can anybody help me with this?
i think i found the equation alright, y'=3x^2-3/2y
but that's all i know how to do

Critical point is when the derivative is 0 or undefined.
y'= (3/2)(x^2-1)/y
= 3/2(x-1)(x+1)/y
So x=1 x=-1 y=0 are the locations where the function is possibly critical. So plug those in original equation and solve.
At x=1 the function is undefined and therefore not a critical point
At x=-1:
y= + or - sqrt(2)
At y= 0:
x=0
x= + or - sqrt(3)
That's 5 critical points with 3 having the same y and 2 having the same x.

I can explain more if needed. But on phone atm

This post was edited by kasey21 on Mar 27 2019 06:44pm

for the second time this semester you've saved my life brother
i understand, but the only part where im confused is where you got x= + or - sqrt (3)
what did you plug in to get that?

original equation is:
y^2 = x^3 - 3x

which can be written as:
y^2 = x(x^2 - 3)
plug in y = 0 (one of the critical points)

0 = x (x^2 - 3)
so we have
x = 0 (critical point: 0,0)
and
x^2 - 3 = 0
solve for x:
x^2 = 3
x = + or - sqrt(3)

or you could factor (x^2 -3) to:
(x-sqrt(3)) * (x+sqrt(3)).
Which of course you get the same answer.

This post was edited by kasey21 on Mar 27 2019 09:04pm