A problem about view windows of graphs

Question

We have the following problem in the problem set we are using:

Avery and Sasha were comparing parabola graphs on their calculators. Avery had drawn $y = 0.001x^2$ in the window $-1000 \le x \le 1000$ and $0 \le y \le 1000$ , and Sasha had drawn $y = x^2$ in the window $-k \le x \le k$ and $0 \le y \le k$. Except for scale markings on the axes, the graphs looked the same! What was the value of $k$?

I myself had no clue. But one student guessed the value of $k$ be $1$ and justified his answer using a graphing calculator. Another student explained that since the point $(1000, 1000)$ of the parabola $y = 0.001x^2$ lies at the top-right corner of the view window defined by $-1000 \le x \le 1000$ and $0 \le y \le 1000$, $k$ must be $1$ so that the point $(1, 1)$ of the parabola $y = x^2$ also lies at the top-right corner of the view window defined by $-k \le x \le k$ and $0 \le y \le k$. Both answers makes some sense. However, the rest of us are still begging for a more intuitive explanation. Someone can help?

Answer 1

Your two students' answers are how I myself solved the problem.

A more intuitive explanation? How about the observation that with the appropriate $x$- and $y$- scale, every upward-facing parabola looks identical; likewise, every downward-facing parabola? In other words, to sketch a quadratic function, I might start by sketching a prototypical upward/downward- facing parabola (depending on the leading coefficient's sign) before drawing in the reference frame (the axes) at the appropriate position.

The given problem is analogous: each curve is being drawn before determining its location (the scale markings).

Answer 2

The original parabola ($y=.001x^2$) touches the upper right and upper left corners of the given graphing area. For the second parabola, we need the same behavior; but the upper right corner should be $(k,k)$. For this to occur, you need $k^2=k$. So $k = 0$ or $1$; but $k=0$ doesn't work. So $k=1$.