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

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(C) $1$. Since the parabola opens downwards,the coefficient of $x^2$ must be negative,so $a < 0$.$2$. The curve intersects the $y$-axis at $(0, c)$. F

Vedclass · Accepted Answer

$c > 0$

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(C) $1$. Since the parabola opens downwards,the coefficient of $x^2$ must be negative,so $a $2$. The curve intersects the $y$-axis at $(0, c)$. From the figure,the intersection point is above the $x$-axis,so $c > 0$.$3$. The vertex of the parabola is at $x = -b/(2a)$. From the figure,the vertex is to the left of the $y$-axis,so $-b/(2a)  0$. Since $a $4$. The curve intersects the $x$-axis at two distinct points,so the discriminant $D = b^2 - 4ac > 0$.Comparing these with the options,$c > 0$ is correct.

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

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