The diagram shows the curve y=x^2 intersecting the straight line AC (y+x=6) at B. Find

Question

The diagram shows the curve y=x^2 intersecting the straight line AC at B. Find

a.     The coordinates of B

b.     The area of the shaded region P

c.     The area of the shaded region Q

Answer 1

y=x²

y+ x=6

y=6-x

x²=6-x

x² + x - 6= 0

(x+ 3) ( x - 2 ) =0

x= -3 or x=2

for x= -3, y=9 , the point of intersection is (-3,9)

for x= 2, y=4, the second point of intersection is ( 2,4)

Since the diagram is not shown, I will assume you need to find the area enclosed between the line y= 6-x and the curve y=x² from x =-3 to x = 2.To find this area

x=2 x=2 x=2

∫ (y₁ - y _{2 ) dx =} ∫ ( 6 - x - x² ) dx = 6x - x² /2 - x³ /3 Ι

x= -3 x= -3 x=-3

=( 12 - 2 - 8/3 ) - ( -18 - 9/2 + 27 /3)

=10 -8/3 +9 +9/2 = 125/ 6

Answer 2

y = x^2 = 6-x

x^2 +x-6 = 0

( x + 3)(x - 2 )=0

x+3=0 --> x = -3

x-2 =0 -->x = 2

(2,4) or (-3,9)