Question

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

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

Answer 1

The correct option is B $b<0$

$\because a<0$
Hence $x_2+x_1<0$ (from figure)
$\Rightarrow \frac{-a}{b}<0\Rightarrow \frac{a}{b}\Rightarrow b<0(\because a<0)$.
The curve $y=a{x}^2+bx+c$ meets $y-axis$ at $x=0$.
Therefore $y=c$ From the figure $y>0\Rightarrow c>0$