If the figure shows the graph of y = ax^2 + b | vedclass

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(D) $1$. The graph is a downward-opening parabola,which implies that the coefficient of $x^2$ must be negative,so $a < 0$.$2$. The graph intersects th

Vedclass · Accepted Answer

None of these.

Vedclass · Answer

(D) $1$. The graph is a downward-opening parabola,which implies that the coefficient of $x^2$ must be negative,so $a $2$. The graph intersects the $x$-axis at two distinct points $(x_1, 0)$ and $(x_2, 0)$,which means the quadratic equation $ax^2 + bx + c = 0$ has two distinct real roots. Therefore,the discriminant $D = b^2 - 4ac > 0$.$3$. The graph intersects the $y$-axis below the origin,which means the $y$-intercept $c$ is negative,so $c $4$. Comparing these findings with the given options,none of the conditions $a > 0$,$b^2 - 4ac  0$ are satisfied. Thus,the correct option is $D$.

If the figure shows the graph of $y = ax^2 + bx + c$,then . . . . . .

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