The adjoining figure shows the graph of y = a | vedclass

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(C) From the graph,the parabola opens downwards,which implies that the coefficient of ${x^2}$ must be negative. Therefore,$a < 0$.The roots of the qua

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Both $(a)$ and $(b)$

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(C) From the graph,the parabola opens downwards,which implies that the coefficient of ${x^2}$ must be negative. Therefore,$a The roots of the quadratic equation $y = a{x^2} + bx + c$ are ${x_1}$ and ${x_2}$,where both ${x_1} > 0$ and ${x_2} > 0$. The sum of the roots is given by ${x_1} + {x_2} = -\frac{b}{a}$.Since both roots are positive,their sum is positive,so $-\frac{b}{a} > 0$,which implies $\frac{b}{a} This means that $a$ and $b$ must have opposite signs.Since the parabola intersects the $X$-axis at two distinct points,the discriminant $D = {b^2} - 4ac > 0$,so ${b^2} > 4ac$.Thus,both $(a)$ and $(b)$ are correct.

The adjoining figure shows the graph of $y = a{x^2} + bx + c$. Then:

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