Thread: How to find f(x) - I suspect an error in the command

Originally Posted by Viou

Hello
I want to solve a question, I don't understand why my answer is false so i think there might be an error in the order.

The quadratic function takes the value of 41 at x=-2 and the value of 20 at x=5. The function is minimalized at x=2.
I'm asked to give :
- the A,B,C of y=Ax^2+Bx+C
- the minimum value of this function, D

The appex have coordinates (2;D)
2=-B/2A
B=-4A

I then replaced B in

41=A(-2)^2 +B(-2)+C
20=A(5)^2 +B(5)+C

In the end I've got f(x)= (-3/7)x^2-(12/7)x+(275/7)

It can't be right because A can't be an negative number if the function has a minimum. Also in this case the appex is not in x=2 because D=239/7 (it's higher than f(-2)=20)
But when I calculate
f(-2)= (-3/7)(-2)^2-(12/7)(-2)+(275/7)
AND
f(5)=(-3/7)5^2-(12/7)5+(275/7)
It takes the value of 41 and 20.

Is it me confusing everything or there is really an error in the command?

see this post ...

http://www.mathhelpforum.com/math-he...le-175161.html

Do you understand that a quadratic function can be written as where is the x coordinate of the vertex (minimum point)? Knowing that the minimum occurs at x= 2 tells you that you can write the quadratic function as . Putting x= -2 and y= 41 gives 41= a(-2-2)^2+ b= 16a+ b. putting x= 5 and y= 20 gives 20= a(5- 2)^2+ b= 9a+ b. You can solve those two equations for a and b.

Ok thank you very much! It worked out nice
And sorry for putting it in the wrong section