Find the intercepts of $y=2x-x^2$ .

A) x -intercept: (0,0) y -intercepts: (0,0) and (2,0)
B) x -intercepts: (0,0) and (2,0) y -intercept: (0,0)
C) x -intercepts: (0,0) and $\left(\frac{1}{2},0\right)$ y -intercept: (0,0)
D) x -intercepts: (0,0) and $(-2,0)$ y -intercept: $(0,0)$
E) x -intercepts: $(-2,0)$ and $(2,0)$ y -intercept: $(0,0)$

To find the x-intercepts, we set y = 0 and solve for x. Thus,
0 = 2x - x^2
x(2-x) = 0
So, x = 0 or x = 2. These are the x-intercepts. To find the y-intercept, we set x = 0 and solve for y. Thus,
y = 2(0) - 0^2
y = 0
So the only y-intercept is (0,0). Choice B is the only option that includes both x-intercepts (0,0) and (2,0) and the y-intercept (0,0).