The number of point(s) on the curve `y^(3)=12y-3x^(2)` where a tangents is vertical is/are

Answer

To find the number of points on the curve \( y^3 = 12y - 3x^2 \) where the tangent is vertical, we need to determine when the derivative \( \frac{dy}{dx} \) is undefined (which corresponds to a vertical tangent).

### Step-by-Step Solution:

1. **Differentiate the given equation implicitly**:
We start with the equation of the curve:
\[ y^3 = 12y - 3x^2 \]
We differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(y^3) = \frac{d}{dx}(12y - 3x^2) \]
Using the chain rule on the left side and the product rule on the right side, we get:
\[ 3y^2 \frac{dy}{dx} = 12 \frac{dy}{dx} - 6x \]

2. **Rearranging the equation**:
We can rearrange the equation to isolate \( \frac{dy}{dx} \):
\[ 3y^2 \frac{dy}{dx} - 12 \frac{dy}{dx} = -6x \]
Factoring out \( \frac{dy}{dx} \):
\[ \frac{dy}{dx}(3y^2 - 12) = -6x \]

3. **Solving for \( \frac{dy}{dx} \)**:
Now, we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{-6x}{3y^2 - 12} \]
A vertical tangent occurs when \( \frac{dy}{dx} \) is undefined, which happens when the denominator is zero:
\[ 3y^2 - 12 = 0 \]

4. **Finding the values of \( y \)**:
Solving for \( y \):
\[ 3y^2 = 12 \implies y^2 = 4 \implies y = \pm 2 \]

5. **Finding corresponding \( x \) values**:
Now we substitute \( y = 2 \) and \( y = -2 \) back into the original curve equation to find the corresponding \( x \) values.

For \( y = 2 \):
\[ (2)^3 = 12(2) - 3x^2 \implies 8 = 24 - 3x^2 \implies 3x^2 = 16 \implies x^2 = \frac{16}{3} \implies x = \pm \frac{4}{\sqrt{3}} \]

For \( y = -2 \):
\[ (-2)^3 = 12(-2) - 3x^2 \implies -8 = -24 - 3x^2 \implies 3x^2 = 16 \implies x^2 = \frac{16}{3} \implies x = \pm \frac{4}{\sqrt{3}} \]

6. **Counting the points**:
Thus, for each value of \( y \) (2 and -2), there are two corresponding \( x \) values. Therefore, we have:
- For \( y = 2 \): \( \left( \frac{4}{\sqrt{3}}, 2 \right) \) and \( \left( -\frac{4}{\sqrt{3}}, 2 \right) \)
- For \( y = -2 \): \( \left( \frac{4}{\sqrt{3}}, -2 \right) \) and \( \left( -\frac{4}{\sqrt{3}}, -2 \right) \)

This gives us a total of 4 points where the tangent is vertical.

### Final Answer:
The number of points on the curve where the tangent is vertical is **4**.