In y = p^3,
if p is constant, then y = p^3 should be a horizontal line only....not an "S"-curve, or cubic curve, as shown in the diagram.
In the figure below, ABCD is a rectangle. Points A and C lie on the graph of , where p is a constant. If the area of ABCD is 4, what is the value of p?
Sorry for the sloppy graph! Can you please explain how you got the answer?
Thanks in advance!!
- - - -599
So it is y = p(x^3). Now it makes sense.
Given, area of ABCD = 4 sq.units.
AD = BC = |-1/2| +|1/2| = 1 unit
So, AB = CD = 4/1 = 4 units ----------**
At point A(-1/2,a)
y = px^3
a = p(-1/2)^3 = -p/8
At point B(-1/2,b)
b = p(-1/2)^3 = -p/8
So a = b = -p/8
That means "a" and b are equidistant from the x-axis.
Since AB = 4 = (a +b), then,
a +b = 4
|-p/8| +|-p/8| = 4
2|-p/8| = 4
|-p/8| = 2
Square both sides to eliminate the absolute sign,
p^2 / 64 = 4
p^2 = 4*64
Take the square roots of both sides,
p = +,-2*8
p = +,-16 -----------answer.
EDIT:
The rectangle corner points distracted me.
The p is concerned about the cubic curve only, I forgot.
The p = sqrt(4*64) = 2*8 = 16 only. Not +,-16.