Find the points of intersection fo y = |x^2-4| and y = 2x+1

Answer 1

|x^2-4| is a parabola, with x intercepts -2 and 2

but with the "inside" part flipped upside down, reflected over the y axis, for -2<x<2

the x intercepts are the minimum points. the "inside" part of the parabola is y= -x^2+4, the "outside" part is y= x^2-4

2x+1 is a straight line with slope =2, y intercept =1 x intercept =-1/2

they intersect twice at (1,3) and (1+sqr6, 2+2sqr6) = about (3.45,9.9)

for "inside" from -2<x<2

-x^2 +4 =2x+1

x^2+2x-3 =0

(x+3)(x-1) =0

x =1, y =2(1)+1 =3 (ignore x=-3 as it's outside the domain)

intersection point is (1,3)

for x<-2 and x>2

x^2-4 =2x+1

x^2 -2x -5 =0

x = 2/2+(1/2)sqr(2^2+4(5)) =1+sqr24/2 = 1+sqr6 = about 3.45

y=3+2sqr6 = about 9.9

it helps to sketch a graph of the parabola and line