Question

Consider the graph of the quadratic polynomial $y=a{x}^2+bx+c$ as shown below.

Which among the following is/are correct?

• A. $\frac{a-b+c}{abc}=0$
• B. $abc(9a+3b+c)<0$
• C. $\frac{a+3b+9c}{abc}<0$
• D. $abc(a-3b+9c)>0$

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{a-b+c}{abc}=0$
C $\frac{a+3b+9c}{abc}<0$


From the given graph,
$a<0,\frac{-b}{2a}>0\Rightarrow b>0f\left(0\right)=c>0\therefore abc<0$

Now,
$f(-1)=0\Rightarrow a-b+c=0,f\left(3\right)=0\Rightarrow 9a+3b+c=0$

Clearly,
$f\left(\frac{1}{3}\right)>0$
$\Rightarrow a+3b+9c>0$
And
$f\left(\frac{-1}{3}\right)>0$
$\Rightarrow \frac{a}{9}-\frac{b}{3}+c>0\Rightarrow a-3b+9c>0$

So,
$\frac{a-b+c}{abc}=0abc(9a+3b+c)=0\frac{a+3b+9c}{abc}<0abc(a-3b+9c)<0$