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 ____.

2. ## Re: Quadratic function (HELP)

Write it as $\displaystyle \displaystyle y = a\,x^2 + b\,x + c$, then $\displaystyle \displaystyle \frac{dy}{dx} = 2a\,x + b$.

You know that when x = 2, [tex]\displaystyle \frac{dy}{dx} = 0[/math] 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.

3. ## Re: Quadratic function (HELP)

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

4. ## Re: Quadratic function (HELP)

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.

5. ## Re: Quadratic function (HELP)

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?