If the figure shows the graph of y = ax^2 + b | vedclass

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

Question

(D) $1$. The graph is a downward-opening parabola,which implies that the coefficient of $x^2$ must be negative,so $a < 0$.$2$. The graph intersects th

Vedclass · Accepted Answer

None of these.

Vedclass · Answer

(D) $1$. The graph is a downward-opening parabola,which implies that the coefficient of $x^2$ must be negative,so $a $2$. The graph intersects the $x$-axis at two distinct points $(x_1, 0)$ and $(x_2, 0)$,which means the quadratic equation $ax^2 + bx + c = 0$ has two distinct real roots. Therefore,the discriminant $D = b^2 - 4ac > 0$.$3$. The graph intersects the $y$-axis below the origin,which means the $y$-intercept $c$ is negative,so $c $4$. Comparing these findings with the given options,none of the conditions $a > 0$,$b^2 - 4ac  0$ are satisfied. Thus,the correct option is $D$.

If the figure shows the graph of $y = ax^2 + bx + c$,then . . . . . .

Similar Questions

If $a, b, c$ are real numbers such that $a+b+c=0$ and $a^2+b^2+c^2=1$,then $(3a+5b-8c)^2+(-8a+3b+5c)^2+(5a-8b+3c)^2$ is equal to

If $\alpha, \beta$ are the real roots of $x^2+p x+q=0$ and $\alpha^4, \beta^4$ are the roots of $x^2-r x+s=0$,then the equation $x^2-4 q x+2 q^2-r=0$ has always

$\alpha$ is the maximum value of $1-2x-5x^2$ and $\beta$ is the minimum value of $x^2-2x+r$. If $5\alpha x^2+\beta x+6>0$ for all real values of $x$,then the interval in which $r$ lies is

If $a^2+b^2+c^2=1$,where $a, b, c \in \mathbb{R}$,then the set of extreme values of $ab+bc+ca$ is

For real values of $x$,the range of $\frac{x^2+2x+1}{x^2+2x-1}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module