To rotate the graph of $y = 2x$ by $180^\circ$ clockwise around the point $(0, 4)$, follow these steps:

Step-by-Step Process:

1. Understand Rotation by $180^\circ$: A $180^\circ$ clockwise rotation is equivalent to flipping every point $(x, y)$ to $(-x, -y)$ with respect to the rotation center.

2. Change the Reference Point: Shift the graph so the rotation center, $(0, 4)$, becomes the origin. To do this, subtract $4$ from all $y$-values: $y - 4 = 2x$ Simplify: $y' = 2x \quad \text{(where } y' = y - 4 \text{)}.$

3. Apply $180^\circ$ Rotation: Under a $180^\circ$ rotation about the origin, the new coordinates of any point $(x, y')$ become $(-x, -y')$. Thus: $x' = -x, \quad y'' = -y'.$

   Substitute $y' = 2x$ into $-y'$: $y'' = -2(-x) = 2x.$ So the new line becomes: $y'' = 2(-x).$ Simplify: $y'' = -2x.$

4. Return to the Original Coordinate System: To reverse the earlier shift, add $4$ back to $y$-values: $y = -2x + 4.$

Final Result:

The equation of the rotated line is: $y = -2x + 4$

This is the graph of the line after a $180^\circ$ clockwise rotation around the point $(0, 4)$.

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Would you like a graph to visualize this transformation?

Related Questions:

1. What happens to the graph of $y = 2x$ if rotated $90^\circ$ clockwise around $(0, 4)$?
2. How would the result change if the rotation center was $(2, 4)$ instead?
3. Can you derive the same rotated equation by matrix transformation?
4. What is the effect of a $180^\circ$ rotation on the slope of a line?
5. How would you rotate a parabolic curve, such as $y = x^2$, about a point?

Tip:

When rotating objects, always simplify the problem by translating the rotation center to the origin! This reduces errors and simplifies the computation.