Question

The adjoining figure shows the graph of $y=a{x}^2+bx+c$. Then

• A. $a>0$
• B. $b>0$
• C. $c>0$
• D. $b^2<4ac$

Answer 1

Answer: (B), (C)

The correct options are
B $c>0$
C $b>0$

Given equation : $y=a{x}^2+bx+c$

From the figure, it is clear that $y>0$ at $x=0$.

Hence we get,

$a{\left(0\right)}^2+b\left(0\right)+c>0$ $\Rightarrow$ $c>0$.

From the figure, we get $x_1<0$ and $x_2>0$.

Hence, product of roots $={x_{1}x}_2<0$ $\Rightarrow$ $\frac{c}{a}<0$.

But $c>0$ $\Rightarrow$ $a<0$.

From the figure, it is also clear that ${x_1+x}_2>0$.
$\Rightarrow$ Sum of roots $=\frac{-b}{a}>0$ $\Rightarrow b>0$

Since, both the roots are real, the discriminant is greater than zero.

Hence, options (B) and (C) are correct choices.