Question

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

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

Answer 1

The correct option is C $b<0$

$\because a<0$

Hence, $x_2+x_1<0$ (from figure)
$\Rightarrow \frac{-a}{b}<0\Rightarrow b<0(\because a<0)$

The curve $y=a{x}^2+bx+c$ meets $y-$axis at $x=0$.

Therefore $y=c$

From the figure $y>0\Rightarrow c>0$