Hi,

Is it possible to manipulate a quadratic function so that it has a gradient of 35.5 at the point (1,33) while forcing it to have an x-intercept of 10 and the other x-intercept being anything less then 10. I have no idea how to approach this...
Any help will be greatly appreciated.

Thanks,
John

2. Originally Posted by deovolante
Hi,

Is it possible to manipulate a quadratic function

Let's call it $ax^2+bx+c$

so that it has a gradient of 35.5 at the point (1,33)

So $a+b+c=33\,\,\,and\,\,\,also\,\,\,2a+b=35.5$

while forcing it to have an x-intercept of 10 and the other x-intercept being anything less then 10

This means $100a+10b+c=0$ and also that there must exist $\alpha<10\,\,\,s.t.\,\,\,a\alpha^2+b\alpha+c=0$

. I have no idea how to approach this...
Any help will be greatly appreciated.

Thanks,
John

Now solve the linear system formed by the first three eq.'s above (it has a unique, pretty nasty, solution) , and then check ( for example, with

the help of Viete equations) if the coefficients a,b,c you get can fulfill the last condition together with the other ones.

Tonio