WitCH 14: Stretching the Truth

The easy WitCH below comes courtesy of the Evil Mathologer. It is a worked example from Cambridge’s Essential Mathematics Year 9 (2019), in a section introducing parabolic graphs.

Update

The problem, as commenters have indicated below, is that there is no parabola with the indicated turning point and intercepts. Normally, we’d write this off as a funny but meaningless error. But, coming at the very beginning of the introduction to the parabola, it most definitely qualifies as crap.

4 Replies to “WitCH 14: Stretching the Truth”

  1. There is no quadratic function which can have a vertex and axial intercepts where pictured.

    If we ignore the x-intercepts, the graph has form y=a(x+3)^2 +5. Using (0,-5) gives a=-10/9

    If, on the other hand we ignore the turning point the graph has form y=a(x+1)(x+5). Using (0,-5) gives a=-1.

    Which is close, but not the same as -10/9.

    Have I missed anything?

  2. What a hoot. To fit the three intercepts and the maximum (-3,5) it would have to be a cubic function, whose graph is not symmetrical. To be symmetrical – so that the graph passes through (-6,-5) – it would have to be a quartic function. My guess is that the displayed graph is the latter, but then of course no part of the graph of a quartic function is a parabola.

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