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I believe the book is wrong (again), but need to confirm my sanity on this parabola problem.

Given the Vertex is $(4,-9)$ and 2 other points: $(1,6)$ and $(8,4).$ Write the equation.

Right away I'm thinking this is strange. I only need the Vertex and ONE other point. So, I use $f(x)=a(x-4)^2 -9$ and I use the point $(1,6)$ to find the a value.

$6=9a-9, a=\frac{5}{3}$

Of course the rub comes when I check this by finding the a value using the other point and get $a=\frac{13}{16}$.

I don't think these 2 points are on this parabola. How can I prove I'm correct?

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    `I only need the Vertex and ONE other point` That's only if you assume that the parabola axis is vertical. – 2017-02-13
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    Good point. However, in this IM3 class we have ONLY studied vertical axis parabolas. – 2017-02-13
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    You can determine the (unique) parabola with vertical axis through those $3$ points, and verify that $(4,-9)$ is not its vertex, so something is missing from this picture. – 2017-02-13
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    Is it sufficient to show that if (4,-9) is vertex then it is NOT possible to obtain 2 different $a$ values to prove that (4,-9) is NOT the vertex? – 2017-02-13
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    Yes, or just show that the $a$ value determined from one of the points gives a parabola which does *not* pass through the other point. – 2017-02-13
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    hate it when the book gives wrong answers and you begin to doubt what you know! – 2017-02-13

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