Question

Find the signs of a, b and c in given graph of polynomial,$f\left(x\right)=a{x}^2+bx+c,wherea\ne 0.$

• A.
  $a>0,b=0andc<0$
• B.
  $a>0,b>0andc<0$
• C.
  $a<0,b=0andc<0$
• D.
  $a<0,b=0andc<0$

Answer 1

The correct option is A

$a>0,b=0andc<0$
The graph of $f\left(x\right)=y$ opens upward,
$\therefore a>0.$

$\text{The vertex}(\frac{-b}{2a},\frac{-D}{4a})\text{lies on y-axis,}\therefore x=0\Rightarrow \frac{-b}{2a}=0\Rightarrow b=0\text{The graph cuts negative y-axis}\therefore c<0$

So, option a is correct.