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$$y=ax^2+bx+c$$ passes through $(1,1)$, $(2,m)$ and $(m,2)$ where $m \ne 2$. Determine an interval for $m$ such that $y(x)$ has a minimum.

Progress:

If it has a minimum, then $a > 0$.

Also, by the points given: $$a+b+c=1$$ $$m=4a+2b+c$$ $$am^2+bm+c=2$$ Using the fact that $a$ is positive, I got the expressions: $$a+b+c=1 \implies b+c < 1$$ $$am^2+bm+c=2 \implies bm+c < 2$$ $$m=4a+2b+c \implies 2b+c < m$$ But I don't know how to solve it.

Edit: I would solve a linear inequality system by graphing, but in this case there are three variables, so it wouldn't be a practical method.

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    you must solve your equation system – 2017-02-19
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    You shouldn't say that "an equation passes through points". It is the corresponding curve which passes through these points – 2017-02-19

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solving your system of equations we get $$a=\frac{m}{1-m},b=\frac{m^2+m+1}{m-1},c=\frac{m^2+m-2}{m-1}$$ $a>0$ means that $$\frac{m}{1-m}>0$$ case work yiels $$0