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A triangle is formed with one vertex at the vertex of the parabola \(y=x^2-1\) and the other two vertices at the intersections of the line \(y=r\) and the parabola. If the area of the triangle is between 8 and 64 inclusive, find all possible values of r. Express your answer in interval notation.

Jul 7, 2022

#2
+129465
+2

To make this problem easier, let's just shift the parabola up one unit so the we have the parabola  y =  x^2

The height  of our triangle = x^2  and the base =  2x

To have an area of 8  we need to  solve this

(1/2)* 2x  * x^2   = 8

x^3  =  8

x = 2

At x = 2 the base will be  2(2)  = 4  and the height will be  2^2 =  4  = y

And similarly   for and area of  64  we need  to  solve this

(1/2) * 2x  * x^2 =  64

x^4 =  64

x =  4

At x = 4, the base =  2(4) = 8   and the height = 4^2  = 16 = y

So....we  need to  shift everything down  by one  unit

y = "r"  will range from   (4-1) =  3  to  (16 -1)  =   15

So  r =   [ 3 , 15 ]

Jul 8, 2022

#1
0

Jul 7, 2022
#2
+129465
+2

To make this problem easier, let's just shift the parabola up one unit so the we have the parabola  y =  x^2

The height  of our triangle = x^2  and the base =  2x

To have an area of 8  we need to  solve this

(1/2)* 2x  * x^2   = 8

x^3  =  8

x = 2

At x = 2 the base will be  2(2)  = 4  and the height will be  2^2 =  4  = y

And similarly   for and area of  64  we need  to  solve this

(1/2) * 2x  * x^2 =  64

x^4 =  64

x =  4

At x = 4, the base =  2(4) = 8   and the height = 4^2  = 16 = y

So....we  need to  shift everything down  by one  unit

y = "r"  will range from   (4-1) =  3  to  (16 -1)  =   15

So  r =   [ 3 , 15 ]

CPhill Jul 8, 2022