If the graph of y = ax^2 + bx + c (a, b, c in | vedclass

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(B) From the graph,the parabola opens upwards,so $a > 0$.Since the vertex is on the left side of the $y$-axis,the $x$-coordinate of the vertex $-\frac

Vedclass · Accepted Answer

$ac^2bD < 0$

Vedclass · Answer

(B) From the graph,the parabola opens upwards,so $a > 0$.Since the vertex is on the left side of the $y$-axis,the $x$-coordinate of the vertex $-\frac{b}{2a}  0$,we must have $b > 0$.The graph intersects the $y$-axis below the $x$-axis,so $c The graph intersects the $x$-axis at two distinct points,so $D = b^2 - 4ac > 0$.Now,let's check the options:$A$: $abc = (+)(+)(-) = - $B$: $ac^2bD = (+)(+)(+)(+) = + > 0$. Thus,$ac^2bD $C$: $\frac{a^2c}{b^2D} = \frac{(+)(+)(-)}{(+)(+)} = - $D$: $bD = (+)(+) = + > 0$. This is correct.Therefore,the incorrect statement is $B$.

If the graph of $y = ax^2 + bx + c$ $(a, b, c \in R)$ is as shown in the figure,where $D = b^2 - 4ac$,which of the following is incorrect?

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