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,so the coefficient of $x^2$ must be negative. Thus,$a < 0$.$2$. The $y$-intercept is the value of $y$ when $x =

Vedclass · Accepted Answer

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

Vedclass · Answer

(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,$a < 0, b < 0, c < 0$.

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

Similar Questions

$STATEMENT-1$: The curve $y = -\frac{x^2}{2} + x + 1$ is symmetric with respect to the line $x = 1$. Because
$STATEMENT-2$: $A$ parabola is symmetric about its axis.

The length of the normal chord to the parabola $y^2 = 4x$,which subtends a right angle at the vertex,is

If a point $P$ moves such that its distances from the point $A(1, 1)$ and the line $x+y+2=0$ are equal,then the locus of $P$ is

The equation of the directrix of the parabola $x^2+8x+12y+4=0$ is

If a normal to the parabola $y^2=12x$ at $A(3,-6)$ cuts the parabola again at $P$,then the equation of the tangent at $P$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module