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

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(A) $1$. The parabola opens downwards,which implies that the coefficient of $x^2$ is negative. Therefore,$a < 0$.$2$. The $y$-intercept of the graph i

Vedclass · Accepted Answer

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

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(A) $1$. The parabola opens downwards,which implies that the coefficient of $x^2$ is negative. Therefore,$a $2$. The $y$-intercept of the graph is where $x = 0$. Substituting $x = 0$ into the equation $y = ax^2 - bx + c$,we get $y = c$. From the graph,the $y$-intercept is below the $x$-axis,so $c $3$. The $x$-coordinate of the vertex of the parabola $y = ax^2 + Bx + C$ is given by $-B / (2A)$. In our equation $y = ax^2 - bx + c$,the coefficient of $x$ is $-b$. Thus,the $x$-coordinate of the vertex is $-(-b) / (2a) = b / (2a)$.$4$. From the graph,the vertex lies to the right of the $y$-axis,meaning the $x$-coordinate of the vertex is positive. Therefore,$b / (2a) > 0$.$5$. Since we already know $a $6$. Combining these,we have $a < 0, b < 0, c < 0$.

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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