Contains points (1,10) and (2,4), axis of symmetry is vertical, vertex is on line4x-3y=6
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Hint : $\displaystyle y=ax^2+bx+c$ $\displaystyle V\left(\frac{-b}{2a},\frac{-D}{4a}\right)$ $\displaystyle D=b^2-4ac$
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