juriemagic

Hi Melody,

According to the handbook, "b" can never be negative. It can only be either a fraction between 0 and 1, or greater than 1. Okay, so this I accept. There is an example in the book on a similar sum, but it is not clear to me. By the way it looks, it appears the "a" is disregarded and "b" is calculated. I followed that aproach and found "b" to be \(1 \over2\),

I then substituted this also into the equation and found "a" to be -1. But I'm not sure if that was the right way to go about it?

Thank you Alan...appreciated..

Heureka,

got it, got it, got it.......thank you kindly...I do appreciate!

Hello Heureka,

i simplified the equation of the parabola before I set the two equations equal to one another...oh golly...I see now...that was wrong..because the one is equal to the other, and if you change one of them, then obviously they will not be equal to one another anymore....I see that...

okay, so then it is \(x^3-4{x^2}-5x=-8\)

it really is from here that I just do not know how to get the point of intersection in the 3rd quadrant...can you please help me here?...

Hello Heureka,

i simplified the equation of the parabola before I set the two equations equal to one another. 

Rom and Heureka,

absolutely right, I honestly do apologise for writing this wrong...I was relying on my memory and failed. please forgive me. the quadrant was the 3rd quadrant, and the value of the hyperbola was \(f(g)={16 \over x}\)

would you please show me how to go about the calculations?..so we have:

\(x^3-4{x^2}-5x=16\)

How do I go from here?

Thank you Melody!,,2 lessons learnt!!...stay blessed!

thank you Melody,

please forgive my stupidness...I have now studied your second step....I don't understand how we can just multiply the bottom part by 10. which math law is at work here?...from step 3 onwards I fully understand, it really is just step 2...would you mind teaching me please?

Melody!!!!!..always to the rescue..lol....thank you!!

You know..this must be telepathy of some sort because I also came across a different approach for this sum in which there was this piece:

\((10-2)*10^{m-1}\)

and I was about to ask here for an explanation as to how this step is achieved, and walla!!..you have it in your response!!..thank you again Melody!

goodness me!!,

Rom, every year the math is becoming more and more complicated...this is for a grade 11 pupil...I cannot believe the complexity of the sums nowadays...thank you very much Rom, I definitely would never have been able!..thank you!!