Why do polynomial terms with an even exponent bounce off the x-axis? Like if it's $f(x) = (x-5)^2(x+6)$
Why, at $x=5$, does the graph reflect off the x-axis?
 A: Well... at $x=5$, it is clear that 
$$f(5)=(5-5)^2(5+6)=0^2*11=0$$
So from here, you need to observe that if you pick some number greater than $5$ for $x$ (we can write this as $x=5+x_0$, where $x_0 >0$) then we get 
$$f(5+x_0) = ((5+x_0)-5)^2((5+x_0)+6)=(x_0)^2(11+x_0)>0$$
The $>$ above is true because $(x_0)^2$ is always positive and $11+ x_0$ is positive because $x_0>0$. But this is true for any $x_0 >0$ and it only gets bigger for larger $x_0$, so this is why it looks the way it does.
A: The expanded form of the polynomial is a cubic with leading coefficient of $1$.  Hence, $\displaystyle \lim_{x \rightarrow \infty} f(x) = \infty$ and $\displaystyle \lim_{x \rightarrow -\infty} f(x) = -\infty$.
The polynomial has only two zeros: $x = 5$ and $x= -6$.  It cannot touch the $x$-axis anywhere else.
Once you begin sketching the graph, it should become obvious why the graph must "bounce" off the $x$-axis at $x = 5$.  
You should be able to generalize the idea from here.  :)
A: Normally, at $x=k$ where $f(x)=(x-k)...$ k is where the value of $(x-k)$ changes from a positive to a negative or vice-versa. This changes the entire function because that sign change changes the sign of the entire function. When $(x-k)^2$ crosses k, it retains its sign because of the squared, because any number squared is positive, which means that $(x-k)$ changing signs is irrelevant to total function sign change.
