The diagram shows the graph of y = ax^2 + bx | vedclass

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(C) $1$. The parabola opens downwards,which implies that the coefficient of $x^2$ is negative,so $a < 0$.$2$. The curve intersects the $y$-axis at $x

Vedclass · Accepted Answer

$c > 0$

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(C) $1$. The parabola opens downwards,which implies that the coefficient of $x^2$ is negative,so $a $2$. The curve intersects the $y$-axis at $x = 0$. Substituting $x = 0$ into the equation $y = ax^2 + bx + c$,we get $y = c$. From the figure,the intersection point on the $y$-axis is above the origin,so $c > 0$.$3$. The roots of the quadratic equation are $x_1$ and $x_2$. From the figure,$x_1 > 0$ and $x_2  0$. Since $a $4$. Since the curve intersects the $x$-axis at two distinct points,the discriminant $D = b^2 - 4ac > 0$.Comparing these findings with the options,$c > 0$ is a correct statement.

The diagram shows the graph of $y = ax^2 + bx + c$. Then:

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