x-axis, but I would think that the curve would start approaching it around 8,000 or 9,000.

This is actually the curve that I'm trying to find the equation for. Not only that, but then instead of having it in the format of y = ... , I would need it in the format x = ... If you could show me how to find that, I would appreciate it very much.

Answer:

The curve looks like the top right of a circle as I visualise it. You will find it difficult though to work out the exact equation of the curve.

Perhaps this may help put it in the form x=ay^2+bx+c:

y=ax^2+bx+c

y-c=ax^2+bx

You then have to complete the square:

y-c=sq( rt(a)x+(b/(2rt(a)) ) - b^2/4a (test it by multiplying it out. I may have made a mistake)

y-c+b^2/4a= sq( rt(a)x + (b/2rt(a)) )

rt( y-c+b^2/4a ) =rt(a)x + (b/2rt(a))

rt( y-c+b^2/4a)- (b/2rt(a)) = rt(a)x

<rt( y-c+b^2/4a ) - (b/2rt(a)) >/ <rt(a)>= x

A bit difficult to follow, so suggest writing it out on paper. Have used sq to mean ^2 at times and obviously rt(x) to mean the square-root of x. May be some mistakes here so apologise in advance if so, but gives the general idea on how to isolate x

home FAQ How can I help feedback Contact