The points at which the tangents to the curve `y = 3^(2) - 12 x + 18` are parallel to x-axis are

Options

A. (2, - 2), (-2, - 34)
B. (2, 34), (-2, 0)
C. (0, 34), (-2, 0)
D. (2, 2), (-2, 34)

Answer

To find the points at which the tangents to the curve \( y = 3x^2 - 12x + 18 \) are parallel to the x-axis, we need to follow these steps:

### Step 1: Differentiate the function
We start by differentiating the function \( y = 3x^2 - 12x + 18 \) with respect to \( x \) to find the slope of the tangent line at any point on the curve.

\[ \frac{dy}{dx} = \frac{d}{dx}(3x^2 - 12x + 18) \]

Using the power rule of differentiation, we get:

\[ \frac{dy}{dx} = 6x - 12 \]

### Step 2: Set the derivative equal to zero
For the tangent to be parallel to the x-axis, the slope must be zero. Therefore, we set the derivative equal to zero:

\[ 6x - 12 = 0 \]

### Step 3: Solve for \( x \)
Now, we solve for \( x \):

\[ 6x = 12 \]
\[ x = 2 \]

### Step 4: Find the corresponding \( y \) value
Next, we substitute \( x = 2 \) back into the original equation to find the corresponding \( y \) value:

\[ y = 3(2^2) - 12(2) + 18 \]
\[ y = 3(4) - 24 + 18 \]
\[ y = 12 - 24 + 18 \]
\[ y = 6 \]

### Step 5: Conclusion
Thus, the point at which the tangent to the curve is parallel to the x-axis is \( (2, 6) \).

### Final Answer
The point is \( (2, 6) \).

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