How do I graph x^2y - y^3 - 5x^2 + 5y^2 = 0 ?

I factored it to get (y-5)(x-y)(x+y) = 0 but would it be correct to graph each factor individually?

Question

How do I graph x^2y - y^3 - 5x^2 + 5y^2 = 0 ?

I factored it to get (y-5)(x-y)(x+y) = 0 but would it be correct to graph each factor individually?

anonymous · Answer

Good one:-)

y= plus minus x or 5, lol

anonymous · Answer

Did u try just doing it as a table? U know, set y at 0, -1,1, etc?

anonymous · Answer

plugging values for y seems to create the same graph as if i had graphed the factors individually, except that it doesn't extend over to quadrants II and III... if i did that correctly...

anonymous · Answer

Wolframs version.. http://www.wolframalpha.com/input/?i=x^2y+-+y^3+-+5x^2+%2B+5y^2+%3D+0

anonymous · Answer

oh ok so would graphing the factors individually work for other equations as well? or are there exceptions?

anonymous · Answer

I think someone was having a little joke with u when they gave u this one..:-)

If u just think of a quadratic equation it is clear that graphing the factors is not correct.

That doesn't mean that the factors don't provide useful info, though.

anonymous · Answer

haha ok thanks :)

anonymous · Answer

ur welcome.

anonymous · Answer

I came across this question in a Pre-calculus book. The answer was actually very simple.

$$x ^{2}y-y ^{3}-5x ^{2}+5y ^{2}=0$$

Then move your isolate your variables like this:

$$x ^{2}y-5x ^{2}=y ^{3}-5y^{2}$$

$$x ^{2}(y-5)=y ^{2}(y-5)$$

Finally, simplify the equation to get:

$$x ^{2}=y ^{2}$$

x=y

anonymous · Answer

Cubic in y means 3 roots (not 1) for y, y= plus or minus x and y= 5