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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

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

Vedclass · Answer

(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:

Similar Questions

The quadratic equation whose one root is $\frac{1}{2 + \sqrt{5}}$ will be

The roots of the given equation $(p - q){x^2} + (q - r)x + (r - p) = 0$ are

If $k \in ( - \infty , - 2) \cup (2, \infty ),$ then the roots of the equation $x^2 + 2kx + 4 = 0$ are

If $8, 2$ are the roots of ${x^2} + ax + \beta = 0$ and $3, 3$ are the roots of ${x^2} + \alpha x + b = 0$,then the roots of ${x^2} + ax + b = 0$ are

If $\alpha$ and $\beta$ are roots of the equation $x^2 - 4\sqrt{2}kx + 2e^{4\ln k} - 1 = 0$ for some $k$, and $\alpha^2 + \beta^2 = 66$, then $\alpha^3 + \beta^3$ is equal to: (in $\sqrt{2}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module