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

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
mousse
Posts: 2
Joined: Tue Jun 02, 2009 10:19 am
Contact:

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

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!

stapel_eliz
Posts: 1670
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
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?

mousse
Posts: 2
Joined: Tue Jun 02, 2009 10:19 am
Contact:

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

no other information..

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

stapel_eliz
Posts: 1670
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
That will be the solution if b = -1. If b has some other value, then the solution will be something else.

Return to “Intermediate Algebra”