Predicting what the graph of a function looks like? How can you predict what the function
$$f(x) = \frac{(x - 5)(x + 4)(x - 3)^2(x)}{(x-5)(x)}$$
looks like before you plot it?
 A: Observations:
$$f(x) = \frac{(x - 5)(x + 4)(x - 3)^2(x)}{(x-5)(x)} = (x+4)(x-3)^2,\;\;x\neq 5, x\neq 0$$The function is not defined at $x = 0, x= 5$. Why not? These are called, however, removable discontinuities, and do not impact the graph of the function $f(x) = (x + 4)(x-3)^2, $ save for there being "holes" at $x = 5,\text{ and } x = 0$
$f(x)$ intersects the x axis at $x = -4, x = 3$.  Why? We call these values of $x$ "zeros" of the function. 
The polynomial we are left with is a cubic polynomial.
We can also take the derivative to find critical points; this will help immensely when graphing: it allows us to determine where, if anywhere, there are local minimums, local maximums, etc. To find these, we compute $f'(x)$ and solve for $x$ when $f'(x) = 0$.
A: Hint: When is the denominator $0$? When is the numerator $0$ and what does this mean for the graph (you should consider limits of $f$ as $x$ tends to these zeros)? What are the turning points of a function and how can you find them, and their characteristics (minimum, maximum, saddle point)? Before you do any of this though, you should make sure nothing cancels (removable discontinuities).
