• Mar 27th 2013, 12:21 AM
Renrie

The Quadratic function which takes the value 41 at x=-2 and the value 20 at x = 5 and is minimized at x = 2 is

y = __x2 - __x +__

the minimum value of this function is ____.

• Mar 27th 2013, 12:29 AM
Prove It
Write it as $\displaystyle y = a\,x^2 + b\,x + c$, then $\displaystyle \frac{dy}{dx} = 2a\,x + b$.

You know that when x = 2, $$\displaystyle \frac{dy}{dx} = 0$$ due to the function being minimised there. You also know that when x = -2, y = 41, and when x = 5, y = 20. Substitute all this information into the equations so that you get equations in terms of a,b,c to solve simultaneously. See how you go.
• Mar 27th 2013, 06:09 AM
Renrie
I try the first part is this one is correct?
y=0, x=2

0=2ax +b
0 = 2(2)x + b
0 =4x + b
b =4x

and when I try the other this is my outcome and I don't know what to do next :(
y= 41, x=-2

41=2(-2)x+b
41= -4x + b
41= -4x + 4x (I'd replaced it to 4x since I got b=4x)
41 = 0 (and then it turn out to 0)

next is

y=20, x=5

20 = 2(5)x +b
20 = 10x +b
20 = 10x + 4x
20 = 14x

could you pls help me more. I'm so sorry If I don't get it right :(
• Mar 27th 2013, 06:12 AM
Prove It
Why do you still have x's in your equations when you have let them equal the numerical values? Remember, you are trying to find a,b,c NOT x.
• Mar 27th 2013, 06:23 AM
Renrie
oh yes, you're right!

I done the first part and I ended up here:

0=2ax +b
0 = 2(2)a + b
0 =4a + b
b = -4a

y=41 x=-2
41=2(-2)a+b
41= -4a + b
41= -4a - 4a (I'd replaced it to 4a since I got b= -4a)
41 = -8a
-41/8 = a

is this correct?
• Mar 30th 2013, 11:39 PM
vicwilson123