The graph below shows the polynomial: $$f(x) = x^2(x-2)^k$$ Use the slider to change the exponent $k\text{,}$ and observe the changes to the graph. The factor $x^2$ corresponds to the $x$-intercept $(0,0)\text{.}$ Since the exponent is even, the graph always ? crosses over bounces off the $x$-axis at that point. When $k = 2\text{,}$ the end-behavior of $f(x)$ is the same as ? x x^2 x^3 x^4 . When $k = 5\text{,}$ the end-behavior of $f(x)$ is the same as ? x^5 x^6 x^7 x^8 , and the graph ? crosses over bounces off the $x$-axis at the point $(2,0)\text{,}$ and it is nearly ? vertical horizontal there.