I'm puzzled on this one. I've studied James Stewart's graph translations, but I think his techniques don't obviously solve this one. Or what am I missing?
The problem. The problem (which is not from Stewart's book) shows a table of values of the functions $f(n) = n$, $g(n) = n^2$, $h(n) = \sqrt{n}$. The problem then asks some easy questions such as --- which one grows more slowly? (Surely $h$.) Which one has constant growth? (Surely $f$.) Then the problem asks me to build a table of values for a function $y$ that maps $n$ to $n^2 - n$. It asks me to draw the graph. I do it. Then it asks me --- is it possible to draw the graph of $y$ by using just the graph of $n^2$ and $n$? Honestly, I don't know.
The graph itself seems very simple. It surely is still a curve that looks like a parabola, but I'm not even sure I can call it a parabola. (To me the definition of a parabola is that curve whose any point on it is far away from its vertex as much as it is far away from the directrix.) Anyway, how can I draw the graph of $x^2 - x$ from $x^2$ and $x$? (Is that even possible --- at least in this case?)
First, a spoiler:
$y = x^2 - x$ is indeed a parabola. We can write it as $y = (x - \frac{1}{2})^2 - \frac{1}{4}$, so it is just the parabola $y = x^2$ shifted right by $\frac{1}{2}$ and down by $\frac{1}{4}$.
As for how you can intuit its graph from the graphs of $y = x$ and $y = x^2$, you can pick out some of the following features:
The graphs intersect at $(0, 0)$ and $(1, 1)$, so $y = x^2 - x$ must have $x$-intercepts at $x = 0, 1$.
$y = x$ takes negative values to the left of $x = 0$ so $y = x^2 - x$ must be above the graph of $y = x^2$ on that side, and similarly it must be below the graph of $y = x^2$ to the right of $x = 0$.
As you noted, $y = x^2$ grows faster than $y = x$ when $x$ increases (and similarly it grows faster than $y = -x$ when $x$ decreases), so the curve of $y = x^2 - x$ will be dominated by the graph of $y = x^2$ for large (positive or negative) $x$.
You could probably also, with some effort, intuit something about it having a single minimum between $x = 0$ and $x = 1$, and it having reflection symmetry about $x = \frac{1}{2}$, but once you get familiar with the algebraic manipulation it will probably be much easier to do those things by converting the formula to another standard form.