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

Question

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

Daniel Valdez · Accepted Answer

Step 1The original parabola   y  =  .001      x    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&#039;t work. So   k  =  1.

vrotterigzl · Accepted Answer

Step 1Your two students&#039; 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&#039;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).

Avery and Sasha were comparing parabola graphs on their calculators. Avery had drawn y = 0

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