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Question Number 216593 by Tawa11 last updated on 11/Feb/25

Answered by A5T last updated on 11/Feb/25

$$\mathrm{At}\mathrm{vertex},\frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{ax}+\mathrm{b}=0\Rightarrow 18\mathrm{a}+\mathrm{b}=0\Rightarrow \mathrm{b}=-18\mathrm{a}$$$$-14=81\mathrm{a}+9\mathrm{b}+\mathrm{c}$$$$\Rightarrow \mathrm{c}=-14+81\mathrm{a}$$$$\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{a}-18\mathrm{a}-14+81\mathrm{a}=64\mathrm{a}-14$$$$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=2\mathrm{a}>0\Rightarrow \mathrm{a}>0$$$$64\mathrm{a}-14=\mathrm{v}\Rightarrow \mathrm{a}=\frac{\mathrm{v}+14}{64}>0\Rightarrow \mathrm{v}>-14$$$$\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{v}>-14\Rightarrow \mathrm{D})$$

Commented by Tawa11 last updated on 12/Feb/25

$$\mathrm{Thanks}\mathrm{sir},\mathrm{I}\mathrm{appreciate}.$$

Answered by mr W last updated on 12/Feb/25

$$y={ax}^2+bx+c=a{(x+\frac{b}{2a})}^2+c-\frac{b^2}{4a}$$$$a>0(⌣)$$$$\frac{b}{2a}=-9\Rightarrow b=-18a$$$$c-\frac{b^2}{4a}=-14\Rightarrow c=-14+a{\left(\frac{b}{2a}\right)}^2=-14+81a$$$$k=a+b+c$$$$=a-18a-14+81a$$$$=64a-14$$$$>-14$$$$\Rightarrow onlyD)-12ispossible$$

Commented by Tawa11 last updated on 12/Feb/25

$$\mathrm{Thanks}\mathrm{sir},\mathrm{I}\mathrm{appreciate}.$$

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