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 $x$ is real,then the value of $\frac{x^2 + 34x - 71}{x^2 + 2x - 7}$ does not lie between

Which of the following is incorrect? If $a \equiv b \pmod{m}$ and $x$ is an integer,then

If the roots of $\sqrt{\frac{1-y}{y}}+\sqrt{\frac{y}{1-y}}=\frac{5}{2}$ are $\alpha$ and $\beta$ $(\beta > \alpha)$ and the equation $(\alpha+\beta) x^4-25 \alpha \beta x^2+(\gamma+\beta-\alpha)=0$ has real roots,then a possible value of $\gamma$ is

Let $S = \{x \in [-6, 3] \setminus \{-2, 2\} : \frac{|x+3|-1}{|x|-2} \geq 0\}$ and $T = \{x \in \mathbb{Z} : x^2 - 7|x| + 9 \leq 0\}$. Then the number of elements in $S \cap T$ 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

Vedclass Test Series

Exam Paper Generator

Online Exam Module