Find the coordinates of points B and C.

Points B and C are the x-intercepts, where the graph crosses the x-axis. At these points, $y=0$.
Set $y=0$ in the equation:

Factor out $x$:

This gives two possible values for $x$:

So, the x-intercepts are $(0, 0)$ and $(4, 0)$. From the graph, point B is the origin and point C is the other intercept.
The coordinates are:

• B: $\boxed{(0, 0)}$
• C: $\boxed{(4, 0)}$

The equation $y = x^2 - 4x$ is a parabola opening upwards (since the coefficient of $x^2$ is positive). Its minimum value occurs at the vertex (turning point).
The x-coordinate of the vertex for a quadratic equation $y = ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$.
In our equation, $a=1$ and $b=-4$.

Now, substitute this x-value back into the equation to find the minimum y-value:

The minimum value of $y$ is $\boxed{-4}$.

The turning point is the vertex of the parabola. From Step 2, we found the x-coordinate of the vertex is $x=2$ and the corresponding y-value (minimum value) is $y=-4$.
The coordinates of the turning point are $\boxed{(2, -4)}$.