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

Jan 6, 2017

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

#### Explanation:

To make y real, $x \in \left[- 1 , 1\right]$

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

I have managed to find approximation tothe square of the two

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

So, the turning points are at $\left(\pm \sqrt{0.56} , 1.30\right) = \left(\pm 0.75 , 1.30\right)$,

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]}