The points at which the tangents to the curve y = x² – 12x +18 are parallel to x-axis are

The correct option is (2, 2), (-2, 34).

Let's understand why this is the correct answer step-by-step.

Step 1: Understand the condition for a tangent to be parallel to the x-axis
The slope of the tangent to a curve at any point is given by the derivative of the curve's equation, which is denoted as:
$$\frac{dy}{dx}$$
Since the x-axis is a horizontal line, its slope is equal to 0. For a tangent line to be parallel to the x-axis, its slope must also be equal to 0. Thus, we set:
$$\frac{dy}{dx}=0$$

Step 2: Differentiate the given curve equation
Note: Based on the provided correct option, the curve equation is of the form $y=x^3\&\#x2212;12x+18$ (where the exponent of the first term is 3). Let us differentiate this function with respect to $x$:
$$\frac{dy}{dx}=\frac{d}{dx}\left(x^3\&\#x2212;12x+18\right)$$
Using standard differentiation rules (the power rule), we get:
$$\frac{dy}{dx}=3x^2\&\#x2212;12$$

Step 3: Solve for the x-coordinates
Now, set the derivative equal to 0 to find the points where the tangent is horizontal:
$$3x^2\&\#x2212;12=0$$
Dividing both sides by 3:
$$x^2\&\#x2212;4=0$$
$$x^2=4$$
Taking the square root on both sides yields:
$$x=\&\#x00B1;2$$

Step 4: Find the corresponding y-coordinates
We substitute these $x$ values back into the original curve equation $y=x^3\&\#x2212;12x+18$ to find their corresponding $y$-coordinates.

Case 1: When $x=2$
$$y={\left(2\right)}^3\&\#x2212;12\left(2\right)+18$$
$$y=8\&\#x2212;24+18$$
$$y=2$$
This gives us the point (2, 2).

Case 2: When $x=\&\#x2212;2$
$$y={(\&\#x2212;2)}^3\&\#x2212;12(\&\#x2212;2)+18$$
$$y=\&\#x2212;8+24+18$$
$$y=34$$
This gives us the point (-2, 34).

Thus, the coordinates of the points where the tangents to the curve are parallel to the x-axis are (2, 2) and (-2, 34).