How do i find the x intercepts of r(x)=-2/3x^3+x^2

I need help finding the x intercepts as well as the y intercept. Thankyou for your time.

You can find the y-intercept by setting x equal to zero.
r(0) = -2/3(0)^3 + (0)^2
r(0) = 0
So, the y-intercept is at the origin (0,0)

The x-intercepts are where the graph crosses the x-axis, or where y=0.
So, set r(x) equal to zero and solve for x

-2/3x^3 + x^2 = 0 -- Factor out an x^2, as it is common between the two terms
x^2 * (-2/3x +1) = 0 -- as either of these terms can be equal to zero to make the entire equation zero, set both terms separately equal to zero.
x^2 = 0 -2/3x + 1 = 0
x = 0 -2/3x = -1
x = 3/2

So, r(x) equals zero when x = 0 and x = 3/2, making these the x-intercepts.

for the y intercept, set x = 0: r(0) = 0
for the x intercepts set y = 0

0 = x² - (2/3)x³ = x²(1- 2x/3)
and the x intercepts
x = 0, 0, 3/2