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Conclusion To Quadratic Equations


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

The 21st Century learning professional: Training isn't learning - TrainingZone.co.uk


TrainingZone.co.uk

The 21st Century learning professional: Training isn't learning
TrainingZone.co.uk, UK - Jun 18, 2008
... only be said to be 'trained' when they have demonstrated that they have applied what they learned (remember the last time you used a quadratic equation? ...


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Test3 April 19th, 2008 A Real Version for the Square root of -1 i above was conceived from the first letter of Imaginary, because the number representing the square root of -1 doesn’t exist on the real number line. Pythagorus and other great Mathematicians were dumbfounded by the meaning of i, yet in todays higher level Maths, i is now taken for granted. However, it’s true meaning seems to have been bypassed as our interpretation of it still allows history to dominate our thinking basi


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