The Purplemath Forums

by **mousse** on Tue Jun 02, 2009 10:48 am

For function $f(x)=ax^2+bx+c$ , it is known that $f\left(\frac{1}{2}\right)=\frac{5}{4}$ and $f(-1)=2$ . If so, what is $f\left(\sqrt{7}-1\right)?$

thanx!

**Sponsor**

by **stapel_eliz** on Tue Jun 02, 2009 8:25 pm

You are given only two data points, so you can't find the exact form of the function's expression. And I'm not seeing any particular way to relate the square root of seven to either of the listed values. The closest I can get is:

$f(\sqrt{7}\, -\, 1)\, =\, (14 - 3\sqrt{7})b\, +\, 9\, -\, 2\sqrt{7}$

Is there any other information?

by **mousse** on Tue Jun 02, 2009 8:36 pm

no other information..

the solution is sqrt(7) - 5 if that helps somehow

by **stapel_eliz** on Wed Jun 03, 2009 12:03 am

That will be the solution if b = -1. If b has some other value, then the solution will be something else.

The Purplemath Forums

for f=ax^2+bx+c, f(1/2)=5/4, f(-1)=2, what is f(sqrt[7]-1)?

for f=ax^2+bx+c, f(1/2)=5/4, f(-1)=2, what is f(sqrt[7]-1)?

Re: for f=ax^2+bx+c, f(1/2)=5/4, f(-1)=2, what is f(sqrt[7]-1)?