# Thread: Calculate the area of a part of a function

1. ## Calculate the area of a part of a function

Hi!
y=x^2-12x+36.36
Find all squares that has 2 points on this function and 2 points on the x axis.

I thought this task with using simply y=x^2
To find at which point there is a square I thought a simple equation: 2x=y ; 2x=x^2
2x because the distance is the same on both sides.

Equation gave me result 0 and 2.

x^2-12x+36.36=2x didn't give me the right result.
Any other ideas?

2. In order that the four points form a rectangle, they must be symmmetric about the vertex. Your method worked for $\displaystyle y= x^2$ has vertex (0, 0).

Complete the square in $\displaystyle y=x^2-12x+36.36$ to determine the vertex, $\displaystyle (x_0, y_0)$. You want to find "h" so that the distance from $\displaystyle x_0- h$ to $\displaystyle x_0+ h$, 2h, is the same as the height $\displaystyle y= (x_0+ h)^2-12(x_0+ h)+36.36$

3. y=x^2-12x+36.36
x0=-b/2a = 12/2=6

2h = (6+h)^2 - 12(6+h) + 36.36
Microsoft Math says that h = 1.8 and 0.2

So the area will be (2h)^2
Thanks!