What is the graph of the Cartesian equation #y = 0.75 x^(2/3) +- sqrt(1 - x^2)#?

1 Answer
Jan 6, 2017

Answer:

See the second graph. The first is for turning points, from y' = 0.

Explanation:

To make y real, #x in [-1, 1]#

If (x. y) is on the graph, so is (-x, y). So, the graph is symmetrical

about y-axis.

I have managed to find approximation tothe square of the two

[zeros](https://socratic.org/precalculus/polynomial-functions-of-

higher-degree/zeros) of y' as 0.56, nearly.

So, the turning points are at #(+-sqrt 0.56, 1.30)=(+-0.75, 1.30)#,

nearly.

See the first ad hoc graph.

The second is for the given function.

graph{x^4+x^3-3x^2+3x-1 [0.55, 0.56, 0, .100]}
.
graph{(y-x^(2/3))^2+x^2-1=0 [-5, 5, -2.5, 2.5]}