# Can anyone teach me, step by step, how to solve this problem? Solving for Y. (x^2+y^2-1)^3-x^2y^3=0I am supposed to graph it. I can't figure out how to solve for X. (or to get Y by itself so i...

Can anyone teach me, step by step, how to solve this problem? Solving for Y.

(x^2+y^2-1)^3-x^2y^3=0

I am supposed to graph it.

I can't figure out how to solve for X. (or to get Y by itself so i can put in the amounts I am supposed to solve for)

The figures I am trying to solve for are: -1.5, -1, -0.5, 0, 0.5, 1, 1.5

### 1 Answer | Add Yours

Graph `(x^2+y^2-1)^3-x^2y^3=0`

This is not a function, and is difficult to write as 2 different functions. One way to graph this is to create a table:

(1) If x=0 then you have `(y^2-1)^3=0` which implies that `y^2-1=0=>y=+-1`

Thus we have the points (0,1) and (0,-1) on the graph

(2) If x=1 we have `(y^2)^3-y^3=0` or `y^6-y^3=0` . Factoring we get `y^3(y^3-1)=0=>y^3=0 "or" y^3-1=0` . Thus y=0 or y=1.

By symmetry we notice that the same holds true for x=-1 (In fact, x and -x will always give the same y-values -- the graph is symmetric across the y-axis)

So we have the points (1,0),(1,1),(-1,0),(-1,1)

(3) If `x=+-.5` we have `(y^2-3/4)^3-y^3/4=0` . Then `(y^2-3/4)^3=y^3/4` . Taking the cube root of both sides yields `y^2-3/4=y/(root(3)(4))` or `y^2-1/(root(3)(4))y-3/4=0` . Using the quadratic formula we get `y~~(4^(-1/3)+-1.843)/2` or `y~~1.23,y~~-.607`

We now have the points (.5,1.23),(.5,-.607),(-.5,1.23),(-.5,-.607)

(4) If `x=+-1.5` you will find that there are no real solutions for y.

x=1.5 implies `(y^2+1.25)^3-2.25y^3=0`

(5) **So we have the points (-1,1),(-1,0),(-.5,1.23),(-.5,-.607),(0,1),(0,-1),(.5,1.23),(.5,-.607),(1,1),(1,0). We also know that the graph is symmetric across the y-axis.**

The graph is shaped like a cardioid(heart) with a vertical axis of symmetry. The tip of the "heart" is at (0,-1), and the point where the sides come together is (0,1).

I've included a link to the graph.

**Sources:**