How to Sketch the Curve y^2 = x^3?

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The equation y^2 = x^3 leads to the function y = ±x^(3/2), indicating that the graph has two branches, one above and one below the x-axis. The shape resembles that of a parabola but is steeper for x > 1 and lower for 0 < x < 1. The derivative dy/dx = ±1.5x^(1/2) confirms the presence of two branches in the slope as well. To sketch the curve accurately, both branches should be included, reflecting the symmetry about the x-axis. Understanding these characteristics is crucial for proper graphing of the equation.
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Homework Statement




Hi everyone, I am not sure how to sketch this:
y^2 = X^3
I have a feeling it will form a parabola.

If I find the square root of both sides I get:
y = X^3/2
Differentiating gives me
dy/dx = 1.5X^1/2. Kind of stuck from here. Please help.

Homework Equations





The Attempt at a Solution

 
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Think about the shape of the graph for y=x2, then y=x3, y=x4 etc. only for values x>0. The shape is roughly the same, no? The only difference is that for x>1 the graph is steeper and for 0<x<1 the graph is lower for larger powers.
So y=x3/2 will be roughly the same as y=x2 for x>0
But remember that when you take the square root of both sides, you need to remember that \pm so your final result of y^2=x^3 or y=\pm x^{3/2} is the graph of y=x3/2 for values of x>0 and then exactly the same thing but reflected in the x-axis for y=-x3/2.
 
aurao2003 said:

Homework Statement




Hi everyone, I am not sure how to sketch this:
y^2 = X^3
I have a feeling it will form a parabola.

If I find the square root of both sides I get:
y = X^3/2
Actually, it's
y = \pm x^{3/2}
There are two branches.
aurao2003 said:
Differentiating gives me
dy/dx = 1.5X^1/2. Kind of stuck from here. Please help.
The derivative also has two branches.
dy/dx = \pm 1.5x^{1/2}
aurao2003 said:

Homework Equations





The Attempt at a Solution

 
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