The curve 3y^2- 3xy +2x^3 = 7 has vertical tangents when:

a. y=x b. y=2x c. 2y=x d. y=3x e. 3y=x

6y dy/dx -3x dy/dx - 3y + 6x^2 = 0

dy/dx (6y - 3x) = 3y - 6x^2
dy/dx = (3y-6x^2)/(6y-3x)
the tangent is undefined when the denominator of dy/dx is zero, that is,
6y - 3x = 0
2y = x

The curve 3y^2 - 3xy + 2x^3 = 7 has vertical tangents when

To find the values of x where the given curve has vertical tangents, we need to determine the slope of the curve at different points. Vertical tangents occur when the slope of the curve is undefined, meaning the derivative of y with respect to x is either infinite or does not exist.

First, let's find the derivative of y with respect to x for the given curve:

3y^2 - 3xy + 2x^3 = 7

Differentiating both sides with respect to x:

6yy' - 3xy' + 6x^2 = 0

Rearranging the equation:

y'(6y - 3x) = -6x^2

y' = -6x^2 / (6y - 3x)
= -x^2 / (y - (x/2))

To find the x-values where y' approaches infinity or does not exist, we need to determine when the denominator of y' becomes zero.

Setting the denominator equal to zero:

y - (x/2) = 0
y = x/2

Therefore, at the points where y = x/2, the curve has vertical tangents. Checking the given options:

a. y = x: Not equivalent to y = x/2. Incorrect.
b. y = 2x: Not equivalent to y = x/2. Incorrect.
c. 2y = x: Equivalent to y = x/2. Correct!
d. y = 3x: Not equivalent to y = x/2. Incorrect.
e. 3y = x: Not equivalent to y = x/2. Incorrect.

Therefore, the correct answer is c. 2y = x