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.
+128408 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 ]
+128408 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 ]
