Question

Consider the graph of quadratic polynomial $y=a{x}^2+bx+c$ as shown below. Which of 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. $ab(a-3b+9c)>0$

Answer 1

Answer: (A), (C)

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

As $x=-1$and $x=3$ are the roots of the equation and graph is concave downwards, therefore a is negative.

So equation is,

$y=-(x+1)(x-3)$=$-x^2+2x+3$

So,$a=-1,b=2,c=3$

Now, $\frac{a-b+c}{abc}$=$\frac{-1-2+3}{(-1)\left(2\right)\left(3\right)}$=0

$abc(9a+3b+c)$=$(-1)\left(2\right)\left(3\right)(-9+6+3)$=0

$\frac{a+3b+9c}{abc}$=$\frac{-1+6+27}{(-1)\left(2\right)\left(3\right)}$<0

$abc(a-3b+9c)$=$(-1)\left(2\right)\left(3\right)(-1-6+27)<0$

Hence, options 'A' and 'C' are correct.