If the graph of y = ax^2 - bx + c is as shown | vedclass

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(A) $1$. The parabola opens downwards,so the coefficient of $x^2$ must be negative. Thus,$a < 0$.$2$. The $y$-intercept is the value of $y$ when $x =

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$a < 0, b < 0, c < 0$

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(A) $1$. The parabola opens downwards,so the coefficient of $x^2$ must be negative. Thus,$a $2$. The $y$-intercept is the value of $y$ when $x = 0$. From the graph,the parabola intersects the $y$-axis below the origin,so $c $3$. The $x$-coordinate of the vertex is given by $x = -(-b) / (2a) = b / (2a)$.$4$. From the graph,the vertex lies to the right of the $y$-axis,so the $x$-coordinate of the vertex is positive,i.e.,$b / (2a) > 0$.$5$. Since $a $6$. Therefore,$a < 0, b < 0, c < 0$.

$A. \ 2x-5=0$	$I. \ \text{Vertex}$
$B. \ (\frac{3}{2}, -1)$	$II. \ \text{Focus}$
$C. \ y+1=0$	$III. \ \text{Equation of directrix}$
$D. \ (2, -1)$	$IV. \ \text{Equation of the axis}$
	$V. \ \text{Equation of the Latus rectum}$

If the graph of $y = ax^2 - bx + c$ is as shown below,then the signs of $a$,$b$,and $c$ are:

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