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# Help with 2 problems

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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 $\pi$.

Guest Feb 17, 2018
#1
+88898
+1

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

CPhill  Feb 17, 2018
#2
+88898
+1

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

CPhill  Feb 17, 2018