The equation of the curve is

Question

Y=x^3+3/2x^2
Find the coordinates of the stationary point

Answer 1

To find the coordinates of the stationary point, we need to find where the derivative of the curve is equal to 0.

First, let's find the derivative of the curve:
Y' = 3x^2 + 3x

Setting Y' to 0:
3x^2 + 3x = 0
3x(x + 1) = 0
x = 0 or x = -1

Now we need to find the corresponding y-values for these x-values:
When x = 0:
Y = (0)^3 + 3/2(0)^2
Y = 0

When x = -1:
Y = (-1)^3 + 3/2(-1)^2
Y = -1 + 3/2
Y = 1/2

Therefore, the coordinates of the stationary points are (0, 0) and (-1, 1/2).