The problem provides the equation of a function $y = -x(x-2)^2$ and its graph. The question does not explicitly ask for anything but based on the format of the request, I will describe the properties of the function.

1. Problem Description

The problem provides the equation of a function $y = -x(x-2)^2$ and its graph. The question does not explicitly ask for anything but based on the format of the request, I will describe the properties of the function.

2. Solution Steps

The equation is $y = -x(x-2)^2$. To understand the function, we should find its roots. The roots are the values of $x$ for which $y=0$.

$0 = -x(x-2)^2$

This equation is satisfied if $x=0$ or $(x-2)^2 = 0$.

If $(x-2)^2 = 0$, then $x-2 = 0$, which means $x=2$.

Thus, the roots of the function are $x=0$ and $x=2$. The root at $x=2$ is a repeated root, meaning the graph touches the x-axis at $x=2$ but does not cross it. The root at $x=0$ is a single root, meaning the graph crosses the x-axis at $x=0$.

Next, we can expand the expression.

$y = -x(x^2 - 4x + 4)$

$y = -x^3 + 4x^2 - 4x$

This is a cubic polynomial. Since the leading coefficient is negative, the graph will tend to $+\infty$ as $x$ tends to $-\infty$ and it will tend to $-\infty$ as $x$ tends to $+\infty$.

To find the y-intercept, we plug in $x=0$, so $y = -0(0-2)^2 = 0$.

The derivative is $y' = -3x^2 + 8x - 4$.

To find critical points, we set $y' = 0$:

$-3x^2 + 8x - 4 = 0$

$3x^2 - 8x + 4 = 0$

$(3x-2)(x-2) = 0$

$x = 2/3$ or $x = 2$

These are the locations of the critical points (local max or min).

3. Final Answer

The function is $y = -x(x-2)^2$. It has roots at $x=0$ and $x=2$. The graph crosses the x-axis at $x=0$ and touches the x-axis at $x=2$. As $x \to -\infty$, $y \to +\infty$. As $x \to +\infty$, $y \to -\infty$. The y-intercept is

0. The critical points occur at $x=2/3$ and $x=2$.