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So I was trying to solve the monbukagakusho maths exam from the year of 2010 that you can find in this link, but I can't understand the questions, I don't know what do they actually want.

Here's one of the questions:

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

$$ y=Ax^2-Bx+C $$

The minimum value of this function is $D$.

I don't know if I have to find $A$, $B$, and $C$. And what's $D$?

Hagen von Eitzen
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Yuran Pereira
  • 123

2 Answers2

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Hint: A quadratic function that has its minumum (or extremum) at $x=x_0$ is of the form $y=a(x-x_0)^2 +b$.

Hagen von Eitzen
  • 374,180
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first of all you should find $A$, $B$ and $C$ with initial $y(-2)=41$, $y(5)=20$ and quadratic function is minimum at $x_{min}=-\frac{-B}{2A}=\frac{B}{2A}$.
so you have three equations and three :
$41=4A+2B+C$
$20=25A-5B+C$
$2=\frac{B}{2A}$
Solve this for $A$, $B$, $C$ and then you can find $D=y_{min}=y(x_{min})$ or $D=y_{min}=-\frac{\Delta}{4A}=-\frac{B^2-4AC}{4A}$

L P
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