1.The parabola with equation $y=ax^2+bx+c$ is graphed below:
https://latex.artofproblemsolving.com/0/7/6/07675a24920f72edd65e30b6b78914ddf70403be.png
The zeros of the quadratic $ax^2 + bx + c$ are at $x=m$ and $x=n$, where $m>n$. What is $m-n$?
2.Three of the four vertices of a rectangle are $(5, 11)$, $(16, 11)$ and $(16, -2)$. What is the area of the intersection of this rectangular region and the region inside the graph of the equation $(x - 5)^2 + (y + 2)^2 = 9$? Express your answer as a common fraction in terms of $\pi$.
+128407 Using the form
y = a (x - h)^2 + k where the vertex is (h, k) = (2, -4)
We know that, since (4,12) is on the graph
12 = a ( 4 - 2)^2 - 4
16 = 4a ⇒ a = 4
So we have
y = 4(x - 2)^2 - 4
Expanding this we have
y = 4x^2 - 16x + 12
To find the zeroes
0 = 4x^2 - 16x + 12 divide through by 4
0 = x^2 - 4x + 3 factor
0 = (x - 3) ( x - 1)
Setting each factor to 0 and soving for x we get that the roots are x = 3 and x = 1
So... m - n = 3 - 1 = 2
+128407 For the second one....the center of the circle forms the remaining vertex of the rectangle......and the circle will have a radius of 9
The intersection of the rectangular region and the inside of this circle will just be the area of 1/4 of this circle....so it will be :
(pi/4) (9)^2 = [81/4] pi units^2
